TITLE: Is there a non-abelian group where $|ab|=5$ such that $a^5 b^5 = e$
QUESTION [2 upvotes]: I was told that if $|ab|=5$ then $a^5 b^5 = e$ but only if we know the group is abelian. I would like a possible example of a nonabelian group where this equality holds.

REPLY [1 votes]: I suppose it's (somewhat vacuously) true for all non-abelian groups $G$ with orders indivisible by $5$, so let's skip those cases.

Result:  There is a unique smallest non-abelian group of order divisible by $5$ for which $|ab|=5$ implies $a^5 b^5=\mathrm{id}$, namely $\mathbb{Z}_5 \times S_3$

In a sense, this is a cop-out, since we take an abelian group $\mathbb{Z}_5$, which must satisfy the property (since it's abelian), and we append $S_3$, which makes it non-abelian, but does not affect the property, since $|S_3|=6$ is coprime to $5$.
I used both Mace4 and GAP, to find there are no examples of orders 5,10,15,20, and 25.  I also checked, of the three non-abelian groups of order $30$, the only one that satisfies the desired property is $\mathbb{Z}_5 \times S_3$.  We can also prove the property holds as follows:
If $(u,x), (v,y) \in \mathbb{Z}_5 \times S_3$ and $|(u,x)(v,y)|=5$, then, since $|S_3|=6$ is coprime to $5$, we must have $xy=\mathrm{id}$.
Hence
\begin{align}
(u,x)^5 (v,y)^5 &= (u^5 v^5,x^5 x^{-5}) & \text{since } y=x^{-1} \\
 &= ((uv)^5,(xx^{-1})^5) & \text{since } u,v \in \mathbb{Z}_5, \text{ and } x \text{ and } x^{-1} \text{ commute} \\
 &= ((u,x)(v,x^{-1}))^5 \\
 &= \mathrm{id} & \text{since } |(u,x)(v,y)|=5. \\
\end{align}
Actually, the same proof works to generate infinitely many examples:

Result:  If $G$ is an abelian group and $H$ is a non-abelian group of order $|H| \not\equiv 0 \pmod 5$, then $G \times H$ is a non-abelian group for which $|ab|=5$ implies $a^5 b^5=\mathrm{id}$.

The smallest examples this result does not cover have order $125$ (found by GAP), which are:

$\mathbb{Z}_{25} \ltimes \mathbb{Z}_5$,
$(\mathbb{Z}_5 \times \mathbb{Z}_5) \ltimes \mathbb{Z}_5$; which is isomorphic to the example given by Geoff Robinson.


TITLE: Equation of a plane through intersection of two and parallel to other
QUESTION [1 upvotes]: I have got two planes $$x+y+z=1$$ and $$2x+3y-z+4=0$$ . I am required to find a plane by intersection of two and parallel to $x$ axis . I think the plane parallel to $x$ axis
so Simply $P1+kP2=0$ hence
$$x(1+2k)+y(1+3k)+z(1-k)-1+4k=0$$
and how to create a relation by which this plane is parallel to $x$ axis and with  any other given plane ?

REPLY [2 votes]: You are right in taking the required plane to be $P_1+k P_2 = 0$ (Intersection of $P_1$ and $P_2$). Since the required plane is parallel to the $x$-axis, the co-efficient of $x$ in it's equation should be $0$. $2k+1 = 0 \implies k = \frac{-1}{2}$. The required plane is:
$$
\frac{-y}{2} + \frac{3z}{2} - 3 = 0
$$
BTW, the  $z$ co-efficient is $1-k$ and not $1-4k$.


TITLE: Why is this not a counter-example of the Hardy-Littlewood tauberian theorem?
QUESTION [0 upvotes]: I am confused about the Hardy-Littlewood tauberian theorem. If we apply it with the sequence $a_n$ whose first few terms are given by $1,-1,1,1,-1,-1,1,1,1,1,-1,-1,-1,-1...$ where we put $2^n$ $1$s followed by $2^n$ $-1$s going on indefinitely, increasing $n$.
At the end of a period of $-1$s, there have been equally as many $+1$s and $-1$s meaning that $\frac{1}{n}\sum_{k=1}^na_k=0$. At the end of a period of $1$s, about $\frac{1}{3}$ of the numbers that have appeared thus far has been this last streak of $1$s and since all of the previous terms average out to zero this means that $\frac{1}{n}\sum_{k=1}^na_k\approx\frac{1}{3}$. Overall, this means that this sequence has no defined Cesàro mean.
On the other hand, when examining $(1-x)\sum_{n=1}^{\infty}a_nx^n$ as $x\to1^-$ using python I have noticed that it seems to converge to a specific value that is $\approx.170$.
This would contradic the Hardy Littlewood tauberian theorem since an asymptotic relation $(1-x)\sum_{n=1}^{\infty}a_nx^n\sim C$ would imply the relation $\frac{1}{n}\sum_{k=1}^na_k\sim C$ which does not seem to exist.
This case is an example of a more general issue I have with the Hardy-Littlewood tauberian theorem. Using L'hopitals we can easily get that for bounded sequences $a_n$ we  have
$$\lim_{x\to1^-}\left[\left(1-x\right)\sum_{n=0}^{\infty}a_{n}x^{n}-\left(1-x\right)^{2}\sum_{n=0}^{\infty}na_{n}x^{n-1}\right]=0$$
and so by differentiating $f(x)=(1-x)\sum_{n=1}^{\infty}a_nx^n$ we get that $f'(x)=o\left(\frac{1}{1-x}\right)$. If $\lim_{x\to1^-}f(x)$ doesn't converge then the difference between consecutive minima and maxima is $>O(1-x)$ and thus my the mean value theorem there is a point between the two with $f'(x)>O\left(\frac{1}{1-x}\right)$. Since there would be infinitely many point like this, we have a contradiction based off the growth rate of $f'(x)$.
This means that $f(x)$ converges as $x\to1^-$, but this implies that $\lim_{n\to\infty}\sum_{k=1}^n a_k$ converges by the Hardy Littlewood tauberian theorem which obviously is not true for a general bounded sequence $a_n$.
Can anyone explain to me what I am doing wrong?

REPLY [1 votes]: This is more of a comment that it got too long; the post above is very sloppy and confusing with a bunch of mistaken statements:
for example "$f(x)=o\left(\frac{1}{1-x}\right)$" is trivial when $f(x)=(1-x)\sum{a_nx^n}, |a_n| \le C$ since $f(x)$ is bounded at $1$
(for $a_n$ real it is just trivial that for $0<x<1, -C/(1-x) \le \sum{a_nx^n} \le C/(1-x)$ and for complex $a_n$ separation in real and imaginary parts will do)
However, this doesn't imply in any way shape or form that $f(x) \to C, x \to 1, x<1$, while also the fact that an analytic function is bounded in the unit disc doesn't imply any growth condition for the derivative (there are infinite Blaschke products $|B(z)| <1, |z|<1, |B(e^{it})|=1$ a.e. s.t $B'$ is not $O(1/(1-z)^a)$ for any $a>0$ however large,
Similarly the last statement is wrong as the H-L Tauberian theorem implies only $a_n \to C$ or the Caesaro mean $(a_0+..a_n)/(n+1) \to C$ in the context of $a_n$ bounded
As for the example given, by a simple computation $g(x)=(1-x)\sum a_nx^n=\sum b_nx^n$
where $b_n$ is a $4/3$ gap sequence since $b_n=a_n-a_{n-1}, n \ge 1$ so $b_n \ne 0$ only when the $1$ transitions to $-1$ and viceversa, so roughly at $2^{n+1}, 3.2^n$ where $b_n = \pm 2$  (except for $b_0=1$) - hence the gaps are $3/2, 4/3$ respectively so overall $4/3$ (or maybe $4/3-\epsilon$) works
But then it is a classic theorem that a gap power series satisfies $g(x) \to C, x \to 1$ iff $\sum {b_n}=C$ (we need no Tauberian conditions in other words), theorem that follows easily from the result that if $|g(x)| \le M, 0 \le x <1$ then $|\sum_{k \le n} {a_k}| \le C_{\rho}M$ for all $n$ where $\rho >1$ is a gap for the coefficients of $g$, namely, $b_{n_k} \ne 0$ implies $n_{k+1} \ge \rho n_k$ (eventually). In particular since for the $g$ in the example, $b_n$ doesn't converge to zero, $g(x)$ cannot converge and it oscillates.
To try and find oscillating sequences, try $x_n=1-1/2^n, y_n=1-1/(3 \times 2^n)$ as $g(x_n), g(y_n)$ should converge to different values (one can do estimates and show that theoretically too, but would check computationally first).
So no contradiction with classical theory anywhere. As for mkistakes, the statement about the derivative is wrong and the computations done were not done at the right numbers or maybe the diffeerent convergence is too slow (though I think for $n$ about $20$ so for $x,y$ in the $1-1/10^6$ range those numbers should be different already when truncating at $n$ abut $8 \times 10^6$)


TITLE: How did this apparatus actually work - discovery of induced nuclear fission?
QUESTION [3 upvotes]: Despite all the ugliness threaded through this story, I'd like to understand the physics of this apparatus of Otto Hahn, Lise Meitner and Fritz Strassmann (in alphabetical order). The system shown represents the first measurement of fission induced by neutron radiation.
note: While this NYTimes Obituary of Otto Hahn does not explain it - the link in the answer below does.
It looks like the electronics in the back might be a power supply, and the two identical electronic systems on the left and right could be amplifiers - is it somehow detecting coincidences of both fission fragments? The boxes labeled PERTRIX seem to be 90V batteries used in many radios in the 1930s. The fact that there are so many more in series on the shelf below suggests a Geiger–Müller tube or at least some kind of gas ionization detector, but I don't see any in the photo.
Question: How did this thing work?

Above: Image from the NYTimes Obituary credited there to Hulton Archive/Getty Images.

REPLY [1 votes]: Here is a pretty good explainer image from the Nuclear Fission Deutsches Museum.
This grouping was divided across 3 rooms for the actual 1938 experiment.


TITLE: Van Kampen theorem on n-manifold remove one point
QUESTION [0 upvotes]: I'm trying to prove for a ($n \geq 3$)-dimensional path connected manifold M, if remove a point in M, then use Van Kampen theorem can somehow show the fundamental group of original $M$ and $M-{x}$, and $M-U_x$ where $U_x$ is the open neighbourhood of $x$ are the same.
I know the open neighbourhood $U_x$ is homeomorphic to open n-ball in $\mathbb{R}^n$, I can feel it is true as then intersect $U_x$ with $(M-{x})$ is homeomorphic to $S^{n-1}$. But I just cannot get my head around with all the details and how the logic actually goes.
Can anybody please try to help me in details? I really wish to understand the details in between as I do have rough intuition, it is details missing here.
Also, what is a good counter example for $n=2$?
Thanks a lot for helping.

REPLY [2 votes]: Let $U=M-x$, $V=U_x$, then $M=U \cup V \cong R^n$, $U \cap V \cong S^{n-1}$
Then use Van Kampen's theorem. Actually since $\pi_1(S^{n-1})=0$, that normal subgroup $N$ must be $0$. So  $\pi_1(M)=\pi_1(U)$.
Similarly $\pi(M)=\pi(M-U_x)$. Just let $V$ be a larger ball.
For n=2. The problem is $\pi_1(S^{1})=\mathbb{Z}$. Think about the torus.
The fundamental group of torus minus one point is $\mathbb{Z} * \mathbb{Z}$.


TITLE: Finding an unknown angle in a right-angled triangle
QUESTION [0 upvotes]: I've used  cosine & sine rule, and after playing around with them, I'll always end up with an equation involving $\sin\theta,\cos\theta$ and other weird (and wonderful expressions). I've approached the question in many different ways: by showing $AD=x$ (since $\angle DAC=\angle DCA=30^\circ$ it forms an isosceles triangle),  also tried to show $\sin\theta=1/2$ ... but I never got the correct answer.
This has been bothering me for weeks.
Thanks in advance]1

REPLY [1 votes]: In right $\Delta BAC$ $$\sin\theta=\frac{AB}{BC}=\frac{x}{BC}\implies BC=x\csc\theta$$
$$BD=BC-CD=x\csc\theta-x \quad \angle ADB=\theta+30^\circ$$
Using Sine rule in $\Delta ADB$ as folows
$$\frac{\sin(\theta+30^\circ)}{x}=\frac{\sin60^\circ}{x\csc\theta-x}$$
$$2\sin(\theta+30^\circ)(1-\sin\theta)=\sqrt3 \sin\theta$$
Solving above equation for $0<\theta<90^\circ$, we get $\color{blue}{\theta=30^\circ}$


TITLE: Reason for negligible chemical potentials of different particles in early universe
QUESTION [2 upvotes]: For the early Universe at high temperatures, the chemical potential is assumed to be zero for all types of particles is negligible. Why is this true?

REPLY [1 votes]: In the current universe baryon and lepton number are conserved. This means that there are two chemical potentials, $\mu_B$ and $\mu_L$. We believe that the universe has a net-baryon density, and $\mu_B\neq 0$. We don't know the net lepton number of the universe, because the net lepton number of the neutrino background cannot be measured.
The net-baryon number of the universe is non-zero, but small. The current baryon-to-photon ratio is of order a few times $10^{-10}$. Compared to baryon (proton and neutron) rest mass the current temperature is essentially zero, so the current baryon chemical potential is a tiny amount larger than the baryon rest mass.
We can take this information and extrapolate to the past. In the past the universe was much hotter, but baryon and lepton number are conserved. At temperatures above $T\simeq 200$ MeV the baryon number is no longer in protons/neutrons, but in approximately massless quarks, which carry 1/3 baryon number. Since all species of particles (leptons, quarks, photons) are approximately massless, the total densities of quarks, leptons and photons are now comparable. This means that the conserved net-quark density is much smaller than the total density of quarks. This implies that $\mu_B<<T$, and the effect of $\mu_B$ on the pressure and energy density of the early universe can be ignored.


TITLE: Find the sum $|\cos \frac{2\pi}{7} + \cos \frac{ 4\pi}{7} +...\cos \frac{12\pi}{7}|$
QUESTION [0 upvotes]: The expression can be reduced as
$$|\cos \frac{2\pi}{7} -\cos \frac {\pi}{7} +\cos \frac{4\pi}{7} -\cos \frac{3\pi}{7}....|$$
$$=|2\sin \frac{\pi}{14}|| \sin \frac{3\pi}{14} + \sin \frac{7\pi}{14} +\sin \frac {11\pi}{14}|$$
$$ =2\sin \frac{\pi}{14} | 1 + 2\sin \frac{\pi}{2} \cos \frac{8\pi}{14}|$$
$$=2[\sin \frac {9\pi}{14} + \sin \frac {\pi}{14}]$$
I don’t think it’s reducible any further, but the given answer is 1. Where am I going wrong?

REPLY [2 votes]: The error is that $\sin \frac{3\pi}{14} + \sin \frac{7\pi}{14} +\sin \frac {11\pi}{14} \ne 1 + 2\sin \frac{\pi}{2} \cos \frac{8\pi}{14}$.
$\frac{3+11}{2} = 7$ instead of $8$, thus $\sin \frac{3 \pi}{14} + \sin \frac{11 \pi}{14} = 2 \sin \frac{7 \pi}{14} \cos \frac{4 \pi}{14}$, so the whole thing should equal $1 + 2 \cos \frac{2 \pi}{7}$ instead.