## Math sunday

I had a chill day thinking about math today without any pressure whatsoever. First I figured out, calculating inductively, that the order of $GL_n(\mathbb{F}_p)$ is $(p^n - 1)(p^n - p)(p^n - p^2)\cdots (p^n - p^{n-1})$. You calculate the number of $k$-tuples of column vectors linear independent and from there derive $p^k$ as the number of vectors that cannot be appended if linear independence is to be preserved. A Sylow $p$-group of that is the group of upper triangular matrices with ones on the diagonal, which has the order $p^{n(n-1)/2}$ that we want.

I also find the proof of the first Sylow theorem much easier to understand now, the inspiration of it. I had always remembered that the Sylow $p$-group we are looking for can be the stabilizer subgroup of some set of $p^k$ elements of the group where $p^k$ divides the order of the group. By the pigeonhole principle, there can be no more than $p^k$ elements in it. The part to prove that kept boggling my mind was the reverse inequality via orbits. It turns out that that can be viewed in a way that makes its logic feel much more natural than it did before, when like many a proof not understood, seems to spring out of the blue.

We wish to show that the number of times, letting $p^r$ be the largest $p$th power dividing $n$, that the order of some orbit is divided by $p$ is no more than $r-k$. To do that it suffices to show that the sum of the orders of the orbits, $\binom{n}{p^k}$ is divided by $p$ no more than that many times. To show that is very mechanical. Write out as $m\displaystyle\prod_{j = 1}^{p^k-1} \frac{p^k m - j}{p^k - j}$ and divide out each element of the product on both the numerator and denominator by $p$ to the number of times $j$ divides it. With this, the denominator of the product is not a multiple of $p$, which means the number of times $p$ divides the sum of the orders of the orbits is the number of times it divides $m$, which is $r-k$.

Following this, Brian Bi told me about this problem, starred in Artin, which means it was considered by the author to be difficult, that he was stuck on. To my great surprise, I managed to solve it under half an hour. The problem is:

Let $H$ be a proper subgroup of a finite group $G$. Prove that the conjugate subgroups of $H$ don’t cover $G$.

For this, I remembered the relation $|G| = |N(H)||Cl(H)|$, where $Cl(H)$ denotes the number of conjugate subgroups of $H$, which is a special case of the orbit-stabilizer theorem, as conjugation is a group action after all. With this, given that $|N(H)| \geq |H|$ and that conjugate subgroups share the identity, the union of them has less than $|G|$ elements.

I remember Jonah Sinick’s once saying that finite group theory is one of the most g-loaded parts of math. I’m not sure what his rationale is for that exactly. I’ll say that I have a taste for finite group theory though I can’t say I’m a freak at it, unlike Aschbacher, but I guess I’m not bad at it either. Sure, it requires some form of pattern recognition and abstraction visualization that is not so loaded on the prior knowledge front. Brian Bi keeps telling me about how hard finite group theory is, relative to the continuous version of group theory, the Lie groups, which I know next to nothing about at present.

Oleg Olegovich, who told me today that he had proved “some generalization of something to semi-simple groups,” but needs a bit more to earn the label of Permanent Head Damage, suggested upon my asking him what he considers as good mathematics that I look into Arnold’s classic on classical mechanics, which was first to come to mind on his response of “stuff that is geometric and springs out of classical mechanics.” I found a PDF of it online and browsed through it but did not feel it was that tasteful, perhaps because I’m been a bit immersed lately in the number theoretic and abstract algebraic side of math that intersects not with physics, though I had before an inclination towards more physicsy math. I thought of possibly learning PDEs and some physics as a byproduct of it, but I’m also worried about lack of focus. Maybe eventually I can do that casually without having to try too hard as I have done lately for number theory. At least, I have not the right combination of brainpower and interest sufficient for that in my current state of mind.

## Composition series

My friend after some time in industry is back in school, currently taking graduate algebra. I was today looking at one of his homework and in particular, I thought about and worked out one of the problems, which is to prove the uniqueness part of the Jordan-Hölder theorem. Formally, if $G$ is a finite group and

$1 = N_0 \trianglelefteq N_1 \trianglelefteq \cdots \trianglelefteq N_r = G$ and $1 = N_0' \trianglelefteq N_1' \trianglelefteq \cdots \trianglelefteq N_s' = G$

are composition series of $G$, then $r = s$ and there exists $\sigma \in S_r$ and isomorphisms $N_{i+1} / N_i \cong N_{\sigma(i)+1} / N_{\sigma(i)}$.

Suppose WLOG that $s \geq r$ and as a base case $s = 2$. Then clearly, $s = r$ and if $N_1 \neq N_1'$, $N_1 \cap N_1' = 1$. $N_1 N_1' = G$ must hold as it is normal in $G$. Now, remember there is a theorem which states that if $H, K$ are normal subgroups of $G = HK$ with $H \cap K = 1$, then $G \cong H \times K$. (This follows from $(hkh^{-1})k^{-1} = h(kh^{-1}k^{-1})$, which shows the commutator to be the identity). Thus there are no other normal proper subgroups other than $H$ and $K$.

For the inductive step, take $H = N_{r-1} \cap N_{s-1}'$. By the second isomorphism theorem, $N_{r-1} / H \cong G / N_{s-1}'$. Take any composition series for $H$ to construct another for $G$ via $N_{r-1}$. This shows on application of the inductive hypothesis that $r = s$. One can do the same for $N_{s-1}'$. With both our composition series linked to two intermediary ones that differ only between $G$ and the common $H$ with factors swapped in between those two, our induction proof completes.

## Automorphisms of quaternion group

I learned this morning from Brian Bi that the automorphism group of the quaternion group is in fact $S_4$. Why? The quaternion group is generated by any two of $i,j,k$ all of which have order $4$. $\pm i, \pm j, \pm k$ correspond to the six faces of a cube. Remember that the symmetries orientation preserving of cube form $S_4$ with the objects permuted the space diagonals. Now what do the space diagonals correspond to? Triplet bases $(i,j,k), (-i,j,-k), (j,i,-k), (-j,i,k)$, which correspond to four different corners of the cube, no two of which are joined by a space diagonal. We send both our generators $i,j$ to two of $\pm i, \pm j, \pm k$; there are $6\cdot 4 = 24$ choices. There are by the same logic $24$ triplets $(x,y,z)$ of quaternions such that $xy = z$. We define an equivalence relation with $(x,y,z) \sim (-x,-y,z)$ and $(x,y,z) \sim (y,z,x) \sim (z,x,y)$ that is such that if two elements are in the same equivalence class, then results of the application of any automorphism on those two elements will be as well. Furthermore, no two classes are mapped to the same class. Combined, this shows that every automorphism is a bijection on the equivalence classes.

## A recurrence relation

I noticed that

$(x_1 - x_k)\displaystyle\sum_{i_1+\cdots+i_k=n} x_1^{i_1}\cdots x_k^{i_k} = \displaystyle\sum_{i_1+\cdots+i_{k-1}=n+1} x_1^{i_1}\cdots x_{k-1}^{i_{k-1}} - \displaystyle\sum_{i_2+\cdots+i_k=n+1} x_2^{i_2}\cdots x_k^{i_k}.$

In the difference on the RHS, it is apparent that terms without $x_1$ or $x_k$ will vanish. Thus, all the negative terms which are not cancelled out have a $x_k$ and all such positive terms have a $x_1$. Combinatorially, all terms of degree $n+1$ with $x_k$ can be generated by multiplying $x_k$ on all terms of degree $n$. Analogous holds for the positive terms. The terms with only $x_1$ and $x_k$ are cancelled out with the exception of the $x_1^{n+1} - x_k^{n+1}$ that remains.

This recurrence appears in calculation of the determinant of the Vandermonde matrix.

## Galois group of x^10+x^5+1

This was a problem from an old qualifying exam, that I solved today, with a few pointers. First of all, is it reducible? It actually is. Note that $x^{15} - 1 = (x^5-1)(x^{10}+x^5+1) = (x^3-1)(x^{12} + x^9 + x^6 + x^3 + 1)$. $1 + x + x^2$, as a prime element of $\mathbb{Q}[x]$ that divides not $x^5-1$ must divide the polynomial, the Galois group of which we are looking for. The other factor of it corresponds to the multiplicative group of $\mathbb{F}_{15}$, which has $8$ elements. Seeing that it has $3$ elements of order $2$ and $4$ elements of order $4$ and is abelian, it must be $C_2 \times C_4$. Thus, the answer is $C_2 \times C_2 \times C_4$.