text:
\begin{document}

\begin{abstract}
For every group of order at most 14 we determine  the values taken  by its
group determinant when its variables are integers.

\end{abstract}

\maketitle

\section{Introduction}\label{secIntroduction}
For a finite group $G=\{g_1,\ldots ,g_n\}$ of order $n,$ we assign a  variable
$x_g$ for each element $g\in G$ and define its {\it group determinant}
$\mathscr{D}_G(x_{g_1},\ldots ,x_{g_n})$ to be the determinant of the $n\times
n$ matrix whose $(i,j)$th entry is $x_{g_i g_j^{-1}}$.
In the case of the cyclic group of order $n$, the group determinant becomes an
$n\times n$ {\it circulant determinant}, where
each row is obtained from the previous one by a cyclic shift one step to the
right. At the meeting of the American Mathematical Society in Hayward,
California, in April 1977,  Olga Taussky-Todd asked which integers could be obtained
as an $n\times n$ circulant
determinant when the entries are all integers.  Of course we can ask for a
complete description of the group determinants over the integers for any group $G$,
not just for the cyclic groups. Thus our problem is to determine the set
\[
  \S(G)=\{  \mathscr{D}_G(x_{g_1},\ldots ,x_{g_n})\midd x_{g_1},\ldots ,x_{g_n}\in\Z \}.
   \]
For the additive cyclic group $\mathbb Z_n$ of order $n$,
  Laquer \cite{Laquer} and Newman \cite{Newman1,Newman2} gave
divisibility conditions on the integers that can be group determinants, as well as sets of
achievable values; for example any integer coprime to $n$ or a multiple of
$n^2$
will be a group determinant, if $m$ is a determinant then so is $-m$, and if
$p\mid m$ and $p^{\alpha}\parallel n$ then $p^{\alpha+1}\mid m$. Conditions like
these enabled them to obtain a complete description  of the values for certain
cyclic groups.
For example   Laquer \cite{Laquer} and Newman
\cite{Newman1} showed that for a prime $p$
\be \label{Zp}  \S(\Z_p)=\{p^a m_p\midd a=0, a\geq 2\}, \ee
while for  $p$ an odd prime, Laquer \cite{Laquer} showed that
\be \label{Z2p}  \S(\Z_{2p})=\{2^a p^b m_{2p}\midd  a=0, a\geq 2, \;\; b=0,
 b\geq 2\}. \ee
Newman \cite{Newman2} determined $\S(\Z_9)$ as
\be \label{Z9}   \S(\Z_9)=\{3^am_3\midd  a=0, a\geq 3\},     \ee
with upper and lower set  inclusions for general $\mathbb Z_{p^2}$.  In the above, and henceforth, $m_t$ denotes an arbitrary integer coprime to $t$.

For a polynomial $F(x_1,\ldots ,x_r)$ in $\mathbb Z [x_1,\ldots ,x_r]$, the
traditional logarithmic Mahler measure $m(F)$ can  be defined by
$$ m(F)=\log M(F) = \int_0^1 \cdots \int_0^1 \log |F(e^{2\pi i x_1},\ldots ,
e^{2\pi i x_r})| dx_1 \cdots dx_r. $$
In 2005 Lind \cite{Lind} viewed the traditional Mahler measure as a measure on
the circle group $(\mathbb R /\mathbb Z)^r$ and generalised the concept to an
arbitrary compact abelian group. In particular for a finite group
\be \label{finiteG}  G=\mathbb Z_{n_1}\times \cdots \times \mathbb Z_{n_r} \ee
one can define the logarithmic measure of an $F(x_1,\ldots ,x_r)$ in $\mathbb Z
[x_1,\ldots ,x_r]$ relative to $G$ to be $$m_G(f)=\frac{1}{|G|} \log |M_G(F)|,
$$ where
$$ M_G(F)=\prod_{j_1=1}^{n_1} \cdots \prod_{j_r=1}^{n_r}F(\om_{n_1}^{j_1},\ldots
,\om_{n_r}^{j_r}),\;\; \;\;\; \om_n:=e^{2\pi i/n}. $$

Curiously, the Lind variant of the Mahler measure for $\Z_n$ had essentially appeared in a 1916 paper of Pierce \cite{Pierce}, and also in the
famous paper of Lehmer \cite{Lehmer}, in the form $\Delta_n:=\prod_{i=1}^r(\alpha_i^n-1)=(-1)^{rn} M_{\Z_n}(F)$, for a monic integer one-variable polynomial $F$ with roots $\alpha_1,\ldots ,\alpha_r$.

As observed by Dedekind, the group of characters $\hat{G}$ of a finite abelian
group $G$ can be used to factor its group determinant as
\be \label{abelianfactor} \mathscr{D}_G(x_{g_1},\ldots ,x_{g_n})=\prod_{\chi \in
\hat{G}} \left(\chi(g_1)x_{g_1}+\cdots + \chi(g_n)x_{g_n}\right). \ee
On making the characters  explicit, it is readily seen that for a group $G$ of the form
\eqref{finiteG} we have
\be \label{connection}  \mathscr{D}_G(a_{g_1},\ldots ,a_{g_n})= M_G(F), \ee
where
\be \label{defF}  F(x_1,\ldots ,x_r) =\sum_{g=(t_1,\ldots t_r)\in G} a_g
x_1^{t_1}\cdots x_r^{t_r}, \ee
a connection observed by Vipismakul \cite{Cid1} in his thesis. Of course any
polynomial in $\mathbb Z[x_1,\ldots ,x_r]$ can be reduced to \eqref{defF}  by
working in the ring $\mathbb Z[x_1,\ldots ,x_r]/\langle x_1^{n_1}-1,\ldots
,x_r^{n_r}-1\rangle.$

Kaiblinger \cite{Norbert2} used the Lind measure approach to obtain
\be \label{Z4}   \S(\mathbb Z_4)=\{2^a m_2\midd  a=0, a\geq 4\} \ee
and
\be \label{Z8}   \S(\mathbb Z_8)=\{2^a m_2 \midd  a=0, a\geq 5\},    \ee
with upper and lower set inclusions for the other $\mathbb Z_{2^k}$. Defining
$\lambda(G)$ to be the smallest non-trivial determinant value
$$ \lambda(G):= \min \{ |s|\midd s\in\S(G),\; s\neq 0, \pm 1\}, $$
Kaiblinger  \cite{Norbert}  obtained $\lambda (\mathbb Z_n)$ when $420\nmid n$;
 this was extended to all $n$ with $892371480 \nmid n$
by Pigno and Pinner \cite{Pigno1}. Values of $\lambda(G)$ for non-cyclic abelian $G$  were
considered in \cite{dilum,pgroups,Cid2, Stian}.

As explored in Boerkoel and Pinner \cite{dihedral}, the connection \eqref{connection} between Lind
measures and group determinants suggests a way to extend the concept of Lind
measure to
non-abelian finite groups, and to measures on (not necessarily commutative) polynomial rings modulo
appropriate group relations. See Dasbach and Lal\'\i n\cite{Lalin} for another approach.
As observed
by Frobenius, see for example \cite{Formanek,Conrad},
the counterpart to \eqref{abelianfactor} for a non-abelian group will involve
non-linear factors and the set of irreducible
representations $\hat{G}$ for $G$. Specifically,
$$ \mathscr{D}_G(x_{g_1},\ldots ,x_{g_n})=\prod_{\rho\in \hat{G}} \det\left(
\sum_{g\in G} x_g \rho(g)\right)^{\deg(\rho)}. $$
For example, for the dihedral
group $G=D_{2n}$ of order $2n$, one can define the measure $M_G(F)$ of  an $F$ in
$ \mathbb Z[x,y]/\langle x^n-1,y^2-1, xy-yx^{-1}\rangle,$ reduced to the form
$$ F(x)=f(x)+yg(x),\;\; \;\; f(x)=\sum_{j=0}^{n-1} a_j
x^j,\;\;\;g(x)=\sum_{j=0}^{n-1} b_j x^j, $$
by
\[
  M_G( F)  =\mathscr{D}_G(a_0,\ldots ,a_{n-1},b_0,\ldots
,b_{n-1}).
\]
This was shown in \cite[Section 2]{dihedral} to equal
\be\label{E-D}
M_{\mathbb Z_n}\left(f(x)f(x^{-1})-g(x)g(x^{-1})\right),
\ee
and was used in \cite{dihedral} to determine $\S(D_{2p})$ for
$p$ an odd prime as
\be \label{D2p}  \S(D_{2p})=\{2^a p^b m_{2p}\midd  a=0,  a\geq 2,\;\; b=0,
 b\geq 3\},   \ee
(which includes $S_3$ under the guise of $D_6$). Also
$$\S(D_{4p})=\S_{\odd}(D_{4p})\cup \S_{\even}(D_{4p}), $$
 where
\be \S_{\odd}(D_{4p})=\{m\equiv 1 \bmod 4\midd p\nmid m, p^3\mid m\}.
\ee
and
\be \label{D4p}  \S_{\even}(D_{4p})=\{2^ap^b m_{2p}\midd  a=4, a\geq 6,\;\; b=0,
 b\geq 3\}.  \ee
For $G=\mathbb Z_2 \times \mathbb Z_2$, viewed as $D_4:$
\be \label{Z2Z2} \S(\Z_2\times\Z_2)=\{4m+1, \;\; 2^4(2m+1),\;\; 2^6m\midd m\in \mathbb Z\}, \ee
for  $G=D_8$:
\be \label{D8}  \S(D_8)=\{4m+1,\;\; 2^8m \midd m\in \mathbb Z\},   \ee
for  $G=D_{16}$:
\be \label{D16}   \S(D_{16})=\{4m+1,\;\; 2^{10}m \midd m\in \mathbb Z\},  \ee
with upper and lower set inclusions for the other $D_{2^k}$. For
$G=D_{2p^2}$ when  $p=3,5$ or $7$:
$$ S(D_{2p^2})=\{2^a p^b m_{2p} \midd  a=0, a\geq 2,\;\; b=0,
b\geq 5\}. $$
Also, the value of $\lambda(D_k)$ was determined for $k<3.79 \times 10^{47}$ in \cite{dihedral}.

Many of the small groups are of one of the above forms. Indeed for the groups of
order at most 14 this just leaves out the groups
$G=Q_8,\mathbb Z_2 \times\mathbb Z_4$, $\mathbb Z_2^3$, $\mathbb Z_3 \times
\mathbb Z_3$, $A_4, Q_{12}$, $\mathbb Z_{12}$  and $\mathbb Z_6 \times \mathbb
Z_2$, where $Q_{4n}$ denotes the dicyclic group of order $4n$.
Our goal here is to determine $\S(G)$ for
these  remaining groups $G$. As we shall see, for example for $G=\mathbb Z_6\times \mathbb Z_2$,
these can become complicated
very quickly, so developing a general theory for dealing with all finite groups is probably not feasible.
While it is known \cite{Formanek}  that the group determinant polynomial, $\mathscr{D}_G(x_{g_1},\ldots ,x_{g_n})$, determines the group, it remains open whether the set of integer values taken, $\S(G)$, also determines the group.

We shall make frequent use of the multiplication property
\be \label{mult} \mathscr{D}_G(a_{g_1},\ldots
,a_{g_n})\mathscr{D}_G(b_{g_1},\ldots ,b_{g_n})=\mathscr{D}_G(c_{g_1},\ldots
,c_{g_n}),\;\; c_g:=\sum_{uv=g}a_ub_v, \ee
corresponding to multiplication $\left(\sum_{g\in G} a_g g
\right)\left(\sum_{g\in G}  b_g g \right)= \sum_{g\in G}  c_g g$ in
$\mathbb Z[G]$  (or multiplication and reduction of polynomials subject to the
relations). Thus $\S(G)$ is a semigroup.

We shall work interchangeably with the group determinants
$\mathscr{D}_G(x_{g_1},\ldots ,x_{g_n})$ and the polynomial measures
$M_G(F)$. We begin by expressing the group determinant for the dicyclic group
$Q_{4n}$ as a $\mathbb Z_{2n}$  Lind measure of an
associated polynomial.

\section{ Dicyclic Groups}

We write the dicyclic group of order $4n$ in the form
$$ Q_{4n} =\langle a,b \; :\; a^{2n}=1,b^2=a^n,ab=ba^{-1}\rangle, $$
and order the elements $1,a,a^2,\ldots ,a^{2n-1},b,ba,\ldots ,ba^{2n-1}$.

Our polynomial measures will be defined on $\mathbb Z[x,y]/\langle x^{2n}-1,
y^2-x^n,xy-yx^{2n-1} \rangle,$ where we can assume that $F$ in $\mathbb Z[x,y]$
has been reduced to the form
\be \label{reducedform}  F(x,y)=f(x)+yg(x), \;\;\;\; \; f(x)=\sum_{j=0}^{2n-1}
a_j x^j,\;\;\; g(x)=\sum_{j=0}^{2n-1} b_j x^j. \ee
The case $n=2$ gives us the classical quaternion group
$$ Q_8=\{1,-1,i,-i,j,-j,k,-k\},\;\;\; i^2=j^2=k^2=ijk=-1, $$
under the correspondence $(1,a,a^2,a^3,b,ba,ba^2,ba^3)=(1,i,-1-i,j,-k,-j,k)$.

Our determinant $\mathscr{D}_{Q_{4n}}(a_0,\ldots ,a_{2n-1},b_0,\ldots
,b_{2n-1})$ will have four linear factors, corresponding to the characters $\chi
(a)=1$ and $\chi(b)=\pm 1$,  and $\chi(a)=-1$ with
$\chi(b)=\pm 1$ if $n$ is even and $\chi (b)=\pm i$ if $n$ is odd,
\begin{align*}
\Big(f(1)+g(1)\Big)\Big(f(1)-g(1)\Big)\Big(f(-1)+g(-1)\Big)\Big(f(-1)-g(-1)\Big)
, & \;\;\; \text{ $n$ even,} \\
\Big(f(1)+g(1)\Big)\Big(f(1)-g(1)\Big)\Big(f(-1)+ig(-1)\Big)\Big(f(-1)-ig(-1)\Big), & \;\;\; \text{ $n$ odd.}
\end{align*}
For the remaining complex $2n$th roots of unity  $\om=\om_{2n}^j$, $1\leq j \leq
n-1$,
 (complex conjugates give the same factors) we have $n-1$ two-dimensional
representations
$$    \rho (a)=\left( \begin{matrix} \om & 0 \\ 0 &
\om^{-1}\end{matrix}\right),\;\;\;\rho(b) = \left( \begin{matrix} 0 & \om^n \\ 1
& 0 \end{matrix} \right),    $$
leading to  the squares of $n-1$ quadratic factors
\begin{align*}
\det\left( \rho\left( \sum_{j=0}^{2n-1}  a_j a^j + \sum_{j=0}^{2n-1} b_j  b
a^j\right)\right)
 & = \det \left( \begin{matrix} f(\omega) & \om^n g(\om^{-1}) \\ g(\om) & f(\om^{-1})
\end{matrix} \right) \\
 & =f(\om)f(\om^{-1}) -\om^n g(\om)g(\om^{-1}). \end{align*}
Hence we can write
$$ \mathscr{D}_{Q_{4n}}(a_0,\ldots ,a_{2n-1},b_0,\ldots
,b_{2n-1})=\prod_{j=0}^{2n-1} (f(\om_{2n}^j)f(\om_{2n}^{-j})
-\om_{2n}^{jn}g(\om_{2n}^j)g(\om_{2n}^{-j}). $$
We take this to be the dicyclic  measure of an $F(x,y)$ in $\mathbb Z[x,y]$,
reduced to the form  \eqref{reducedform},
\be\label{E-Q} M_{Q_{4n}}(F) = M_{\mathbb Z_{2n}}\Big( f(x)f(x^{-1})-x^n g(x)g(x^{-1})\Big).
\ee
We observe for future reference that
\begin{align}\label{E-ggff}
M_{\mathbb Z_{2n}}\Big( g(x)g(x^{-1})-x^n& f(x)f(x^{-1})\Big) \nonumber\\
&=(-1)^nM_{\mathbb Z_{2n}}\Big( f(x)f(x^{-1})-x^n g(x)g(x^{-1})\Big),
\end{align}
so that $\S(Q_{4n})=-\S(Q_{4n})$ when $n$ is odd.

\section{Groups of order 8}\label{8gps}

In this section we determine $\S(G)$ for the
five  groups of order eight: $G= \mathbb Z_8$, $D_8$,  $Q_8,$ $\mathbb Z_4 \times\mathbb
Z_2$ and $\Z_2^3$.

As mentioned in the introduction, $\S(G)$ is already known for $G=\mathbb
Z_8$ and $D_8$, namely
 $$ \S(\Z_8)=\{2m+1 \text{ and } 32m \midd m\in \mathbb Z\} $$
and
$$\S(D_8)= \{4m+1  \text{ and } 2^8m \midd  m\in \mathbb Z\}. $$

For the groups
 $G=D_8, Q_8$ and $\mathbb Z_4 \times \mathbb Z_2$, the  group determinants
correspond to  Lind-Mahler measures on the two-variable polynomials $F(x,y)\in
\mathbb Z[x,y]$  reduced to
$$ F(x,y)=f(x)+yg(x), \text{ where } f(x) = \sum_{j=0}^3 a_jx^j,\;\; g(x)=\sum_{j=0}^3
b_jx^j. $$
These are given in terms of Lind-Mahler measures of cyclic groups by \eqref{E-D} for $D_8$,
by \eqref{E-Q} for $Q_8$, and by
\be\label{E-Z42}
M_{\Z_4\times\Z_2}(F)=M_{\Z_4}(f(x)+g(x))\cdot M_{\Z_4}(f(x)-g(x))
\ee
for $\Z_4\times\Z_2$.

For $G=\mathbb Z_2^3$ the determinants correspond to measures of polynomials in
$\mathbb Z[x,y,z]$, reducible mod $\langle x^2-1,y^2-1,z^2-1\rangle$  to
$$ F(x,y,z)=\sum_{i,j,k\in \{0,1\}} a_{i,j,k}x^iy^jz^k \in \mathbb Z[x,y,z],\;\;\; M_G(F)=\prod_{x,y,z=\pm 1} F(x,y,z). $$

\begin{theorem} \label{Q8} We have
$$\S(\Z_4\times\Z_2)= \{8m+1  \text{ and } 2^8m \midd  m\in \mathbb Z\},$$
$$\S(Q_8)= \S(\Z_4\times\Z_2)\cup \{(8m-3)p^2\midd m\in\Z, \; p \equiv 3\bmod 4 \text{ prime}\},$$
and
$$\S(\Z_2^3)=\{8m+1 \text{ and }2^8(4m+1)  \text{ and } 2^{12}m \midd m\in\Z\}. $$
\end{theorem}

Note  that
$$\S(\Z_2^3)\subsetneq \S(\Z_4\times\Z_2) \subsetneq \S(Q_8) \subsetneq\S(D_8)\subsetneq \S(\Z_8). $$

The Theorem immediately gives us the minimum non-trivial measure for the groups of order 8.
\begin{corollary}
$$\lambda(\mathbb Z_4\times \mathbb Z_2)=\lambda(\mathbb Z_2^3)=\lambda(Q_8)=7 \text{ and }  \lambda (D_8)=\lambda(\mathbb Z_8)=3.
$$
\end{corollary}

\begin{comment}
 CJS reworded Thm., and combined Th 3.1 and 3.2.
For $G=\mathbb Z_8$ the determinants correspond to measures of one variable
polynomials $f(x)$ in $\mathbb Z[x]$  reduced mod $\langle x^8-1 \rangle$ to the
form
$$ f(x)=\sum_{0\leq j\leq 7} a_jx^j \in \mathbb Z[x]. $$

\begin{theorem} \label{Z8}
For $G=\mathbb Z_8$ the values taken by $\mathscr{D}_{G}(a_0,\ldots ,a_7)$  for
integers $a_0,\ldots ,a_7$ are the odd integers and the multiples of 32.
\end{theorem}

For $G=\mathbb Z_2^3$ the determinants correspond to measures of polynomials in
$\mathbb Z[x,y,z]$, reducible mod $\langle x^2-1,y^2-1,z^2-1\rangle$  to
$$ f(x)=\sum_{i,j,k\in \{0,1\}} a_{(i,j,k)}x^iy^jz^k \in \mathbb Z[x,y,z]. $$

\begin{theorem}\label{Z222} For $G=\mathbb Z_2^3$ the values taken by
$\mathscr{D}_G(a_{(0,0,0)},\ldots ,a_{(1,1,1)})$ for integer variables
$a_{(i,j,k)}$ are the  integers of the form $8m+1$, $2^8(4m+1)$ and $2^{12}m$
for some integer $m$.
\end{theorem}
\end{comment}

\section{The Alternating Group $A_4$}\label{S-A4}

Taking the two generators
$$ \alpha =(123),\;\;\; \beta=(12)(34),  $$
we order the elements $(g_1,g_2,\ldots ,g_{12})$ of $A_4$ as
\begin{align*}
& ( 1, (12)(34), (13)(24), (14)(23), (123), (243), (142), (134), (132),(143),
(234), (124) )\\
 & \;\;\;\; =(1, \beta, \alpha^2\beta \alpha , \alpha \beta \alpha^2 ,\alpha
,\beta \alpha,\alpha^2 \beta \alpha^2 ,\alpha \beta, \alpha^2, \beta \alpha^2,
\alpha^2 \beta ,\alpha \beta \alpha).
\end{align*}
Now $A_4$ has  four irreducible representations: three linear ones
$\chi_0,\chi_1,\chi_2$ where $\chi(\beta)=1$ and $\chi(\alpha)=1,\om$ or $\om^2$
respectively with $\om :=e^{2\pi i/3}$, and one, $\rho$,  of degree $3$.
This latter representation comes from isometries of the regular
tetrahedron (vertices $1,2,3,4$) relative to the three axes passing through the
centres of opposite pairs of sides. Explicitly,
\[
\rho(\alpha) =\left(\begin{matrix}0&0&1\\1&0&0\\0&1&0\end{matrix}\right)
\qquad\text{ and } \qquad
\rho(\beta) = \left(\begin{matrix} 1&0&0\\0&-1&0\\0&0&-1\end{matrix}\right),
\]
generating the representation:
\begin{align*}
\rho( \alpha^2 \beta \alpha) =
\left(\begin{matrix}-1&0&0\\0&-1&0\\0&0&1\end{matrix}\right), &\;\;\;
\rho( \alpha \beta \alpha^2) =
\left(\begin{matrix}-1&0&0\\0&1&0\\0&0&-1\end{matrix}\right),\;\;\;
\rho( \beta \alpha) =
\left(\begin{matrix}0&0&1\\-1&0&0\\0&-1&0\end{matrix}\right),\;\;\; \\
\rho( \alpha^2\beta \alpha^2) =
\left(\begin{matrix}0&0&-1\\-1&0&0\\0&1&0\end{matrix}\right), & \;\;\;
\rho(\alpha \beta) =
\left(\begin{matrix}0&0&-1\\1&0&0\\0&-1&0\end{matrix}\right),\;\;\;
 \rho( \alpha^2) =
\left(\begin{matrix}0&1&0\\0&0&1\\1&0&0\end{matrix}\right),\;\;\; \\
\rho( \beta \alpha^2) =
\left(\begin{matrix}0&1&0\\0&0&-1\\-1&0&0\end{matrix}\right),& \;\;\;
\rho( \alpha^2 \beta)=
\left(\begin{matrix}0&-1&0\\0&0&-1\\1&0&0\end{matrix}\right),\;\;\;
\rho( \alpha \beta \alpha)=
\left(\begin{matrix}0&-1&0\\0&0&1\\-1&0&0\end{matrix}\right).\;\;\;
\end{align*}
Writing $x=\sum_i a_i g_i$ in $\mathbb Z [G]$,
then for $G=A_4$ the group determinant takes the form
$$ \mathscr{D}_{G}(a_1,a_2,\ldots ,a_{12}) =l_0 l_1 l_2 D^3 $$
where, putting
$$a:=a_1+a_2+a_3+a_4,\;\;\; b:=a_5+a_6+a_7+a_8 \;\; \text{ and } \;\;
c:=a_9+a_{10}+a_{11}+a_{12},$$
we have
$$ l_0 =\chi_0(x)=a+b+c,\;\;\;l_1=\chi_1(x) = a+b\om +c\om^2,\;\;\;l_2=
\chi_2(x)=a+b\om^2+c\om, $$
and
$$ D=\det(\rho(x))=\det\left( \begin{matrix}
a_1+a_2-a_3-a_4&a_9+a_{10}-a_{11}-a_{12}&a_5+a_6-a_7-a_8\\
                     a_5-a_6-a_7+a_8& a_1-a_2-a_3+a_4&a_9-a_{10}-a_{11}+a_{12}\\
a_9-a_{10}+a_{11}-a_{12}&a_5-a_6+a_7-a_8&a_1-a_2+a_3-a_4
\end{matrix}\right). $$

We  can regard $\mathscr{D}_{G}(a_1,a_2,\ldots ,a_{12})$ as the Lind measure
$M_G(F)$ of the generic polynomial
\begin{align*}  F(x,y) = a_1  +  & a_2 y + a_3 x^2 y x+ a_4 x y x^2 + a_5 x+ a_6
yx +a_7 x^2 y x^2 \\
   &  + a_8 xy + a_9 x^2 +a_{10} y x^2 + a_{11} x^2 y + a_{12}x y x,
\end{align*}
in $\mathbb Z[x,y]$ with non-commutative multiplication,  and reduction according to the
relations
$$x^3=1,\;\; y^2=1,\;\; yxy=x^2yx^2\;\; \text{ and }\;\; yx^2y=xyx. $$

\begin{theorem} \label{A4} We have $\S(A_4)=\S(A_4)_{\even}\cup\S(A_4)_{\odd}$, where
\[
\S(A_4)_{\even}  =\{2^a 3^b m_6\midd  a=4, a\geq 8, \; b=0, b\geq 2\}
\]
and
\[
\S(A_4)_{\odd}=\{m\equiv 1 \bmod 4 \midd 3\nmid m, 3^2\mid m\}.
\]
\end{theorem}

\section{Groups of Order 12} \label{12gps}

There are five groups of order twelve: $\mathbb Z_{12},$ $\mathbb Z_6 \times
\mathbb Z_2$, $D_{12}$, $Q_{12}$ and $A_4$ (dealt with in the
previous section).

From \cite{dihedral} we know that
$\S(D_{12})=\S(D_{12})_{\even}\cup\S(D_{12})_{\odd}$, where
\[
\S(D_{12})_{\even}=\{ 2^a3^bm_6 \midd   a=4, a\geq 6,\;\;  b=0, b\geq 3\}
\]
and
\[
\S(D_{12})_{\odd} =\{m\equiv 1\bmod 4 \midd  3\nmid m, 3^3\mid m\}.
\]

For the groups $G=Q_{12}$, $D_{12}$ and $\mathbb Z_6 \times \mathbb Z_2$ we
work with measures of polynomials
$$ F(x,y)=f(x)+yg(x)\in \mathbb Z[x,y],\;\;\; f(x)=\sum_{j=0}^5 a_jx^j,\;\;
g(x)=\sum_{j=0}^5 b_jx^j. $$
These are given in terms of Lind-Mahler measures of cyclic groups by \eqref{E-Q} for $Q_{12}$,
by \eqref{E-D} for $D_{12}$, and by
\be\label{E-Z62}
M_{\Z_6\times\Z_2}(F)=M_{\Z_6}(f(x)+g(x))\cdot M_{\Z_6}(f(x)-g(x)).
\ee

\begin{theorem} \label{Q12}
If  $G=Q_{12}$ the set $\S(Q_{12})$ consists of measures  $M_G(F)$ of the following forms
\begin{align} 2^a 3^b m_6 \midd \; &  a=0, 4, a\geq
6,\;\; b=0,  b\geq 3, \\
2^5 3^b m_6 \midd \; &  b=4, b\geq 6,\\
\label{32m} 2^5 3^bm_6k \midd\; &   b=0, 3, 5,
 \end{align}
where, in \eqref{32m},  $k$ can be a prime $p\equiv 5\bmod 12$ and also the square of a  prime $p\equiv 5 \bmod 6$.

\comments{\vspace{1ex}
\noindent
For $G=D_{12}$ the odd measures are the $m\equiv 1 \bmod 4$ with $3\nmid m$ or
$27\mid m$.
The even measures take the form $2^4m$ with $2\nmid m$ or $2^6m$ with $3\nmid m$
or $3^3\mid m$.
}

\end{theorem}

For $G=\mathbb Z_6\times\mathbb Z_2$ and $\mathbb Z_{12}$ we need to partition the
primes $p\equiv 1\bmod 12$ into two sets
\begin{align}
\P_1:& =\{p\equiv 1 \bmod 12 \; :\; p=(6k+2)^2 + (6t+3)^2 \text{ for some
$k,t\in \mathbb Z$}\},\nonumber \\
\label{defS} \P_2:& =\{p\equiv 1 \bmod 12 \; :\; p=(6k)^2 + (6t+1)^2
\text{ for some $k,t\in \mathbb Z$}\}.
\end{align}

These sets are disjoint by the uniqueness of representation of primes $\equiv 1\bmod 4$ as a sum of two squares. They are probably not describable  by a simple congruence; see  Cox \cite{Cox} for a
class field theory approach  to distinguishing which primes are  of the form
$x^2+36y^2$.

\begin{comment} Your original version below seems to need some extra work to prove disjointness?
If I'm wriong, please reinstate it:

These sets are disjoint by uniqueness of factorisation in $\mathbb Z[i]$  (and
probably not describable  by a simple congruence; see  Cox \cite{Cox} for a
class field theory approach  to distinguishing which primes are  of the form
$x^2+36y^2$).
\end{comment}

\vspace{2ex}
\noindent
\begin{theorem}\label{Z2Z6}  For   $G=\mathbb Z_6 \times \mathbb Z_2$, the set $\S(\mathbb Z_6 \times \mathbb Z_2)$ consists of measures $M_G(F)$ of the following forms:

\vspace{1ex}

\noindent
(a) The measures with $3^3\mid M_G(F)$ take the form
$$ 3^3(4m-1),\;\;\; 2^4\cdot 3^3 (2m-1),
  \;\;\;2^6\cdot 3^3 m. $$

\vspace{1ex}

\noindent
(b) The measures with $3^2 \parallel  M_G(F)$ take the form
$$ 3^2(4m-1)p\;\;\; 2^4 \cdot 3^2(2m-1)p  \;\;\;  2^6 \cdot 3^2 mp  $$
for some prime $p\equiv 7 \bmod 12$, or
$$ 2^8 \cdot 3^2 (4m-1) \;\;\; \hbox{ or } \;\;\; 2^{10}\cdot 3^2 (4m-1) \;\;\;
\hbox{ or }\;\;\; 2^{12} \cdot 3^2 (2m-1) \;\;\; \hbox{ or } \;\;\; 2^{14}\cdot
3^2 m. $$

\vspace{1ex}

\noindent
(c) The measures coprime to $3$ take the form
$$ 12m+1, \;\;\;  2^4(6m+1)  \;\; \; 2^6(3m+1), $$
or
\be \label{type2}  (12m+5)k,  \;\;\; -2^4(6m+1)k,\;\;\; -2^6(3m+1)k \ee
where $k=p$ for  some prime $p\equiv 1\bmod 12$ in $\P_1$, or $k=p^2$ for some
$p\equiv 5\bmod 12$, or $k=p_1p_2$ for some primes $p_1,p_2\equiv 7\bmod 12$, or
$$ 2^8(12m+5), \;\;\;  2^8(12m+5)p, \;\;\;
2^{10}(12m+5),  \;\;\;  2^{10}(12m+5)p, $$
for some $p\equiv 7\bmod 12$, or
$$-2^{12}(6m+1), \;\;\;  -2^{14}(3m+1).$$
In each case $m$ runs through all the integers.

\end{theorem}
Note that in the theorem there are no measures with $3\parallel M_G(F)$.

For the group $G=\mathbb Z_{12}$, we work with measures on polynomials
$F(x)=\sum_{j=0}^{11} a_jx^j$.

\begin{theorem}\label{Z12}
Let $G=\mathbb Z_{12}$.  We separate  $\S(\Z_{12})$ into the odd and even  measures:

\vspace{1ex}
\noindent
(a) The odd measures coprime to $3$ can take any value $m_6$.

\noindent
The odd multiples of $3$ take the form
$9m_6p$ for  primes $p\equiv 5$ and $7\bmod 12$, and $p$ in $\P_1$, and $27m_2$.

\vspace{1ex}
\noindent
(b) The even values divisible by $3$  take the form $2^4 \cdot 3^2 m$ for all
integers $m$.

\noindent
The even values coprime to $3$ take the forms $2^4m_6$, and
$2^5m_6p$ for some prime  $p\equiv 5$ and $7$ mod $12,$ and $p$  in $\P_1$, and $2^6m_3$.

\end{theorem}
Again, in this theorem there are no measures with $3\parallel M_G(F)$.

\

The Lind-Lehmer constant (the minimal non-trivial measure)  is readily
determined for the  groups of order 12:
\begin{corollary}
$$\lambda(D_{12})=\lambda(Q_{12})=\lambda(\mathbb Z_{12})=\lambda(A_4)=5,\;\;  \lambda(\mathbb
Z_6\times\mathbb Z_2)=11
$$
\end{corollary}

\section{The remaining group $\mathbb Z_3 \times \mathbb Z_3$}
\label{oddments}

\begin{theorem}\label{Z3Z3} For $G=\mathbb Z_3 \times \mathbb Z_3$ we have
$$ \S(G)= \{9m\pm 1 \midd m\in\Z\}\cup \{3^6m  \midd m\in\Z\}. $$
\end{theorem}

\section{Proofs for Section \ref{8gps}}

\begin{proof}[Proof of Theorem \ref{Q8}] We first prove the result for $G=Q_8$ and $\Z_4\times\Z_2$. Since it requires no extra work
we also include $D_8$, although it is already covered in \cite{dihedral}. We begin by comparing the  form of  the Lind measure
for  these three groups.
In all three cases we have the same four linear factors in $\mathbb
Z[a_0,a_1,a_2,a_3,b_0,b_1,b_2,b_3]$, namely
\begin{align*}
\ell_1 & :=F(1,1)=(a_0+a_2)+(a_1+a_3) +(b_0+b_2)+(b_1+b_3),\\
\ell_2 & :=F(1,-1)=(a_0+a_2)+(a_1+a_3) - (b_0+b_2)- (b_1+b_3),\\
\ell_3 & :=F(-1,1)=(a_0+a_2)-(a_1+a_3) + (b_0+b_2)-(b_1+b_3),\\
\ell_3 & :=F(-1,-1)=(a_0+a_2)-(a_1+a_3) -(b_0-b_2)+(b_1+b_3).\end{align*}
The remaining factors are  quadratics   in $\mathbb
Z[a_0,a_1,a_2,a_3,b_0,b_1,b_2,b_3]$:
\begin{align*} \mathscr{D}_{Q_8}(a_0,a_1,a_2,a_3,b_0,b_1,b_2,b_3) & = \ell_1
\ell_2\ell_3\ell_4 q_1^2, \\
\mathscr{D}_{D_8}(a_0,a_1,a_2,a_3,b_0,b_1,b_2,b_3)&= \ell_1 \ell_2\ell_3\ell_4
q_2^2,\\
\mathscr{D}_{\mathbb Z_4\times \mathbb Z_2}(a_0,a_1,a_2,a_3,b_0,b_1,b_2,b_3)&=
\ell_1 \ell_2\ell_3\ell_4 q_3q_4,
\end{align*}
where
\begin{align*}
 q_1 & :=|f(i)|^2+|g(i)|^2=(a_0-a_2)^2+(a_1-a_3)^2+(b_0-b_2)^2+(b_1-b_3)^2,\\
 q_2 & :=|f(i)|^2-|g(i)|^2=(a_0-a_2)^2+(a_1-a_3)^2-(b_0-b_2)^2-(b_1-b_3)^2,\\
q_3  & :=|f(i)+g(i)|^2= ((a_0-a_2)+(b_0-b_2))^2+ ((a_1-a_3)+(b_1-b_3))^2,   \\
q_4 & := |f(i)-g(i)|^2=  ((a_0-a_2) -(b_0-b_2))^2 +((a_1-a_3) -(b_1-b_3))^2 .
\end{align*}

Thus for $G=Q_8$, $D_8$ or $\mathbb Z_4 \times \mathbb Z_2$ we have
\begin{align*}
&\mathscr{D}_{G}(m+1,m,m,m,m,m,m,m) =8m+1,\\
&\mathscr{D}_{G}(k+ 2,k,k,k,k,k,k,k) =2^8(4k+1),\\
&\mathscr{D}_{G}(k-1,k+1,k-1,k+1,k+1,k+1,k,k) =-2^8(4k+1),\\
&\mathscr{D}_{G}(k+1,k,k+1,k,k-1,k-1,k,k)  =2^8(2k).
\end{align*}
Equivalently, writing these as polynomial measures we have
\begin{align*}  & M_G\left( 1+ m\frac{x^4-1}{x-1} +
ym\frac{x^4-1}{x-1}\right)=8m+1,\\
& M_G\left( 2+ k\frac{x^4-1}{x-1} + yk\frac{x^4-1}{x-1}\right)=2^8(4k+1),\\
 & M_G\left( (x^2+1)(x-1)+ k\frac{x^4-1}{x-1} + y\left(
(x+1)+k\frac{x^4-1}{x-1}\right)\right)=-2^8(4k+1),\\
 & M_G\left( (x^2+1)+ k\frac{x^4-1}{x-1} +
y\left(-(x+1)+k\frac{x^4-1}{x-1}\right)\right)=2^8(2k).
\end{align*}
Writing $p\equiv 3$ mod 4 as $p=a^2+b^2+c^2+d^2$ with $a$ even, $b,c,d$ odd,
then
\begin{align*}
(a_0,a_1,a_2,a_3) &
=\left(m+\frac{a}{2},m+\frac{(b-1)}{2},m-\frac{a}{2},m-\frac{(b+1)}{2}\right),\\
 (b_0,b_1,b_2,b_3) &
=\left(m+\frac{(c-1)}{2},m+\frac{(d-1)}{2},m-\frac{(c+1)}{2},m-\frac{(d+1)}{2}
\right),
\end{align*}
has $\mathscr{D}_{Q_8}(a_0,a_1,a_2,a_3,b_0,b_1,b_2,b_3)=(8m-3)p^2,$
while
$$\mathscr{D}_{D_8}(m,m,m,m-1,m,m,m-1,m-1)=8m-3. $$

It remains to show that a determinant $\mathscr{D}_G$ takes one of the stated
forms. Suppose first that $\mathscr{D}_G$ is even. Since the $\ell_j\equiv
\ell_1$ mod 2 and the $q_i\equiv \ell_1^2$ mod 2 we know that the $\ell_j$ and
$q_j$ are all even. If $(a_0-a_2)$, $(a_1-a_3)$, $(b_0-b_2)$, $(b_1-b_3)$
are all even  or all odd then  in all cases $4\mid q_j$ and $2\mid \ell_j$ and
$2^8\mid \mathscr{D}_G$.
So suppose two of them are even and two odd. Hence two of $(a_0+a_2)$,
$(a_1+a_3)$, $(b_0+b_2)$, $(b_1+b_3)$
are even and two odd. Call these $A,B,C,D,$ in any order,  then
$$ \ell_1\ell_2\ell_3\ell_4=((A+B)^2-(C+D)^2)((A-B)^2-(C-D)^2). $$
Hence if $A,C$ are odd and $B,D$ even then $(A\pm B)^2-(C\pm D)^2 \equiv 1-1=0$
mod 8, so $2^6\mid  \ell_1\ell_2\ell_3\ell_4$
and $2\mid q_j$, and $2^8\mid \mathscr{D}_G$.

Suppose that $\mathscr{D}_G$ is odd. So either  one or three of the
$(a_0-a_2)$, $(a_1-a_3)$, $(b_0-b_2)$, $(b_1-b_3),$
will be odd (and hence one or three  of  the corresponding  $A,B,C,D$,  will
odd). Suppose that $A$ is odd and $B$ is even and $C,D$ have the same parity.
Plainly $q_1^2,q_2^2\equiv $ 1 mod 8 and
$$ q_4=q_3 -4(a_0-a_2)(b_0-b_2) -4(a_1-a_3)(b_1-b_3).   $$
So if three of $A,B,C,D$ are  even  then  $q_4\equiv q_3$ mod 8 and
$q_3q_4\equiv q_3^2\equiv 1$ mod 8, while if three are odd
we have  $q_4\equiv q_3+4$ mod 8 and $q_3q_4\equiv q_3^2-4\equiv -3$ mod 8.
Similarly
\begin{align*} (A-B)^2 - (C-D)^2  & = (A+B)^2 - (C+D)^2 -4AB+4CD \\
& \equiv (A+B)^2 - (C+D)^2 +4CD \text{ mod 8}.\end{align*}
Hence if $C,D$ are even we have $\ell_1\ell_2\ell_3\ell_4 \equiv
((A+B)^2-(C+D)^2)^2 \equiv 1$ mod 8 and
$\mathscr{D}_G\equiv 1$  mod 8.
If $C,D$ are odd then $\ell_1\ell_2\ell_3\ell_4 \equiv ((A+B)^2-(C+D)^2)^2-4
\equiv -3$ mod 8, and
  $\mathscr{D}_G\equiv 1$ mod 8 for $G=\mathbb Z_4 \times \mathbb Z_2$, and
$\mathscr{D}_G\equiv -3$ mod 8 for $G=Q_8$ or $D_8$, with $q_1\equiv 3$ mod 4
for $G=Q_8$ (and so divisible by at least one $p\equiv 3$ mod 4).
This completes the proof for $G=Q_8$, $\Z_4\times\Z_2$ and $D_8$.

\begin{comment}
\begin{proof}[Proof of Theorem \ref{Z8}]
We have
$ \mathscr{D}_{\mathbb Z_8}=\prod_{j=0}^7 f(w_8^j) = L_1 L_2 HK, $
where
\begin{align*}
 L_1 & =f(1) = (a_0+a_4)+(a_1+a_5)+(a_2+a_6)+(a_3+a_7),\\
 L_2 & =f(-1) = (a_0+a_4)-(a_1+a_5)+(a_2+a_6)-(a_3+a_7),\\
H & = |f(i)|^2= ((a_0+a_4)-(a_2+a_6))^2 + ((a_1+a_5)-(a_3+a_7))^2,\\
K & = |f(w_8)f(-w_8)|^2=((a_0-a_4)^2-(a_2-a_6)^2 +2(a_1-a_5)(a_3-a_7))^2 \\
 & \;\;\;\;\;\;\;\;+ ((a_3-a_7)^2-(a_1-a_5)^2+2(a_0-a_4)(a_2-a_6))^2.
\end{align*}
We have
\begin{align*}
\mathscr{D}_{\mathbb Z_8}(k+1,k-1,k,k,k,k,k,k)& =(2k)2^5,\\
\mathscr{D}_{\mathbb Z_8}(k+1,k+1,k+1,k,k+1,k,k,k)&  =(2k+1)2^5,
\end{align*}
while for odd $j$
$$\mathscr{D}_{\mathbb Z_8}(\overbrace{k+1,\ldots , k+1}^{j},k,\ldots ,k)=(8k+j)
 $$
(the latter construction working in $\mathbb Z_n$ to give $(nk+j)$ for any
$\gcd(n,j)=1$).
Hence it just remains to show that the even $\mathbb Z_8$-measures are multiples
of $2^5$.
Suppose that  $\mathscr{D}$ is even.  Since  $L_2\equiv L_1$ mod 2, $H\equiv
L_1^2$ mod 2 and $K\equiv L_1^4$ mod 2
we know that $L_1,L_2,H,$ and $K$ are all even. If  $A=(a_0+a_4)$ and
$B=(a_2+a_6)$ have the same parity then $C=a_1+a_5$ and $D=a_3+a_7$
must have the same parity and plainly $4\mid H,4\mid K$ and $2^6\mid
\mathscr{D}$.  So suppose that $A,B$ have opposite
parity. Hence $C,D$  also have opposite parity, giving $2||H,K$ but
$L_1L_2=(A+B)^2-(C+D)^2 \equiv 1-1=0\text{ mod } 8,$ and
$2^5\mid \mathscr{D}$.

\end{proof}
\end{comment}

Next, for $G=\mathbb Z_2^3$ we have
$\displaystyle M_G(F)=\prod_{x,y,z=\pm 1} F(x,y,z)$.

We can achieve anything of the form $8m+1$, $2^8(4m+1)$ or $2^{12}m$ using
\begin{align*}
& M_G\left( 1+m(1+x)(1+y)(1+z)\right)  = 8m+1, \\
& M_G\left( 2+m(1+x)(1+y)(1+z)\right)  = 2^8(4m+1), \\
&M_G\left( 3+z+k(1+x)(1+y)(1+z)\right)  = 2^{12}(2k+1), \\
& M_G\left( x+y+z-3+(1-x)(1-y)(1-z) +k(1+x)(1+y)(1+z)\right)  = 2^{12}(2k),
\end{align*}
so it just remains to check that any $M_G(F)$ is  of one of these forms.

We write $F(x,y)=f(x,y)+zg(x,y)$, with $f(x,y)$ and $g(x,y)$ of the form
$$ g(x.y)=a(0,0) + a(1,0)x+a(0,1)y+a(1,1)xy\in \mathbb Z[x,y], $$
so that
$$ M_G(F) = \prod_{x,y=\pm 1} f(x,y)^2-g(x,y)^2. $$
Notice that the $f(\pm 1,\pm 1)\equiv f(1,1)$ mod 2, and $g(\pm 1,\pm 1)\equiv
g(1,1)$ mod 2, and $M_G(F)\equiv (f(1,1)-g(1,1))^4 $ mod 2 is even if $f(1,1)$
and $g(1,1)$ have the same parity and odd otherwise.

Suppose first that $M_G(F)$ is odd. Reversing the roles of $f(x,y)$ and $g(x,y)$
as necessary we suppose that the
$f(\pm 1,\pm 1)$ are all  odd and the $g(\pm 1,\pm 1)$ all even.  Then mod 8 we
have
\begin{align*}  M_G(F)  & \equiv \prod_{x,y=\pm 1} (1-g(x,y)^2 )
  \equiv 1- \sum_{x,y=\pm 1} g(x,y)^2\equiv 1- \left(\sum_{x,y=\pm 1}
g(x,y)\right)^2.
\end{align*}
But
$$\sum_{x=\pm 1} g(x,y)=4a(0,0), $$
and  so $M_G(F)\equiv 1$ mod 8.

Suppose that $M_G(F)$ is even. If  the $f(\pm 1,\pm 1)$ and $g(\pm 1,\pm 1)$ are
all odd then the $f(x,y)^2-g(x,y)^2\equiv 1-1=0$ mod $8$ for each of the four
factors and $2^{12} \mid M_G(F)$. So suppose that they are all even and
$$ M_G(F) = 2^8 \prod_{x,y=\pm 1} (f(x,y)/2)^2 - (g(x,y)/2)^2. $$
If any of the $f(x,y)/2$ and $g(x,y)/2$ have the same parity then 4 divides that
factor and $2^{12}\mid M_G(F)$. So assume that they have opposite  parity for
all $x,y=\pm 1,$ with $(f(x,y)/2)^2 - (g(x,y)/2)^2$ equalling  $1$ mod 4 if
$g(x,y)/2$ is even and $-1$ mod 4 if $g(x,y)/2$ is odd. But the $\sum_{x,y=\pm
1} g(x,y)/2=2a(0,0)$ is even, so we must have an even number of $-1$'s and
$\prod_{x,y=\pm 1} (f(x,y)/2)^2 - (g(x,y)/2)^2\equiv 1$ mod 4.
\end{proof}

\section{Proof of Theorem  \ref{A4}}

Suppose that 3 divides
$$\mathscr{D}_G(a_1,\ldots ,a_{12})=l_0l_1l_2 D^3. $$
If $3\mid l_0 l_1 l_2$ then 3 divides  $l_0$ or $l_1l_2$
and, from  the  congruence
$$ l_1 l_2 \equiv l_0^2 \text{ mod } 3, $$
must divide  both, and $3^2 \mid \mathscr{D}_G(a_1,\ldots ,a_{12})$.
If $3\mid D$ then plainly $3^3\mid \mathscr{D}_G(a_1,\ldots ,a_{12})$.  Hence
$3\mid \mathscr{D}_G(a_1,\ldots ,a_{12})$
implies that $3^2\mid \mathscr{D}_G(a_1,\ldots ,a_{12})$.

It is easy to see that $D\equiv \det \begin{pmatrix}  a & c & b\\ b & a & c \\ c
& b & a   \end{pmatrix}$ mod 2, the circulant determinant equalling $l_0l_1l_2$,
the $\mathbb Z_3$ measure of $a+bx+cx^2$,  but in fact expanding  we have the
stronger congruence
$$ D\equiv l_0l_1l_2 \text{ mod } 4. $$
From this we see that any odd determinant must have $l_0l_1l_2$ odd and
$$ \mathscr{D}_G(a_1,\ldots ,a_{12})\equiv (l_0l_1l_2)^4 \equiv 1 \text{ mod }
4. $$
If the determinant is even then 2 divides $l_0l_1l_2$ or $D$ and from the
congruence must divide both. Moreover if $2\parallel  l_0 l_1 l_2$
then $2\parallel D$  and $2^4 \parallel  \mathscr{D}_G(a_1,\ldots ,a_{12})$,
while if $2^2\mid l_0l_1l_2$ then $2^2\mid D$ and $2^8\mid
\mathscr{D}_G(a_1,\ldots ,a_{12})$.

Hence the determinants must be of the stated form. It remains to show that we
can achieve all these.
From
\begin{align*}
\mathscr{D}_G(1,1,0,0,0,0,0,0, 1,0,0,0) & =9,\\
\mathscr{D}_G(1,1,-1,0, 0,0,0,0, 0,0,0,0) & = -27,
\end{align*}
and  multiplication we can obtain any $\pm 3^b$ with $b\geq 2$ which is $1$ mod
$4$.

For the powers of 2 we have
\begin{align*}
 &\mathscr{D}_G(0,0,0,0, 1,0,0,0, 1,0,0,0)  = 2^4,\\
& \mathscr{D}_G(1,-1,0,0, 1,0,0,0, 1,0,0,0)  = -2^4,
\end{align*}
and
\begin{align*}
& \mathscr{D}_G(k+2,k,k,k, k+1,k,k,k, k+1,k,k,k)  =2^8(1+3k),\\
& \mathscr{D}_G(k+2,k+1, k-1,k,  k+1,k,k,k, k+1,k,k,k)  =-2^8(1+3k),
\end{align*}
with  $k=0,-1,1,-3$ giving $\pm 2^8,\pm 2^9,\pm 2^{10}, \pm 2^{11}.$
Multiplication of these gives all  $\pm 2^a$ with $a=4$ or $a\geq 8$.

The $m\equiv 1$ mod 4 with $(m,6)=1$ can be obtained with
\begin{align*}
& \mathscr{D}_{G}(k+1, k,k,k, k,k,k,k ,k,k,k,k)=1+12k, \\
& \mathscr{D}_{G} (k+1,k,k,k, k+1,k+1,k,k, k+1,k+1,k,k)=5+12k.
\end{align*}
Multiplication of these produces  all  the forms in Theorem \ref{A4}. \qed

\section{Proofs for Section \ref{12gps}}
We write $\om$ for the primitive cube root of unity  $\om_3 =e^{2\pi i/3}$, and observe that  $\om i$ is a primitive
$12$th root of unity.

We shall need some results on factoring in $\mathbb Z [i]$
and $ \mathbb Z[\om]$ and $\mathbb Z[\om_{12}]=\mathbb Z[\om i]=\mathbb Z[\om,i]$. Note that  (see for
example \cite[Chapter 11]{Washington})  all these rings are UFDs, with the primes
splitting in $\mathbb Z[i]$ being $2$ and those $p\equiv 1$ mod 4 and in
$\mathbb Z[\om ]$ being $3$ and  those primes $p$ with
$\left(\frac{-3}{p}\right)=1$,
namely  $p\equiv 1$ or $7\bmod 12$. The primes  $p\equiv 1\bmod 12$ split in both
$\mathbb Z[\om]$ and $\mathbb Z[i]$, being a product of $4$ primes in
$\mathbb Z [\om,i]$.
We write $N_1$ and $N_2$ for the norms for $\mathbb Z[\om]$ and $\mathbb
Z[\om,i]$, so that
\begin{align*} N_1(H(\om))  & =H(\om)H(\om^2)=|H(\om)|^2,\\
 N_2(H(\om,i)) & =H(\om,i)H(\om^2,i)H(\om,-i)H(\om^2,-i).
\end{align*}

\begin{lemma}\label{normform1}
If $\alpha$ in $\mathbb Z [\om]$ has $3\nmid N_1(\alpha)$  then one of $\pm
\alpha$ takes  the form
$$-1+(A+B\om)(1-\om), \;\;\; A,B\in \mathbb Z,$$
while if  $\gcd(N_1(\alpha),6)=1$ then exactly one of $\pm \alpha,\pm \alpha \om, \pm
\alpha \om^2$
takes the form
$$   - 1 + 2(A+B\om)(1-\om),\;\;\; A,B\in \mathbb Z. $$
\end{lemma}

\begin{proof} Suppose that $3\nmid N_1(\alpha)$. Writing $\alpha=a+b(1-\om)$ we
have $3\nmid a,$ else $3\mid N_1(\alpha)=a^2+3ab+3b^2$.
Replacing $\alpha $ by $-\alpha$ as necessary we can assume that $a\equiv -1$
mod 3 and, writing $a=-1+3k$, $3=(1-\om)(2+\om)$
we get
$$\alpha = -1  +(A+B\om)(1-\om).    $$
Suppose also that $2\nmid N_1(\alpha)$. We can't have $A$ even and $B$ odd,
else
$$\alpha = 2w+(A+(B-1)\omega)(1-\omega) \; \Rightarrow \; 2\mid N_1(\alpha). $$

 If $A,B$ are both odd we take
$$ \alpha \om= -1+ ((1-B) + (A-B)\om)(1-\om), $$
and if $A$ is odd and $B$ even we can take
$$ \alpha \om^2 =-1 + ( (B-A +1)+ (1-A)\om)(1-\om). $$
With $A,B$ even the $\alpha \om^j$   cycle through the three possible parity
combinations.
\end{proof}

In \eqref{defS} we partitioned the primes $p\equiv 1\bmod 12$  into two
sets $\P_1$ and $\P_2$.
\begin{lemma}\label{normform}
If $p\equiv 7\bmod 12$ then $p=N_1(\alpha)$ for an
$$ \alpha = -1 + 2(1-\om) (2A+1 + B\om), \;\;\;\; A,B \in \mathbb Z, $$
with $B$ even (also  one  with $B$  odd).

If $p\equiv 1\bmod 12$ then $p=N_1(\alpha)$ for an
\be \label{norm1mod12}  \alpha = -1 + 2(1-\om) (2A + B\om), \;\;\;\; A,B \in
\mathbb Z, \ee
with $B$ even when $p\in\P_2$ and $B$ odd when $p\in\P_1$.

If $p\equiv 1\bmod 12$ then  $p=N_2(\alpha_1)$ for some
\be \label{form1}  \alpha_1 = -1 + (2A+2B\om)(1-\om) +
i\left(2+(C+2D\om)(1-\om)\right),\;\;\; A,B,C,D\in \mathbb Z, \ee
with  $C$ even when $p\in \P_1$ and $C$ odd when $p$ is in $\P_2$. Also
$p=N_2(\alpha_2)$ for some
\be \label{2ndform}  \alpha_2= (A+1+2B\om)(1-\om) +  i \left(1+(C+D\om)(1-\om)\right), \;\;\;
A,B,C,D\in \mathbb Z,\ee
with  $A,C$ and $D$ all  even when $p\in \P_1$ and all  odd when $p\in\P_2$.
\end{lemma}

\begin{proof}
As above for $p\equiv 7\bmod 12$  we can take $\alpha = -1 + 2(a+b\om)(1-\om)$.
Note that $a$ must be odd, since otherwise
$\alpha \equiv -1 +2b\om(1-\om)\equiv -1+2b \text{  mod  } 4$ and
$N(\alpha)\equiv 1$ mod 4. If $b$ is also odd  then we can replace
$\alpha$ by its conjugate $-1+2(a+b\om^2)(1-\om^2)= -1 + 2(1-\om)(a+(a-b)\om)$
with $(a-b)$ even.

Supposing $p\equiv 1\bmod 12$, then $p=N_2(\alpha)$ for some $\alpha
=(a+b(1-\om))+i(c+d(1-\om))$.
We can't have $3\mid a$ and $3\mid c$, else $3\mid N(\alpha)$ and we can assume
that $3\nmid ac$ else we replace $\alpha$ by
$$ (1+i\om)\alpha = (a-c-3d) +(1-\om)(b+c+2d) +i( a+c+3b+(1-\om)(d-a-2b)). $$
Replacing $\alpha$ by $-\alpha$ or $\pm i \alpha$ we can assume that $a,c\equiv
2$ mod  3 and writing $3=(1-\om)(2+\om)$ we can write
$$\alpha = -1+(1-\om)(A+B\om) + i\left(2 + (1-\om)(C+D\om)\right). $$
It remains to show that we can take $A,B,D$ all even. We work first with $A,B$.
We have
\begin{align*}
\om ( -1+(A+B\om)(1-\om)) & = -1 + (1-B+(A-B)\om)(1-\om),\\
\om^2 ( -1+(A+B\om)(1-\om)) & = -1 + (B-A+1+(1-A)\om)(1-\om),
\end{align*}
so if $A,B$ are both odd we can replace  $\alpha$ by $\om \alpha $  and if $A$
is odd and $B$ even by $\om^2\alpha$. If $A$ is even and $B$ odd we write
$$ -1 + (A+B\om)(1-\om)= 2+(1-\om)(A-2+(B-1)\om).$$
Notice this does not occur if $C,D$ are both even else $2\mid \alpha$, so
replacing $\alpha$ by a conjugate of  $-i\alpha$ we reverse the roles of the
$A,B$ and $C,D$ and then work to make the new $A,B$ both even noting that the
process keeps $C,D$ even
\begin{align*}
\om ( 2+(C+D\om)(1-\om)) & = 2 + (-D-2+ (C-D)\om)(1-\om),\\
\om^2 ( 2+(C+D\om)(1-\om)) & = 2 + (D-C-2+(-C-2)\om)(1-\om).
\end{align*}
So suppose that $A,B$ are even. We can't have $C$ even, $D$ odd else $\alpha
\equiv -1 +i \text{ mod } 2$
and $2\mid N(\alpha)$. If $C$ and $D$ are both odd then we can replace $\alpha$
by its conjugate
\begin{align*}  &  -1+ 2(A+B\om^2)(1-\om^2) + i
\left(2+(C+D\om^2)(1-\om^2)\right) \\
 & \hspace{5ex} =  -1+ 2(A+(A-B)\om)(1-\om) + i\left(
2+(C+(C-D)\om)(1-\om)\right),    \end{align*}
so we can assume that $D$ is also even.  Observing that
\begin{align*}
\alpha &=  -1+2(A+B\om)(1-\om)+i(2+(C+2D\om)(1-\om) ),\\
\alpha' &  = -1+2(A+B\om^2)(1-\om^2)+i(2+(C+2D\om^2)(1-\om^2),
\end{align*}
have $\alpha \alpha'=-3\delta + i\rho,$ with
\begin{align*}
\delta & =1+C^2+2A+2C-2CD+4D^2-4(A^2-AB+B^2),\\
\rho  & =  -4-3C+6(2A+2AC+4BD-BC-2AD),
\end{align*}
plainly $C$ even leads to $2\mid \rho$ and an $\P_1$  representation of $p$ and
if $C$ is odd $2\mid \delta$ and $p$ must be in $\P_2$.

With
$$\alpha_1=-1+(2A+2B\om)(1-\om)+i\left( 2 + (C+2D\om)(1-\om)\right)$$
we take  $\alpha_2=(\om^2+i)\alpha_1=\delta_2+i\rho_2$ where
\begin{align*}
\delta_2 & = \left( (2B-2A-C-1) -(2A+2D)\om\right)(1-\om),\\
\rho_2 & = 1 +\left((2A+2D-C-2)+(2B-C-2)\om\right)(1-\om),
\end{align*}
and the second form \eqref{2ndform}  is plain.
Observe that
\begin{align*}
\alpha &=  -1+2(A+B\om)(1-\om)+i(2+(C+2D\om)(1-\om) ),\\
\alpha'' & =  -1+2(A+B\om)(1-\om)-i(2+(C+2D\om)(1-\om)),
\end{align*}
has
$$ \alpha \alpha'' = 5+ C^2(1-\om)^2 +4(1-\om)h(\om). $$
When   $p$ is in $\P_1$ and $C$ is even
$$ \alpha\alpha''=-1+2(1-\om)\Big(2+\om + 2(C/2)^2(1-\om)+2h(\om)\Big),$$
giving \eqref{norm1mod12} with $B$ odd, while  when $p$ is in $\P_2$ and $C$ is
odd
$$ \om \alpha \alpha''
=-1+4(1-\om)\left(\frac{1}{4}(C^2-1)+\frac{1}{2}(C^2+1)\om +\om h(\om)\right),
$$
giving \eqref{norm1mod12} with $B$ even.
\end{proof}

\begin{comment}

Before embarking on the proofs of our theorems, we begin comparing the Lind measure of a polynomial
$F(x,y)=f(x)+yg(x)$ for the groups $G=D_{12}$, $Q_{12}$ and $\mathbb Z_6 \times
\mathbb Z_2$:
$$
M_{D_{12}}(F)  = ab_1 (c d_1)^2, \;\;
M_{Q_{12}}(F)  =ab_2 (c d_2)^2, \;\;
M_{\mathbb Z_6 \times \mathbb Z_{2}}(F)  =ab_1 e_1 e_2 e_3e_4, $$
where  we have the integers
\begin{align*}
 & a:  =f(1)^2-g(1)^2,\\
&  b_1:  =f(-1)^2-g(-1)^2,\;\;\;\; b_2:=f(-1)^2+g(-1)^2, \\
 &    c:  =|f(\omega)|^2-|g(\omega)|^2,\\
& d_1  := |f(-\omega)|^2-|g(-\omega)|^2,\;\;\;\; d_2:=|f(-\omega)|^2+|g(-\omega)|^2,\\
 & e_1:=  |f(\omega)+g(\omega)|^2,     \;\; e_2:=|f(\omega)-g(\omega)|^2 ,\;\;\;
e_1e_2=N_1(f(\om)^2-g(\om)^2),\\
 &  e_3:=  |f(-\omega)+g(-\omega)|^2,     \;\;  e_4:=|f(-\omega)-g(-\omega)|^2,\;\; \;
e_3e_4=N_1(f(-\om)^2-g(-\om)^2).
\end{align*}
Here $f$ and $g$ come from the formulae \eqref{E-D} and \eqref{E-Q} for $M_G(F)$ for $G=D_{12}$ and $G=Q_{12}$ respectively.
\end{comment}

We write
$$ h(x) := \frac{x^6-1}{x-1} =\sum_{j=0}^5 x^j,\;\;\; h(1)=6,\; h(-1)=h(\pm
\om)=h(\pm w^2)=0. $$

\begin{proof}[Proof of Theorem \ref{Q12}]

As before, we let $m_t$ (e.g., $m_2,m_6$) denote an arbitrary integer
coprime to $t$.
From the formula  \eqref{E-Q}  we know that $\S(Q_{12})$ consists of measures $M:= M(F):=M_{Q_{12}}(F)$ of the form
\be \label{E-MQ12} M  =ab (c d)^2, \ee
where, if  $F(x,y)=f(x)+yg(x)$,
\[
a:  =f(1)^2-g(1)^2, \; b:=f(-1)^2+g(-1)^2,\]
\[
c:  =|f(\omega)|^2-|g(\omega)|^2,\;
 d:=|f(-\omega)|^2+|g(-\omega)|^2.
 \]
The proof proceeds by a series of steps, followed by their proofs:

\vspace{1ex}

{\bf  \noindent Step 1.} We have $c\equiv a \bmod 3$,  $d\equiv b \bmod 3$ and
\be \label{27div} 3\nmid M   \;\;\; \hbox{ or }\;\;\; 3^3\mid M. \ee
\vspace{1ex}

Since $\om \equiv 1$ mod $(1-\om)$ in $\mathbb Z[\om]$ and $a$ and $c$ are
integers we have $c\equiv a \bmod 3$. Likewise $d\equiv b \bmod 3$. So
$cd\equiv ab$ mod 3 and  $M\equiv (ab)^3$ mod $3$. Hence   $3\mid
M$ iff $3\mid ab$ and $3\mid cd$, proving Step 1.

\vspace{1ex}

{\bf  \noindent Step 2.} All odd integers of the forms $M=m_6$ and $M=3^3m_2$ belong to $\S(Q_{12})$.

\vspace{1ex}
From Step 1 we know that all odd $M$ are of one of these two forms. Conversely,
we obtain all odd integers that are $1 \bmod 4$ satisfying \eqref{27div} as follows:
\begin{align} & \label{1mod12}   M\left(1+t \:h(x) +yt\: h(x)\right) =
1+12t, \\
 & M\left(1+x(x^3+1)+t\: h(x) +y\left((1+x^3)+ t\:h(x)\right)\right)  =
5+12t,\label{5plus12t} \\
  &   M\left(1+t \; h(x) +y\left((1+x^3)+t\:h(x)\right)\right) =-3^3
(1+4t).\label{27}
\end{align}
Swapping $f$ and $g$ in \eqref{E-Q} changes the sign, by \eqref{E-ggff}, so that these identities give us all odd integers
satisfying \eqref{27div}.

This deals with the odd measures, so we can assume that $M$ is even from now on. In particular, at least one of $ab$ and $cd$ is even.

\vspace{1ex}

{\bf  \noindent Step 3.}  We have $c\equiv d \bmod 2$, $f(1)\equiv f(-1)\bmod 2$, $g(1)\equiv g(-1)\bmod 2$, and $ab$ even iff $f(1)\equiv g(1)\bmod 2$.

\vspace{1ex}

{\bf  \noindent Step 4.} We have $2^4\mid M$, and $2^5\parallel M$ possible only when $c$ and $d$ are both odd, and $f(-1),g(-1)$ are both odd, or both even with $f(-1)/2$, $g(-1)/2$  both odd.

\vspace{1ex}

If $cd$ is even, then from $d\equiv c \bmod 2$ we have that $2^4\parallel (cd)^2$ or $2^6\mid (cd)^2$.

 If $f(1)$
and $g(1)$ are both odd then $2^3\mid a$ and $2\parallel b$.
If both are even then,  writing
\[
f(1)=2\alpha_1,\quad f(-1)=2\alpha_2, \quad g(1)=2\beta_1, \quad g(-1)=2\beta_2,
\]
we get
$a=4(\alpha_1^2-\beta_1^2)$ and $b=4(\alpha_2^2+\beta_2^2)$.
Hence  $2^4 \mid ab$, and  we get  $2^4\mid M$,
with $2^5\parallel M$ only as specified.

\vspace{1ex}

{\bf  \noindent Step 5.} All $M$ with $2^4\parallel M$ with
property \eqref{27div} are possible.

\vspace{1ex}

This is seen  using
\begin{align*}
 & M\left(1+ x^2+t \:h(x) +yt\:h(x)\right) =2^4 (1+6t), \\

 & M\left((1+x^2)+t \:h(x) +y\left((1+x^2)(1+x^3)+t\:h(x)\right)\right) =-3^3
2^4(1+2t),
\end{align*}
recalling that we can use \eqref{E-ggff} to change the sign on the first of these.

\vspace{1ex}

{\bf  \noindent Step 6.} All $M$ with $2^6\mid M$ with
property \eqref{27div} are possible.

\vspace{1ex}

This is seen using \eqref{E-ggff} and
\begin{align*}

& M\left((1+x+x^2+x^4)+t \:h(x) +yt\;h(x)\right) =2^6(1+3t),\\
 & M\left((1+x^2)+t \:h(x) +y\left(-(1+x^3)+t\:h(x)\right)\right) =2^63^3t.
\end{align*}
So the even measures with
$2^4\parallel M$ or $2^6\mid M$  are as claimed.

For the rest of the proof we can assume that $2^5\parallel M$ and that $3^\beta\parallel M$. We know that $\beta=0$ or $\beta\ge 3$.

\vspace{1ex}

{\bf  \noindent Step 7.} If $\beta =4$ or $\beta \ge 6$ then $M=2^53^\beta m_6$.

\vspace{1ex}

We get  $2^53^4$ and $2^53^6$ from
 $$M\left( (1-x-x^5)+h(x)
+y\left((x^4+x^2+1)+(x^2-1)(x+1)^{\delta}\right)\right) =2^5 3^{4+2\delta}. $$
Hence, multiplying by an odd measure, we can
obtain  any $2^53^\beta m_6$ with  $\beta =4$ or $\beta \geq 6$.

This just leaves  $\beta =0,3$ or $5$, which we can now assume holds.

\begin{comment}{
\begin{align*}
& M_G\left( (1-x-x^5)+(2t+1)h(x) +y\left((x^4+x^2+1)+(x^2-1)(x+1)^{\delta} +2t\:
h(x)\right)\right) \\
 & \hspace{5ex}=2^5 3^{4+2\delta}(1+3t) \\
& M_G\left( (x^4+x^2+1)-(x^3+1)+(t+1)h(x) +y\left(x(1+x)-(x^4+x^2+1)+t
\:h(x)\right)\right)\\
  & \hspace{3ex} =2^5 3^{7}(1+2t) \end{align*}
we have any $2^53^bm$, $(m,6)=1$ with  $b=4$ or $b\geq 6$.
This just leaves  $b=0,3$ or $5$.}

\end{comment}
\vspace{1ex}

{\bf  \noindent Step 8.} We have $3\nmid b$.

\vspace{1ex}

 For if $3\mid b=f(-1)^2+g(-1)^2$ then $3\mid f(-1),g(-1)$
with $(f(-1)/3)^2+(g(-1)/3)^2$  coprime to 3 or a multiple of $3^2$,
giving $3^2\parallel b$ or $3^4\mid b$. Then also, from Step 1,
$d\equiv b\equiv 0\bmod 3$,
 with $d$ contributing an even power of $3$ to $bd^2$, so that $3^4\parallel  bd^2$ or $3^6\mid bd^2$. Also, from $a\equiv c\bmod 3$ we have either
$3\nmid a$ or $3^3\mid ac^2$.
 Hence  $3^4\parallel M$ or $3^6\mid M$, contrary to assumption.

\vspace{1ex}

{\bf  \noindent Step 9.} If $3\mid f(-1)g(-1)$ then $p\mid M$ for some prime $p\equiv 5\bmod 12$.

\vspace{1ex}

Supposing first that $3\mid f(-1)g(-1)$, then we have a factor of $M$ of the form
$$f(-1)^2+g(-1)^2 \; \text{ or}\; (f(-1)/2)^2+(g(-1)/2)^2=A^2+(3B)^2=2k $$ with
$k$ odd and $2k\equiv 1$ mod 3. Thus $k\equiv 2$ mod 3 is an odd sum of two
squares, so must contain an odd power of a  prime $p\equiv 2$ mod 3 which  must
be  a sum of two squares, and hence $p\equiv 5\bmod 12$.
Conversely suppose that we have a prime $p\equiv 5\bmod 12$ then $2p$ is a sum of
two squares $2p=a^2+b^2$ where $a$ and $b$ must be odd and one of them a
multiple of 3, $2p=(1+6A)^2+(3+6B)^2$.

\vspace{1ex}

{\bf  \noindent Step 10.} Conversely every  $2^5p$ with $p\equiv 5\bmod 12$  is an $M$.

\vspace{1ex}
This is because if $2p=(1+6A)^2+(3+6B)^2$ then
$$ M\left( -1-Ah(-x) +h(x) + y\left( x^4+x^2+1+ Bh(-x)\right) \right)   =
2^5p. $$

\begin{comment}
{\begin{align*}
& M_G\left( -1-A h(-x) +(2t+1)h(x) + y\left( x^4+x^2+1+Bh(-x)+2t\:h(x)\right)
\right)  \\
 & \hspace{5ex} = 2^5p(1+3t)\\
& M_G\left( (x^5+x-1)+ B\frac{x^6-1}{x+1} + m\:h(x) +
y\left(-1+A\frac{x^6-1}{x+1}+m\:h(x)\right) \right) \\
 & \hspace{5ex} = 2^4 3^3 pm.
\end{align*}
}
\end{comment}

\vspace{1ex}

{\bf  \noindent Step 11.} If $3\nmid f(-1)g(-1)$ then $p^2\mid M$ for some prime $p\equiv 5\bmod 6
$.

\vspace{1ex}

For if $3\nmid f(-1)g(-1)$ then
$$d=|f(-\omega)|^2+|g(-\omega)|^2\equiv f(-1)^2+g(-1)^2 \equiv 2 \text{ mod  } 3$$
 is odd. Hence this term
must be divisible by a prime $p\equiv 5$ mod 6 and $M$ by $p^2$.

\vspace{1ex}

 {\bf  \noindent Step 12.} Conversely every such $2^5p^2$ with $p\equiv 5\bmod 6$  is an $M$.

\vspace{1ex}

We can write a $p\equiv 5$ mod 6 as
a sum of two norms of elements in $\mathbb Z[w]$,
$p=N_1(\alpha)+N_1(\gamma)$; we can do this since any integer not of the form
$9^k(9n+6)$ can be represented by $x^2+xy+y^2+z^2$ -- see Dickson \cite{Dickson}
(in fact  $x^2+xy+y^2+z^2+zw+w^2$ should represents all integers
\cite{290}).
Moreover since $p$ is odd  one of the norms must be even and hence an
element in $2\mathbb Z[w]$ and we can write $p=N_1(\alpha)+4N_1(\beta)$ with
$N_1(\alpha),N_1(\beta)\equiv 1$ mod 3. By Lemma \ref{normform1}  we can assume
that $\alpha = -1+ 2(1-\omega)(A+B\omega)$ and $\beta =-1+(C+D\omega)(1-\omega)$. We take
\begin{align*} f(x)&= -1 -(A-Bx)(x^3-1)(1+x)+ h(x) \\
g(x) & = (x^2+x+1)- (C-1-Dx)(x^3-1)(1+x). \end{align*}
Then $f(1)^2-g(1)^2=2^4$, $f(-1)^2+g(-1)^2=2,$ $|f(\omega)|^2-|g(\omega)|^2=1$,
$|f(-\omega)|^2+|g(-\omega)|^2=N_1(\alpha)+N_1(2\beta)=p$ and $M_G(f(x)+yg(x))=2^5p^2$.

\comments{\begin{align*} f(x)&= -1 -(A-Bx)(x^3-1)(1+x)+ (2m+1)h(x) \\
g(x) & = (x^2+x+1)- (C-1-Dx)(x^3-1)(1+x)+ 2m\: h(x). \end{align*}
Then $f(1)^2-g(1)^2=2^4(1+3m)$, $f(-1)^2+g(-1)^2=2,$ $|f(\omega)|^2-|g(\omega)|^2=1$,
$|f(-\omega)|^2+|g(-\omega)|^2=N_1(\alpha)+N_1(2\beta)=p$ and $M_G(f(x)+yg(x))=2^5mp^2$.
}

\vspace{1ex}

{\bf  \noindent Step 13.} Any $2^53^\beta km_6$, with $k=p$, $p\equiv 5 \bmod 12$,  or $k=p^2$,
$p\equiv 5 \bmod 6$ and $\beta =0$ or $\beta \geq 3$ is an $M$.

\vspace{1ex}

Hence we can get any $2^5k$, where  $k=p$, $p\equiv 5 \bmod 12$,  or $k=p^2$,
$p\equiv 5$ mod 6, and by  multiplicativity any $2^53^\beta km_6$, with
$\beta =0$ or $\beta \geq 3$.
\end{proof}

\

We remark that the evaluation of $\S(D_{12})$, already known from \cite{dihedral}, can be modelled on the first six steps of the above proof for $Q_{12}$. In place of \eqref{E-MQ12}, for $G=D_{12}$  formula \eqref{E-D} gives $M = ab_1 (c d_1)^2$, with $b_1:  =f(-1)^2-g(-1)^2$ and $d_1:= |f(-\omega)|^2-|g(-\omega)|^2$. Step 1 of the proof also holds for $G=D_{12}$, because $d_1\equiv b_1\bmod 3$. Step 2 also follows using the identities \eqref{1mod12}, \eqref{5plus12t} and \eqref{27}. For $D_{12}$ swapping $f$ and $g$ is not necessary because odd $M\equiv (acd_1)^2\equiv 1\bmod 4$. In Step 3 we still have $c\equiv d_1\bmod 2$. In Step 4 we  also  have $2^3\mid b_1$ when $f(1)$ and $g(1)$ are both odd and $b_1=4(\alpha_2^2-\beta_2^2)$, when both are even, giving $2^4\parallel ab_1$ or $2^6\mid ab_1$, and so $2^4\parallel M$ or $2^6\mid M$. For Steps 5 and 6, we use the same four identities, but, without \eqref{E-ggff}
to change  signs, we need the two extra identities
\begin{align*}
& M\left((1+x^2)(1+x(x^3+1))+t \:h(x) +y\left((1+x^3)(1+x^2)+ t\:h(x)\right)\right)  = 2^4(5+6t),\\
& M\left((1-x)+t \: h(x) +y\left((1+x^2)(1+x^3)+t\:h(x)\right)\right)
=-2^6(1+3t),
\end{align*}
to complete the proof.

\begin{proof}[Proof of Theorem \ref{Z2Z6}] Suppose that $G=\mathbb Z_6\times
\mathbb Z_2$. From \eqref{E-Z62} we have
\[
M_G(F)  =ab_1 e_1 e_2 e_3e_4,
\]
for the integers
\begin{align*}
 & a:  =f(1)^2-g(1)^2,\;\;\;  b_1:  =f(-1)^2-g(-1)^2,\\
 & e_1:=  |f(\omega)+g(\omega)|^2,     \;\; e_2:=|f(\omega)-g(\omega)|^2 ,\;\;\;
e_1e_2=N_1(f(\om)^2-g(\om)^2),\\
 &  e_3:=  |f(-\omega)+g(-\omega)|^2,     \;\;  e_4:=|f(-\omega)-g(-\omega)|^2,\;\; \;
e_3e_4=N_1(f(-\om)^2-g(-\om)^2).
\end{align*}

Since $f(\om)\equiv f(\om^2)\equiv f(1)$ mod $(1-\om)$ etc.  in $\mathbb
Z[\om]$, we readily see that $e_1e_2\equiv a^2$ mod 3
and $e_3e_4\equiv b_1^2$ mod $3$ in $\mathbb Z$. Hence we cannot have
$3\parallel  ab_1 e_1e_2e_3e_4$  and
\be \label{3div}  3^{\alpha} \parallel M_G(F) \Rightarrow  \alpha=0 \hbox{ or }
\alpha \geq 2. \ee
Moreover  $3^2\parallel M_G(F)$ implies that $3\parallel ab_1$ and $3\parallel
e_1e_2e_3e_4$.

Since $f(-x)=f(x)+2u_1(x)$ we readily see that all the $e_i\equiv e_1$ mod $2$.
Moreover considering factorisation
in $\mathbb Z[\om]$, or by checking that $2\mid N_1(A+B\om)=A^2-AB+B^2$ only
when  $2\mid A,B$, we see that
a norm of an element in $\mathbb Z[\om]$ is divisible by an even power of 2.
Hence
$$ 2^{\beta}\parallel e_1e_2e_3e_4 \Rightarrow \beta =2t, \; t=0 \hbox{ or }
t\geq 4. $$
From the proof of Theorem \ref{Q12}, and  the comments on $D_{12}$ after it,  we have that
$$ 2^{\beta} \parallel ab_1 \Rightarrow \beta=0,4 \hbox{ or } \beta \geq 6. $$
Hence we certainly have
\be \label{2div}  2^{\beta} \parallel M_G(F) \Rightarrow \beta =0,4 \hbox{ or }
\beta \geq 6. \ee
Similarly, since $f(-x)^2=f(x)^2+4u_2(x)$ we get $b_1\equiv a$ mod $4$ and
$e_3e_4\equiv e_1e_2$ mod 4
and if $ab_1$ is odd then $ab_1\equiv a^2\equiv 1$ mod $4$ and if
$e_1e_2e_3e_4$ is  odd then it is 1 mod 4.
In particular
\be \label{oddity} M_G(F) \text{ odd } \Rightarrow M_G(F)\equiv 1 \text{ mod }
4. \ee

\noindent
{\bf Case (a)} We begin with the multiples of 27. We can achieve all odd
multiples  satisfying \eqref{oddity} using \eqref{27}, and all even multiples
satisfying \eqref{2div} with
\begin{align*}
& M_G\left( (x^2+1)(x^3+1)+ kh(x) + y \left( (x^2+1)+k\:h(x)\right) \right)=-2^4
3^3(1+2k), \\
& M_G\left( 1+ m\:h(x) + y \left( (1-x-x^5) +m\:h(x)\right) \right)=-2^6 3^3m.
\end{align*}

\noindent
{\bf Case (b)} Suppose next that $3^2\parallel  M_G(F)$.
Consider first the case $e:=e_1e_2e_3e_4$ odd. From the above discussion we know
that  $e\equiv 1$ mod 4 and $3\parallel e$. So  $e/3\equiv 3$ mod $4$,
and must be divisible by an odd power of a prime $p\equiv 3$ mod 4 with $p\neq
3$.
Since $e$ is a norm in $\mathbb Z[\om]$ we must have $p\equiv 7\bmod 12$.
Conversely suppose that $p\equiv 7\bmod 12$.
By Lemma \ref{normform} we have $p=N_1(\alpha_1)$ with $\alpha_1= 1-2\om
+4(A+B\om)(1-\om)$. Take
\begin{align*}
f(x) & = (x+1) -(A-Bx)(x^3-1)(x+1) + m\:h(x),\\
g(x) & = x -(A-Bx)(x^3-1)(x+1) + m\: h(x).
\end{align*}
Then $a=3(1+4m)$, $b_1=-1$,
$e_1e_2=|f(\om)^2-g(\om)^2|^2=|\om-\om^2|^2=3,$
$e_3=N_1( \alpha_1)=p,$ and $ e_4=1, $
giving   $M_G(F)=-9(1+4m)p$, and hence all the odd multiples of $9p$ satisfying
\eqref{oddity}.
Likewise
\begin{align*}
f(x) & = (x+1) -(A-Bx)(x^3-1)(x+1) + (m-1)\:h(x),\\
g(x) & = (x^2 +1) - (A-Bx)(x^3-1)(x+1) - m\: h(x),
\end{align*}
has  $a=12(1-2m)$, $b_1=-4$,
$e_1e_2=N_1(\om-\om^2)=3,$
$e_3= N_1(\alpha_1)$ and $e_4=1$,
and $M_G(F)=9\cdot 2^4(2m-1)p$,
 while
\begin{align*}
f(x) & = (x+1) +(x^4+x^2+1)-(A-Bx)(x^3-1)(x+1) + (m-1)\:h(x),\\
g(x) & = x -(A-Bx)(x^3-1)(x+1) + m\:h(x).
\end{align*}
has  $a=-24m$, $b_1=8$,
$e_1e_2=3,$ $e_3e_4=p,$
and $M_G(F)=-3^2\cdot 2^6mp$. Hence we can  achieve all the even multiples of
$9p$ satisfying \eqref{2div}.

Suppose now that $3^2\parallel M_G(F)$ with $e=e_1e_2e_3e_4$ even, then
$2^{8+2t}\parallel e$ and $3\parallel e$.
We can suppose that $M_G(F)$ is not divisible by a prime $p\equiv 7\bmod 12$,
since we already have all such  multiples of these. Hence the odd powers of
primes, other than $3$, dividing
$e$ are all 1 mod 4 and hence $e 2^{-8-2t}3^{-1}$ is 1 mod 4. Also the odd $ab_1$  are
1 mod 4 and the evens are odd multiples of $2^4$ or multiples of $2^6$.
Thus  either $M_G(F)=2^8m$ or $2^{10}m$ with $m\equiv -1$ mod 4, or
$2^{12}\parallel M_G(F)$ or $2^{14}\mid M_G(F)$. We can achieve all these:
\begin{align*}
f(x) & = 2(x^2+1)-(x^4+x^2+1) + m\:h(x),\\
g(x) & = (x^3+1)+ m\; h(x).
\end{align*}
has  $a=-3(1+4m)$, $b_1=1$,
$e_1e_2=N_1(4\om^2-4)=2^4\cdot 3,$ $e_3e_4=N_1(4\om^2)=2^4$
and  $M_G(F)=-3^2\cdot 2^8(1+4m)$, while
\begin{align*}
f(x) & = (1-x^2)-2(x^3+1)+(m+1)\:h(x),\\
g(x) & = -x^4+x^2(x^3+1)+ m\: h(x),
\end{align*}
has  $a=3(1+4m)$, $b_1=-1$,
$e_1=N_1(2(\om^2-1))=2^2\cdot 3,$ $e_2=N_1( 4\om)=2^4,$
$e_3=N_1(2)=2^2$, $e_4=N_1(-2\om^2)=2^2,$
and  $M_G(F)=-3^2\cdot 2^{10}(1+4m)$.

Taking
\begin{align*}
f(x) & = 2(x^2+1)-h(-x)+m\: h(x),\\
g(x) & = (x^3+1)+ m\:h(x),
\end{align*}
has  $a=12(1+2m)$, $b_1=4$,
$e_1e_2=N_1(4\om^2-4)= 2^4\cdot 3,$ $e_3e_4=N_1(4\om^2)=2^4,$
and  $M_G(F)=3^2\cdot 2^{12}(1+2m)$. Similarly
\begin{align*}
f(x) & = 2(x^2+1)-(x^4+x^2+1)+m\;h(x),\\
g(x) & = (x^3+1)-(x^4+x^2+1)+ m\:h(x),
\end{align*}
has  $a=24m$, $b_1=-8$,
$e_1e_2= 2^4\cdot 3,$ $e_3e_4=2^4,$
and  $M_G(F)=-3^2\cdot 2^{14}m$.

\vspace{1ex}
\noindent
{\bf Case (c)} Finally we deal with the measures coprime to 3.  We achieve any
value $1\bmod 12$ with \eqref{1mod12}.
We can also get all odd multiples of $2^4$ or multiples of $2^6$  when  the
multiple is 1 mod 3:
\begin{align*}
& M_G\left( (x^2+1)+ m\: h(x) + y m\: h(x) \right)=2^4(6m+1), \\
& M_G\left( (x^4+x^2+1)+ m\:h(x)+ y \left( 1+m\:h(x)\right) \right)=2^6 (3m+1).
\end{align*}
Since odd measures are 1 mod 4 and even measures are divisible by  exactly $2^4$
or at least $2^6$  this leaves the measures 5 mod 12 or $-2^4(6m+1)$ or
$-2^6(3m+1)$.

Suppose that $p\equiv 1\bmod 12$ is in $\P_1$  then, by Lemma \ref{normform},
$p=N_1(\alpha)$ for some
\be \label{formk}  \alpha =-1-2\om(1-\om) +4(A+B\om)(1-\om). \ee
Suppose $p=12t+5$ then $p=-1-2\om(1-\om) + 4((2t+1)+(t+1)\om)(1-\om)$ and
$p^2=N_1(\alpha)$ for an $\alpha$ of the form
\eqref{formk} and if $p_1,p_2\equiv 7\bmod 12$ then by Lemma \ref{normform}
$p_1p_2=N_1(\alpha)$ for some
\begin{align*} \alpha  & =\big( -1 + 2(1-\om)+4(1-\om)K_1(\om)\big)\big(
1+2(1-\om)(1+\om)+4(1-\om)K_2(\om)\big) \\
 & = -1-2\om(1-\om) +4(1-\om)K_3(\om),
\end{align*}
which is also of the form \eqref{formk}. Hence for $k=p$ with $p$ in $\P_1$ or $k=p^2$
with $p\equiv 5\bmod 12$ or $k=p_1p_2$
with $p_1,p_2\equiv 7\bmod 12$ we can write $k=N_1(\alpha)$ for an $\alpha$ of
the form \eqref{formk} and taking
\begin{align*}
f(x) &= x(x^2+x+1) - (A-Bx)(x^3-1)(1+x)+  m\:h(x)),\\
g(x) & = x(x+1)  - (A-Bx)(x^3-1)(1+x)+m\:h(x),
\end{align*}
or
\begin{align*}
f(x) &= x^2(1-x^3)- (A-Bx)(x^3-1)(1+x)+  m\:h(x),\\
g(x) & = x(x+1)  - (A-Bx)(x^3-1)(1+x)+m\:h(x),
\end{align*}
or
\begin{align*}
f(x) &= x(x^2+x+1)- (A-Bx)(x^3-1)(1+x)+  m\:h(x),\\
g(x) & = x(x+1) -(x^4+x^2+1) - (A-Bx)(x^3-1)(1+x)-m\:h(x),
\end{align*}
we have $e_1e_2= N_1(-1)=1$, $e_3=N_1(\alpha)=k$, $e_4=N_1(-1)=1$
with,
$$ (a,b_1)=(5+12m, 1),\;\;\; (-2^2(6m+1), 2^2),\;\; \text{ and } \;\;
(2^3(3m+1),-2^3),$$
respectively, and $M_G(F)=(5+12m)k,$ $-2^4(6m+1)k$ and $-2^6(3m+1)k$.

So we can achieve the \eqref{type2} and need just consider cases that do not
contain the square of a prime 5 mod 12 or two primes 7 mod 12 or a prime in
$\P_1$. Suppose first that  $e=e_1e_2e_3e_4$ is odd, then since it is 1 mod 4,
it must contain only primes from $\P_2$ and squares of primes $11\bmod 12$. Since
$e\equiv 1$ mod 3, to get a measure $2$ mod 3 we must have one of
$a=f(1)^2-g(1)^2,$ $b_1= f(-1)^2-g(-1)^2\equiv 1$ mod 3 and the other $-1$ mod
3.
Replacing $\pm x$, $\pm f$, $\pm g$ we can assume that
$f(1)\equiv 1$ mod 3, $g(1)\equiv 0$ mod 3, $f(-1)\equiv 0$ mod 3, $g(-1)\equiv
1$ mod 3. Since $f(\pm \om)\equiv f(\pm 1)$ mod $(1-\om)$, $g(\pm\om)\equiv
g(\pm1)$ mod $(1-\om)$, we can write
\begin{align*}  f(\om)+g(\om) & =1+(1-\om)(A_1+B_1\om),\\
 f(\om)-g(\om) & =1+(1-\om)(A_2+B_2\om),\\
 f(-\om)+g(-\om) & =1+(1-\om)(A_3+B_3\om),\\
 f(-\om)-g(-\om) & =-1+(1-\om)(A_4+B_4\om),
\end{align*}
and since $f(-\om)\equiv f(\om)$ mod $2$ and $g(-\om)\equiv g(\om)$ mod 2 and 2
is prime we readily see that the $A_i$ have the same parity and the $B_i$ all
have the same parity.
Multiplying by $x$ or $x^2$ as necessary we can assume that $A_1$ and $B_1$ are
both even and hence all the $A_i$ and $B_i$ are all even. All the primes in
$\P_2$ or $11\bmod 12$ factor in $\mathbb Z[\om]$ into $1+4(A+B\om)(1-\om)$ times
a unit, and (however we factor) an expressions of this type  with $A_i$, $B_i$
even will have $A_i$, $B_i$  both a multiple of 4.
Hence $g(\om)\equiv 0$ mod 2 and $g(-\om)\equiv 1$ mod 2 contradicting
$g(\om)\equiv g(-\om)$ mod 2, so we have no extra measures.
This leaves $e$ even and hence $2^{8+2\ell_1}\parallel e$ and
$2^{\ell_2}\parallel ab_1$
with $\ell_1=0,4$ or $\ell_1\geq 6$ and hence
$2^t \parallel M_G(F)$ with $t=8,10,12$ or $t\geq 14$.
We can get all the multiples of $2^{12}$ and $2^{14}$:
\begin{align*}
f(x) & = (x^2+x+1) + m\:h(x),\\
g(x) & = (x^2-x+1) + m\: h(x),
\end{align*}
and
\begin{align*}
f(x) & =( x^2+x+1)-(x^4+x^2+1) + m\: h(x),\\
g(x) & = (x^2-x+1)-(x^4+x^2+1) - m\: h(x),
\end{align*}
have $e_1e_2=N_1(-4\om^2)=2^4$, $e_3e_4=N_1(4\om^2)=2^4$ with, respectively,
$$a=2^3(1+3m),\;\; b_1=-2^3 \;\;  \text{  and }  \;\; a=-2^2(1+6m), \; b_1=2^2,
$$
giving $M_G(F)=-2^{14}(3m+1)$ and $-2^{12}(6m+1)$.

For $t=8$, 10 we have  $ab_1\equiv 1$ mod 4. If $e$ does not contain a prime  7
mod 12 then
 $e=2^{\ell_2}n'$ with $\ell_2=8$ or $10$ and  $n'\equiv 1$ mod 4 and
$M_G(F)=2^8n$ or $2^{10}n$ with $n\equiv 1$ nod 4, and all these 2 mod 3 are
obtainable:
\begin{align*}
f(x) & =( x^2+x+1)+ m\:h(x),\\
g(x) & = (x^3+1)+  m\:h(x),
\end{align*}
has $e_1e_2=N_1(-2^2)=2^4$, $e_3=e_4=N_1(-2\om)=2^2,$
$a=(5+12m),$ $b_1=1$,
and
\begin{align*}
f(x) & =-1-2x^3- m\:h(x),\\
g(x) & = -1+(x^2+x+1)+m\:h(x),
\end{align*}
has $e_1e_2=N_1(2^3)=2^6$, $e_3=N_1(-2\om)=2^2$, $e_4=N_1(2+2\om)=2^2$ with
$a=(5+12m),$ $b_1= 1, $
giving $M_G(F)=2^8(5+12m)$ and $2^{10}(5+12m)$ respectively.

If $e$ contains a prime $p\equiv 7\bmod 12$ (since it has at most one the
remaining odd primes are in  $\P_2$ or squares of primes 11 mod 12)  we have
$M_G(F)=2^4pn$ or $2^6pn$ with $n\equiv 1$ mod 4 and all are again achievable.
We write
$$p=N_1(1+2(A+B\omega)(1-\omega)), \;\;\;  k(x):=-x(A-Bx)(x^3-1)(1+x),$$
then adding $k(x)$ to both $f(x)$ and $g(x)$  in the previous two examples,
instead of
$e_3=N_1(-2\om)$   we have  $e_3=N_1(-2\om-4w(A+B\omega)(1-\omega))=2^2p$, with the other
values remaining  unchanged, and $M_G(F)=2^8(5+12m)p$ and $2^{10}(5+12m)p$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{Z12}] We can write
$$ F(x)=f(x^2)+xg(x^2)\in \mathbb Z[x],\;\;\; f(x)=\sum_{j=0}^5 a_jx^j,\;\;
g(x)=\sum_{j=0}^5 b_jx^j $$
so that
$$M_G(F)=\prod_{j=0}^{11} F(\om_{12}^j) =a b  s_1s_2$$
where, as before $a=f(1)^2-g(1)^2$, $b=f(-1)^2+g(-1)^2$, and
$$ s_1:=N_1(f(\om)^2-\om g(\om)^2),\;\;s_2:=N_1(f(-\om)^2+\om g(-\om)^2).$$
Observe that $s_1\equiv a^2$ mod 3 and $s_2 \equiv b^2$ mod 3, so either
$3\nmid M_G(F)$ or $3^2\mid M_G(F)$.

\noindent
{\bf Case (i)} Suppose that $M_G(F)$ is odd. Note that $M_G(x)=-1$ giving us
both $\pm m$ for measures $m$.
For the odd values we can get any integer coprime to $6$ using
$ M_G\left(1+ m\:k(x)\right)=1+12m$ and
\[
M_G\left(\frac{x^5-1}{x-1}+
m\:k(x)\right)=5+12m,\;\; k(x):=\frac{x^{12}-1}{x-1},
\]
and any odd multiple of $27$ with
$$ M_G\left(1+x^3+x^6+t \:k(x)\right)=3^3(1+4t). $$
 This leaves  $9m,$  $2,3\nmid m.$
Moreover  we  cannot have $3\mid b$ unless $3\mid f(-1),g(-1)$ and $3^2\mid
b$ and $3^3\mid M_G(F)$. So we assume that $3\nmid s_2b$
 and $3\parallel a$, $3\parallel s_1$.

Suppose first that $3\nmid f(-1)g(-1)$. Then $b\equiv 2$ mod 3 and hence is
divisible by an odd power of some
prime $p\equiv 2$ mod 3 which must itself be a sum of two squares, and $p\equiv
5\bmod 12$. Conversely suppose that
$p\equiv 5\bmod 12$ then $p$ is a sum of two squares both must $\pm 1$ mod 3 with
one even and one odd. So
changing signs as necessary $p=(6A+1)^2+(6B+2)^2$. Taking
\begin{align*}
f(x)= &  (x^2+1) -B(x^4+x^2+1)(x-1) + m\:h(x),    \\
g(x)= &  1 -A(x^4+x^2+1)(x-1) + m\:h(x),
\end{align*}
we have $a=3(1+4m)$, $ b= (6B+2)^2+(6A+1)^2=p$, $s_1=N_1(\om^2-\om)=3$,
$s_2=N_1(\om^2+\om)=1$
and $M_G(F) = 9p(1+4m)$.

Suppose now that $3\mid f(-1)g(-1)$.
Writing
\begin{align}  \label{aandalpha} u & := |f(\om)|^2+|g(\om)|^2,\;\;\; v:=\om
f(\om)g(\om^2)+\om^2f(\om^2)g(\om), \nonumber \\
\alpha & :=|f(-\om)|^2-|g(-\om)|^2,\;\;    \beta:=  \om
f(-\om)g(-\om^2)+\om^2f(-\om^2)g(-\om), \end{align}
then $u,v,\alpha,\beta $ are integers with
\begin{align*} ( f(\om)+\om^2g(\om))(f(\om^2)+\om g(\om^2)) & = u + v, \\
( f(\om)-\om^2g(\om))(f(\om^2)-\om g(\om^2)) & = u -v, \\
 ( f(-\om)+i\om^2g(-\om))(f(-w^2)+i\om g(-\om^2)) & = \alpha + i\beta, \\
( f(-\om)-i\om^2g(-\om))(f(-\om^2)-i\om g(-\om^2)) & = \alpha - i\beta,
\end{align*}
and
$$s_1  =(u+v)(u-v)=u^2-v^2,\;\;\;\;
 s_2  =(\alpha+i\beta )(\alpha -i\beta)=\alpha^2+\beta^2.$$
Writing $f(-\om)=f(\om) +2h_1(\om)$, $f(\om)=f(1)+(1-\om)h_2(\om)$,
$f(-\om)=f(-1)+(1-\om)h_3(\om)$ etc. and observing that since $3\parallel  a$ we
must have $3\nmid f(1)g(1)$, we readily get
$$ u\equiv f(1)^2+g(1)^2 \equiv 2 \text{ mod } 3,\;\;\;  \alpha \equiv
f(-1)^2-g(-1)^2 \equiv \pm 1 \text{ mod } 3, $$
$$ u\equiv \alpha \text{ mod } 2,  \;\;\; v\equiv \beta \text{ mod } 2, $$
$$ v\equiv 2f(1)g(1) \equiv \pm 1 \text{ mod } 3,\;\;\;  \beta \equiv
2f(-1)g(-1) \equiv 0 \text{ mod } 3. $$
Suppose first that $u$ is odd. Then $v$ is even and $s_1=u^2-v^2\equiv 1$ mod 4.
Hence, in addition to the single prime 3,  $s_1$ must
contains an odd power of a prime $p\equiv 3$ mod 4. Since it is a norm  in
$\mathbb Z[\om],$ the prime must be
$1$ or $7\bmod 12$ and $p\equiv 7\bmod 12$.  Suppose that $p\equiv 7\bmod 12$.
From Lemma  \ref{normform} we have $p=N_1(\alpha)=N_1(-\alpha \om)$
where  $\alpha= -1+2(1-\om)(2A+1+2B\om)$ and
$$ -\alpha \om = 1-\om^2 +4(\om^2+A(\om^2-\om)+B(1-\om^2))= 1-\om^2 +4(C+D\om)$$
for some integers $C,D$. Taking
\begin{align*}
f(x) & = x^3+1 +(C+Dx)(1-x)(x^3+1)+ m\:h(x),\\
g(x) & = x +x(C+Dx)(1-x)(x^3+1)+ m\: h(x),\end{align*}
gives $a=3(1+4m)$, $b=1$, $s_2=N_1(1)=1$ and $f(\om)-\om^2g(\om)= 1$ with
$$ f(\om)+\om^2g(\om)= 3+4(1-\om)(C+D\om)=(1-\om)(-\alpha \om), $$
and $s_1=3p$.
Suppose next that $u$ is even, so $\alpha \equiv \pm 1$ mod $3$ is even and
$\beta$ is an odd multiple of $3$. Since we have produced all multiples of
primes $5$ or $7\bmod 12$ we suppose that all the primes dividing
$s_2=\alpha^2+\beta^2$ are 1 or 11 mod 12. Since $\alpha$ and $\beta$ are both
non-zero we have a non-trivial factorization    in $\mathbb Z[i]$ and $s_2$ must
contain at least
one prime $p\equiv 1\bmod 12$. Moreover we can't have all these primes in $\P_2$,
since
$$ (6k +  (6t+1)i)(6k'\pm (6t'+1)i)= 6k''\mp 1 + 6t''i  $$
all the possible factorisations of such an  $n$  in $\mathbb Z[i]$ would produce
an $\alpha + i\beta$ of the form $6k\pm (6t+1)i$
not $(6k+3)+(6t+2)i$,  so at least one of the primes is in $\P_1$. Conversely
suppose that $p\equiv 1\bmod 12$ is in $\P_1$. Then by Lemma \ref{normform} we can
write $p=N_2(\alpha_2)$ with
$$\alpha_2 =(1+2A+2B\om)(1-\om)+i\left(1+2(C+D\om)(1-\om)\right). $$
We take
\begin{align*}
f(x) & = 1+x -(A-Bx)(1+x)(x^3-1)+ m\; h(x),\\
g(x) & = x -x(C-Dx)(1+x)(x^3-1)+ m\;h(x),\end{align*}
which gives $a=3(1+4m)$, $b=1$, $s_1=N_1(w-1)=3$ and
$$ f(-\omega)= (1-\omega)(1+2A+2B\omega), \;\;\; -w^2g(-\omega)=1+(1-\omega)(2C+2D\omega), $$
and $s_2=N_2(f(-\omega)-iw^2g(-\omega))=p$ and $M_G(F)=9(1+4m)p$.

\vspace{1ex}
\noindent
{\bf Case (ii)} Suppose that $M_G(F)$ is even. Observe, as in the proof of
Theorem \ref{Q12}, that $ab$ is odd or a multiple of $2^4$ and since
$s_1\equiv s_2$ mod 2,  with a norm of a $\xi+\eta w$  even only when it is a power
of 4 we have $s_1s_2$
is odd or an odd multiple of $2^4$ or a multiple of $2^6$.
We can achieve all odd multiples of $2^4$ and all multiples of $2^6$  that are
coprime to 3, and multiples of $2^4$ divisible by $3^2$.
\begin{align*}
& M_G\left( 1+x^4 + m\:k(x)\right)= 2^4 (1+6m),\\
& M_G\left( (1+x^4)^2 -\;h(-x^2)+m\:k(x)\right) =2^6 (1+3m),\\
 & M_G\left( x-1 + m\;k(x)\right) = 2^4 3^2 m. \end{align*}

That just leaves $2^5m$ with $(m,6)=1$.
Suppose first that $3\mid f(-1)g(-1)$. As  in the discussion in Theorem
\ref{Q12} we must have  $f(-1)^2+g(-1)^2=2n$ or $8n$,  with  here $n\equiv 2$
mod $3$ odd. Hence $n$ is divisible by an odd power of an odd  prime $p\equiv 2$
mod 3 and since $p$ factors in $\mathbb Z [i]$ must also be 1 mod 4 and $p\equiv
5$  mod 12. Conversely suppose that $p\equiv 5\bmod 12$. Then $2p$  is a sum of
two squares
$2p=r^2 + s^2.$  Since it is $1$ mod $3$, replacing $r$ by $-r$  as necessary,
we can assume  that $3\mid s$ and  $r\equiv 1$ mod 3 and $2p= (1-3A)^2 + (3B)^2.
$
The choice
\begin{align*}  f(x) & =(x^2+1) +  A(x-1)(x^4+x^2+1) +m\:h(x),\\
 g(x)& = Bx(x-1)(x^4+x^2+1) +m\:h(x),
\end{align*}
has $a=4(1+6m)$,  $s_1=s_2=N_1^2(1+\om^2)=1$,
$b= (2-6A)^2+ (6B)^2=8p, $
and $M_G(F)=32(1+6m)p$.

Suppose that $3\nmid f(-1)g(-1)$. Defining $u,v$ and $\alpha,\beta$ as in
\eqref{aandalpha} and observing that since $3\nmid a$ we must have $3\mid f(1)$
or $g(1)$ but not both, we readily get
$$ u\equiv f(1)^2+g(1)^2 \equiv 1 \text{ mod } 3,\;\;\;  \alpha \equiv
f(-1)^2-g(-1)^2 \equiv 0 \text{ mod } 3, $$
$$ u\equiv \alpha \text{ mod } 2,  \;\;\; v\equiv \beta \text{ mod } 2 , $$
$$ v\equiv 2f(1)g(1) \equiv 0\text{ mod } 3,\;\;\;  \beta \equiv 2f(-1)g(-1)
\equiv \pm 1 \text{ mod } 3. $$
Suppose first that $u$ is even. Since $2\nmid s_1=u^2-v^2$ we get $v$ is odd and
$s_1\equiv 3$ mod 4  is a positive integer
so must be divisible by an odd power of a  prime $p\equiv 3$ mod 4. Since $n$ is
the norm of an element of $\mathbb Z[\om]$ which is a UFD the prime $p$ must
split in $\mathbb Z[\om]$ and so $p\equiv 7\bmod 12$. Conversely suppose that
$p\equiv 7\bmod 12$. Then we can, by Lemma \ref{normform}, write
$p=N_1(\alpha)$ with
$\alpha = -1+ 2(2A+1+2B\omega)(1-\omega)$.  We take
\begin{align*}
f(x) &  =(x^2+1) + (A+Bx)(x^3+1)(1-x)+ m\:h(x), \\
g(x)  & =x(1-x) + x(A+Bx)(x^3+1)(1-x)+ m\:h(x),
\end{align*}
then $a=4(1+6m)$, $b=8$,  $s_2=N_1(\om^2+\om)=1$,
with
$$ f(\om)+\om^2g(\om)=-1+2(2A+1+2B\om)(1-\om),   \;\; f(\om)-\om^2g(\om) =  -1,
$$
and $s_1=N_1(\alpha)=p$ and $M_G(F)=32(6m+1)p$.

Suppose that $u$ is odd. Then $\alpha$ is an odd multiple of 3 and $\beta \equiv
\pm 1$ mod 3  is even.
We have already obtained all multiples of primes $5$ or $7$ mod $12$ so we can
assume that $s_2=\alpha^2+\beta^2$  contains
only  primes $1$ or $11\bmod 12$, but the primes $11\bmod 12$ do not split in
$\mathbb Z[i]$ so we can not obtain a factorisation of $s_2$ with $\alpha$ and
$\beta$ both non-zero from just those primes.  Thus $s_2$ contains at least one prime $p\equiv 1$ mod
12 with at least one of these in $\P_1$.
Conversely suppose that $p$ is in $\P_1$. By Lemma \ref{normform}  there is an
$$\alpha_1=-1+2(A+B\om)(1-\om) + i(2+2(C+D\om)(1-\om)).$$
with $p=N_2(\alpha_1)$, and
\begin{align*}
f(x) & = x^2(x^2+1)- (A-Bx)(x^3-1)(1+x)+m\:h(x), \\
g(x) & = x(x^3-1)+ (C-Dx)x(x^3-1)(1+x)+m\:h(x),
\end{align*}
has  $a=4(1+6m)$,  $b=8$, $s_1=N_1(1)=1$, with
$$ f(-\om)+ i\om^2g(-\om)=   -1 +2(A+B\om)(  1-\om) + i\left(  2
+2(C+D\om)(1-\om)    \right) =\alpha_1, $$
so that $s_2=N_2(\alpha_1)=p$ and $M_G(F)=32(6m+1)p.$
\end{proof}

\section{Proof of Theorem \ref{Z3Z3}}

 Note that $M_G(-F)=-M_G(F)$ so we take both signs for each value. In
\cite{dilum} it was shown that the measures for $\mathbb Z_p \times \mathbb Z_p$
coprime to $p$ are exactly the $(p-1)$st  roots of unity mod $p^2$. For $p=3$
these are $\pm 1$ mod 9 achieved with
$$ M_G( \pm 1 + m(x^2+x+1)(y^2+y+1))=9m\pm 1. $$
For the multiples of 3 we note, writing $\om = e^{2\pi i/3}$,  that in $\mathbb
Z [\om]$
$$ F(\om^i,\om^j)\equiv F(1,1) \text{ mod } (1-\om),   $$
in particular if $3\mid M_G(F)$  then $3\mid F(1,1)$ and $(1-\om)\mid
F(\om^i,\om^j)$ in $\mathbb Z[\om]$ for all $i,j$, and $3(1-\omega)^8\mid M_G(F)$ and
$3^5\mid M_G(F)$. In fact more is true. We write our polynomial in the form
\begin{align*}  F(x,y) =(A_0+A_1(x&-1) +   A_2(x-1)^2)&\\   &+ (B_0+B_1(x-1)   +
B_2(x-1)^2)(y-1)&\\
        &\qquad + (C_0+C_1(x-1) + C_2(x-1)^2)(y-1)^2.&
\end{align*}
If $3\mid F(1,1)$ we  have $3\mid A_0$ and hence
$$ F(\om^i,\om^j)\equiv A_1(\om^i-1) + B_0(\om^j-1) \text{ mod } (1-\om)^2. $$
If $3\mid A_1$ or $3\mid B_0$ we have $(1-\om)^2\mid F(x,y)$ for
$(x,y)=(1,\om),(1,\om^2)$ or $(\om,1),(\om^2,1)$ and we gain an additional 3.
Similarly if $A_1\equiv B_0$ mod 3 we have $(1-\om)^2\mid F(x,y)$ for
$(x,y)=(\om^2,\om),(\om,\om^2)$
and if $A_1\equiv -B_0$ mod 3 we have $(1-\om)^2\mid F(x,y)$ for
$(x,y)=(\om,\om),(\om^2,\om^2)$. Hence we must have $3^6\mid M_G(F)$.

Noting that $M_G(-F)=-M_G(F)$  we can obtain all multiples of $3^6$ using
\begin{align*}
& M_G(1+2x+m(x^2+x+1)(y^2+y+1)) = 3^6(1+3m), \\
& M_G(1+2x-x(y^2+y+1)+m(x^2+x+1)(y^2+y+1))  = 3^7m. \qed
\end{align*}


TITLE: Is there a function that is differentiable but not integrable?
QUESTION [4 upvotes]: is there such a function that is defined in [0,1], differentiable in (0,1) but not integrable? Thanks in advance.

REPLY [14 votes]: It's important to know that if a function is continuous on a closed interval, it is integrable on that interval.  This question seems to be getting at how strict that condition is.
If a function is differentiable on an open interval, then it is automatically continuous on that interval.  The question is whether it can be extended to a continuous function on the enclosing closed interval.  And one way this can fail to happen is if the function has an infinite limit at an endpoint.
So that is how we construct the counterexample.  The function
$$
f(x) = \begin{cases}
    0 & x=0 \\
    \frac{1}{x} & x > 0
       \end{cases}
$$
is defined on $[0,1]$, differentiable on $(0,1)$, but not integrable on $[0,1]$.