# Abstract Nonsense

## Actions by p-Groups (Pt. II)

Point of Post: This is a continuation of this post.

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Probably the most significant use of actions by $p$-groups is given by a clever proof of Cauchy’s Theorem given by McKay. Namely:

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Theorem (Cauchy): Let $G$ be any group (not necessarily a $p$-group) and $p$ a prime with $p\mid |G|$. Then, there exists $x\in G$ with $|x|=p$.

Proof: Let $X=\left\{(g_1,\cdots,g_p)\in G^p:g_1\cdots g_p=e\right\}$. Since we can clearly choose the first $p-1$ coordinates of a tuple in $X$ and the last is determined as the inverse of the product of the first $p-1$ we have that $\#(X)=|G|^{p-1}$. It can be easily verified that $H=\mathbb{Z}_p$ acts on $X$ via cyclic shifting (forward) of the coordinates of a tuple in $X$. More explicitly,

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$k\cdot(g_1,\cdots,g_p)=\left(g_{1+k\text{ mod }p},\cdots,g_{p+k\text{ mod }p}\right)$

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Indeed, it clearly suffices to prove that $k\cdot (g_1,\cdots,g_p)\in X$ for all $(g_1,\cdots,g_p)\in X$ since the other properties of a group action are apparent. But, this is clear by conjugation, namely

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\begin{aligned}e &=g_{p-k+1}\cdots g_p e(g_{p+k-1}\cdots g_p)^{-1}\\ &=g_{p-k+1}\cdots g_p g_1\cdots g_p(g_{p+k-1}\cdots g_p)^{-1}\\ &=g_{p+k-1}\cdots g_p g_1\cdots g_k\end{aligned}

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but, the very last expression is the product left-to-right product of $k\cdot (g_1,\cdots,g_p)$ and so $k\cdot (g_1,\cdots,g_p)\in X$ as desired. Note then that by the fundamental theorem we have that $\#(X^H)$ is divisible by $p$ (since $p\mid |G|^{p-1}$ since $p\mid |G|$). But, since $(e,\cdots,e)\in X$ we must have that $\#(X^H)\geqslant p$ and so there exists $(g_1,\cdots,g_p)\in X$ with $(g_1,\cdots,g_p)\ne (1,\cdots,1)$. That said, note that since $(g_1,\cdots,g_p)$ is fixed by every element of $H$ we have that $(g_1,\cdots,g_p)=(g_p,\cdots,g_1)=\cdots=(g_2,\cdots,g_1)$ and so, in particular, $g_1=\cdots=g_p=g$ and by construction $g\ne 1$. But we see then that since $(g,\cdots,g)\in X$ we have that $g^p=1$. It follows then that $|g|\mid p$ and since $|g|\ne 1$ we have that $|g|=p$ as desired. $\blacksquare$

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From where we get the following, fairly relevant corollary:

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Theorem: A finite group $G$ is a $p$-group if and only if $|g|$ is a power of $p$ for every $g\in G$.

Proof: Evidently from Lagrange’s theorem we have that if $G$ is a $p$-group then every element of $G$ has prime power order. Conversely, evidently if $p,q$ are two different primes dividing $|G|$ then by Cauchy’s Theorem we know that there exists $x,y\in G$ with $|x|=p$ and $|x|=q$. It follows then that every element of $G$ has order equal to the power of a single prime. $\blacksquare$

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And  a simple proof that the only finite fields can be of order $p^n$ for some prime $p$. Indeed:
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Theorem: Let $F$ be a finite field, then $|F|=p^n$ for some prime $p$.

Proof:  Note that the map $f_{a,b}:F\to F:x\mapsto ab^{-1}$ is an (additive) isomorphism for each $a,b\in F^\times$ and so $|a|=|b|$ for all $a,b\in F^\times$. Note though that if $|F|$ is divisible by two primes, we could find two elements of different orders. Thus, $|F|$ is a primer power as desired. $\blacksquare$

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What we now wish to do is prove that $\left[G:N_G(H)\right]\equiv \left[G:H\right]\text{ mod }p$ if $H$ is a $p$-group sitting inside the larger group $G$. From where we will be able to prove that if $H\notin\text{Syl}_p(G)$ then $H\ne N_G(H)$.

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Theorem: Let $G$ be any finite group and $H$ a $p$-subgroup. Then $\left[N_G(H):H\right]\equiv \left[G:H\right]\text{ mod }p$.

Proof: Consider the action of $H$ on $X=G/H$ given by left multiplication.  What we’d like to figure out is what $X^H$ really is. Well, note that if $gH\in X^H$ then $hgH=gH$ for all $h\in H$ and so $hg=gh'$ for some $h'\in H$ and so $g^{-1}hg\in H$. Thus, since $h$ was arbitrary we may conclude that $g^{-1}Hg=H$ and so $g\in N_G(H)$ and conversely one can check that if $g\in N_G(H)$ then $gH\in X^H$. Thus, $X^H=\left\{gH:g\in N_G(H)\right)=N_G(H)/H$. Thus, $\#(X^H)=\left[N_G(H):H\right]$. The conclusion then follows from the fundamental theorem. $\blacksquare$

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Corollary: If $p\mid [G:H]$ then $p\mid [N_G(H):H]$, and since $[N_G(H):H]\ne 0$ we may conclude that $\left[N_G(H):H\right]\geqslant p$ and so $N_G(H)\supsetneq H$.

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This tells us that if $|G|=p^n$ and $H\leqslant G$ has $|H|=p^{n-1}$ then $H\unlhd G$, since by the corollary we have that $H\subsetneq N_G(H)$ and so clearly $N_G(H)=G$. That said, we already knew this from the theorem that says if the index of a subgroup is the smallest prime dividing the group then it’s normal. What this theorem is really useful for is:

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Theorem: Let $G$ be a finite $p$-group with $|G|=p^n$. Then, there exists a chain of subgroups

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$\{e\}=H_0\unlhd\cdots\unlhd H_{n-1}\unlhd H_n=G$

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with $|H_j|=p^j$.

Proof: We first use Cauchy’s Theorem to prove that there exists some $x\in G$ with $|x|=p$, we then take $H_1=\langle x\rangle$. We then proceed by ‘induction’. Suppose that we have found a group $H_j$ with $j and $|H_j|=p^j$. We see then that $[G:H_j]=p^{n-j}\geqslant p$ and so by the previous corollary we have that $H_j\subsetneq N_G(H_j)$. We see then that $p\mid |N_G(H_j)/H|$ and so by Cauchy’s theorem we have that there exists $gH_j\in N_G(H_j)$ with $|gH_j|=p$. But, from the correspondence theorem we know that $\langle gH_j\rangle=K/H_j$ for some $K\leqslant N_G(H_j)$ and moreover that $\left[K:H\right]=\left|\langle gH_j\rangle\right|=p$ and so $|K|=p^{j+1}$. To see that $H_j\unlhd K$ we may merely appeal to the paragraph immediately preceding this theorem. Thus, the ‘induction’ is complete. $\blacksquare$

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This allows us to conclude from the first (weak) Sylow’s Theorem, namely the existence of Sylow $p$-subgroups, the stronger result that if $p^m\mid |G|$ where $G$ is any group that $G$ contains a subgroup $H$ with $|H|=p^m$.

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To finish this post let’s give a cool proof of Wilson’s theorem:

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Theorem(Wilson): Let $n\in\mathbb{N}$. Then, $(n-1)!\equiv -1\text{ mod }n$ if and only if $n$ is prime.

Proof: Suppose that $n=p$ is prime. Consider the action of $\mathbb{Z}_p=G$ on $X=\left\{(g_1,\cdots,g_p)\in S_p^p:g_1\cdots g_p=e\right\}$ as previously discussed. We noted that $X^G$ is equipotent to $\left\{g\in S_p:g^p=e\right\}$ and so by the Fundamental Theorem we have that $\{g\in S_p:g^p=e\}\equiv 0\text{ mod }p$. Note that if $g\in S_p$ is non-identity then it’s order is the least common multiple of the cycle lengths of its disjoint cycle decomposition. It clearly follows that the only non-identity elements of $S_p$ of order $p$ are the $p$-cycles. Now, how may distinct $p$-cycles are there? Well, there are $p!$ ways to arrange $1,\cdots,p$,  but each of these arrangements is equivalent to precisely $p$ others (the $p$ cyclic shifts forward) and so the number of distinct $p$-cycles is $(p-1)!$. Thus, since the only $g\in S_p$ for which $g^p=e$ are the elements of order $p$ and $e$ we may conclude that $\#\{g\in S_p:g^p=e\}=(p-1)!+1$, thus as previously noted $(p-1)!+1\equiv 0\text{ mod }p$.

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Conversely, if $n$ has a proper divisor $d$ then we have to have that $d\mid (n-1)!$ but by assumption it would also divide $(n-1)!+1$ and so $d\mid 1$, which is a contradiction. $\blacksquare$