# Abstract Nonsense

## Actions by p-Groups (Pt. I)

Point of Post: In this post we discuss the theory of $p$-groups acting on sets, and some of its ramifications.

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Motivation

I have previously discussed group actions, but being a rush to discuss group theory I skirted over some of the beautiful theory. So, I’d like to take some time to discuss one of the prettier and more powerful branches of the theory, namely when we restrict our attention to group actions by $p$-groups. Not only will we be able to say some prove some fairly substantive  theorems about $p$-group actions explicitly, but will be able to prove some very neat things in more general group theory and in number theory. The interesting fact about the theory we will discuss is that at the root of everything is a ‘fundamental theorem’ whose presence (being the theorem in the case of a particular group action) is the proof that every $p$-group has a non-trivial center.  Namely, we were able to conclude that since the cardinality of any conjugacy class must divide the order of the group, that they must be divisible by $p$. From this and the fact that the sum of the cardinalities of all the distinct conjugacy classes must sum to the order of the group (which is divisible by $p$) that the sum of all the one point conjugacy classes must have cardinality divisible by $p$. Well, the generalization of this idea (which, as I’m sure is pretty clear, can be restated for an arbitrary action with conjugacy class replaced by orbit) will be the main tool I mentioned from which all our other theorems are (not always straight-forward) consequences.

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Actions by $p$-Groups

In all the follows $G$ will be, unless stated otherwise, a $p$-group for a fixed but arbitrary prime $p$ (recall that a (finite) $p$-group is a group of order $p^n$ for some $n\in\mathbb{N}$). If $G$ acts on a set $X$ we will denote the set $\left\{x\in X:gx=x\right\}$ and $X^G$ the set $\left\{x\in X:gx=x\text{ for all }g\in G\right\}$. A transversal  for the action of $G$ on $X$ will be a set $\mathscr{T}$ which consists of precisely one element of each orbit of the action of $G$ on $X$. For $x\in X$ we denote the orbit of $x$ under the action by the notation $\mathcal{O}_x$ or  $\mathcal{O}_x^G$ if there is ambiguity about which group is acting (e.g. if we are considering the action of a group $G$ and a subgroup $H$ on the same set). We denote the stabilizer subgroup of $G$ by $\text{stab}(x)$.  The fundamental theorem we stated is:

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The Fundamental Theorem: Let $G$ act on a set $X$. Then,

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$\#(X)\equiv \#(X^G)\text{ mod }p$

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Proof: Fix some transversal $\mathscr{T}$ for the action of $G$ on $X$. We can obviously partition $\mathscr{T}$ into the two sets $\mathscr{A}$ of all $x\in\mathscr{T}$ with $\#(\mathcal{O}_x)=1$ and $\mathscr{B}=\mathscr{T}-\mathscr{A}$. From the Orbit Decomposition Theorem we know that

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$\displaystyle X=\bigsqcup_{x\in\mathscr{T}}\mathcal{O}_x=\bigsqcup_{x\in\mathscr{A}}\mathcal{O}_x\sqcup\bigsqcup_{x\in\mathscr{B}}\mathcal{O}_x$

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(where $\sqcup$ denotes the union is disjoint). That said, it’s clear by definition that

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$\displaystyle X^G=\bigsqcup_{x\in\mathscr{A}}\mathcal{O}_x$

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from where we may easily conclude that

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$\displaystyle \#(X)-\#(X^G)=\sum_{x\in\mathscr{B}}\#(\mathcal{O}_x)\quad\mathbf{(1)}$

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We know though from the Orbit Stabilizer Theorem that $\#(\mathcal{O}_x)\mid |G|$ and so $\#(\mathcal{O}_x)=p^a$ for some $a$. But, since $\#(\mathcal{O}_x)>1$ for each $x\in\mathscr{B}$ we have that $\#(\mathcal{O}_x)=p^a$ for some $a>1$ and so in particular, the right hand side of $\mathbf{(1)}$ is a sum of things divisible by $p$, and thus itself divisible by $p$. The conclusion follows. $\blacksquare$

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Corollary: If $\#(X)$ is not divisible by $p$ then $\#(X^G)\geqslant 1$. If $\#(X)$ is divisible by $p$ then $\#(X)$ is divisible by $p$.

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From this we get the fact that all $p$-groups have non-trivial centers easily:

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Theorem: Let $N\unlhd G$ then, $N\cap\mathcal{Z}(G)$ is non-trivial.

Proof: Let $X=N$ and let $G$ act on $N$ by conjugation (this is well-defined precisely because $N$ is normal). Note then that $X^G$ are the elements of $N$ which commute with everything in $G$, in other words, $X^G=N\cap\mathcal{Z}(G)$. Since $\#(X)=|N|$ is divisible by $p$ we have by the previous corollary that $p\mid \#(N\cap\mathcal{Z}(G))$. But, since $e\in N\cap\mathcal{Z}(G)$ we must have then that $\#(N\cap\mathcal{Z}(G))\geqslant p$. $\blacksquare$

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Another consequence of the Fundamental Theorem is

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Theorem: Let $V$ be an $n$-dimensional  vector space over $\mathbb{F}_q$ where $q=p^m$, then there exists a non-zero (in fact, at least $p-1$) vector $v\in V$ such that $T(v)=v$ (i.e. is an eigenvector with eigenvalue $1$) for all $T\in\text{GL}(V)$.

Proof: $G=\text{GL}(V)$ naturally acts on $X=V$ by just defining $Tv$ to be $T(v)$. We see then the question is equivalent to the fact that $X^G$ is non-trivial, but since $\#(V)=q^n=p^{mn}$ we have by the Fundamental Theorem that $p\mid \#(X^G)$ and since $0\in X^G$ we may conclude that $\#(X^G)\geqslant p$. The conclusion follows. $\blacksquare$

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200