# Thursday August 22nd

## Group Actions

Let $G$ be a group and $X$ be a set; we say $G$ *acts* on $X$ (or that $X$ is a $G\dash$ set) when there is a map $G\cross X \to X$ such that $ex = x$ and
$$
(gh) \actson x = g \actson (h \actson x)
.$$

We then define the **stabilizer** of $x$ as
$$
\mathrm{Stab}_G(x) = G_x \definedas \theset{g\in G \suchthat g\actson x = x} \leq G,
$$

and the **orbit**
$$
G.x = \mathcal O_x \definedas \theset{g\actson x \suchthat x\in X} \subseteq X.
$$

When $G$ is finite, we have
$$
\abs{G.x}  = \frac{\abs G}{\abs{G_x}}.
$$

We can also consider the **fixed points** of $X$,
$$
X_g = \theset{x\in X \mid g\actson x = x ~~\forall g\in G} \subseteq X
$$

## Burnside's Theorem

**Theorem (Burnside):**
Let $X$ be a $G\dash$set and $v \definedas \abs{X/G}$ be the number of orbits. Then
$$
v \abs{G} = \sum_{g\in G} \abs{X_g}.
$$

*Proof:*
Define
$$
N = \theset{(g,x) \mid g\actson x = x} \subseteq G \cross X
,$$

we then have

\begin{align*}
\abs N &= \sum_{g\in G} \abs{X_g} \\
&= \sum_{x\in X} \abs{G_x} \\
&= \sum_{x\in X} \frac{\abs G}{\abs {G.x}} \quad\text{by Orbit-Stabilizer} \\
&= \abs{G} \left( \sum_{x\in X} \frac{1}{\abs{G.x}} \right) \\
&= \abs{G} \sum_{G.x ~\in~ X/G} \left( \sum_{y ~\in~ G.x} \frac{1}{\abs{G.x}} \right) \\
&= \abs{G} \sum_{G.x ~\in~ X/G} \left(
\abs{G.x} \frac{1}{\abs{G.x}}
\right) \\
&= \abs{G} \sum_{G.x ~\in~ X/G} 1 \\
&= \abs{G} v
.\end{align*}

The last two equalities follow from the following fact: since the orbits partition $X$, say into $X = \displaystyle\disjoint_{i=1}^v \sigma_i$, so let $\sigma = \theset{\sigma_i \mid 1 \leq i \leq v}$.

By abuse of notation, replace each orbit in $\sigma$ with a representative element $x_i\in \sigma_i \subset X$.

We then have
$$
\sum_{x \in \sigma} \frac{1}{\abs{G.x}} = \frac{1}{\abs{G.x}} \abs{\sigma}  = 1.
$$

$\qed$

*Application:*
Consider seating 10 people around a circular table. How many distinct seating arrangements are there?

Let $X$ be the set of configurations, $G = S_{10}$, and let $G\actson X$ by permuting configurations.
Then $v$, the number of orbits under this action, yields the number of distinct seating arrangements.

By Burnside, we have
$$
v = \frac{1}{\abs{G}} \sum_{g\in G} \abs{X_g} = \frac{1}{10} (10!) = 9!
$$

since $X_g = \theset{x\in X \mid g\actson x = x} = \emptyset$ unless $g = e$, and $X_e = X$.

## Sylow Theory

Recall Lagrange's theorem:

If $H \leq G$ and $G$ is finite, then $\abs{H}$ divides $\abs{G}$.

Consider the converse: if $n$ divides $\abs G$, does there exist a subgroup of size $n$?

The answer is **no** in general, and a counterexample is $A_4$ which has $4!/2 = 12$ elements but no subgroup of order 6.

### Class Functions

Let $X$ be a $G\dash$set, and choose orbit representatives $x_1 \cdots x_v$.

Then
$$
\abs{X} = \sum_{i=1}^v \abs{G .x_i}.
$$

We can then separately count all orbits with exactly one element, which is exactly
$$
X_G = \theset{x\in G \mid g\actson x = x ~ \forall g\in G}
$$.

We then have
$$
\abs X = \abs{X_G} + \sum_{i=j}^v \abs{G. x_i}
$$
for some $j$ where $\abs{G.x_i} > 1$ for all $i \geq j$.

**Theorem:**
Let $G$ be a group of order $p^n$ for $p$ a prime.

Then
$$
\abs X = \abs{X_G} \mod p
.$$

*Proof:*
We know that
$$
\abs{G.x_i} = [G : G_{x_i}] \text{ for } j \leq i \leq v \text{ and } \abs{Gx_i} > 1 \implies G.x_i \neq G
,$$
and thus $p$ divides $[G: G x_i]$.
The result follows.

$\qed$

*Application:*
If $\abs G = p^n$, then the center $Z(G)$ is nontrivial.

Let $X=G$ act on itself by conjugation, so $g\actson x = gxg\inv$. Then
$$
X_G = \theset{x\in G \suchthat gxg\inv = x} = \theset{x\in G \suchthat gx = xg} = Z(G)
$$

But then, by the previous theorem, we have
$$
\abs{Z(G)} \equiv \abs{X}\equiv \abs{G} \mod p
,$$
but since $Z(G) \leq G$ we have $\abs{Z(G)} \cong 0 \mod p$.
So in particular, $Z(G) \neq \theset{e}$.

**Definition:**
A group $G$ is a **$p\dash$group** iff every element in $G$ has order $p^k$ for some $k$.
A subgroup is a $p\dash$group exactly when it is a $p\dash$group in its own right.

### Cauchy's Theorem

**Theorem (Cauchy):**
Let $G$ be a finite group, where $p$ is prime and divides $\abs{G}$.
Then $G$ has an element (and thus a subgroup) of order $p$.

*Proof:*
Consider
$$
X = \theset{(g_1, g_2, \cdots , g_p) \in G^{\oplus p} \mid g_1g_2\cdots g_p = e}
.$$

Given any $p-1$ elements, say $g_1 \cdots g_{p-1}$, the remaining element is completely determined by $g_p = (g_1 \cdots g_{p-1})\inv$.

So $\abs X = \abs{G}^{p-1}$.and since $p \divides \abs{G}$, we have $p \divides \abs X$.

Now let $\sigma \in S_p$ the symmetric group act on $X$ by index permutation, i.e.
$$
\sigma \actson (g_1, g_2 \cdots g_p) = (g_{\sigma(1)}, g_{\sigma(2)}, \cdots, g_{\sigma(p)})
.$$

*Exercise*: Check that this gives a well-defined group action.

Let $\sigma = (1~2~\cdots~p) \in S_p$, and note $\generators{\sigma} \leq S_p$ also acts on $X$ where $\abs{\generators{\sigma}} = p$. Therefore we have
$$
\abs{X} = \abs{X_{\generators{\sigma}}} \mod p.
$$

Since $p\divides \abs{X}$, it follows that $\abs{X_{\generators{\sigma}}} = 0 \mod p$, and thus $p \divides \abs{X_{\generators{\sigma}}}$.

If $\generators{\sigma}$ fixes $(g_1, g_2, \cdots g_p)$, then $g_1 = g_2 = \cdots g_p$.

Note that $(e, e, \cdots) \in X_{\generators{\sigma}}$, as is $(a, a, \cdots a)$ since $p \divides \abs{X_{\generators{\sigma}}}$. So there is some $a\in G$ such that $a^p = 1$. Moreover, $\generators{a} \leq G$ is a subgroup of size $p$.

$\qed$

### Normalizers

Let $G$ be a group and $X = S$ be the set of subgroups of $G$.
Let $G$ act on $X$ by $g\actson H = gHg\inv$.
What is the stabilizer?

$$
G_x = G_H = \theset{g\in G \mid gHg\inv = H}
,$$
making $G_H$ the largest subgroup such that $H \normal G_H$.

So we **define** $N_G(H) \definedas G_H$.

*Lemma:*
Let $H$ be a $p\dash$subgroup of $G$ of order $p^n$.
Then
$$
[N_G(H) : H] = [G : H] \mod p
.$$

*Proof:*
Let $S = G/H$ be the set of left $H\dash$cosets in $G$.
Now let $H$ act on $S$ by
$$
H\actson x + H \definedas (hx) + H
.$$

By a previous theorem, $\abs{G/H} = \abs{S} = \abs{S_H} \mod p$, where $\abs{G/H} = [G: H]$.
What is $S_H$?

This is given by
$$
S_H = \theset{x + H \in S \suchthat xHx\inv \in H \forall h\in H}
.$$

Therefore $x\in N_G(H)$.

$\qed$

**Corollary:**
Let $H \leq G$ be a subgroup of order $p^n$.
If $p \divides [G: H]$ then $N_G(H) \neq H$.

*Proof:*
Exercise.
$\qed$

**Theorem:**
Let $G$ be a finite group, then $G$ is a $p\dash$group $\iff \abs{G} = p^n$ for some $n\geq 1$.

*Proof:*
Suppose $\abs{G} = p^n$ and $a \in G$.
Then $\abs{\generators{a}} = p^\alpha$ for some $\alpha$.

Conversely, suppose $G$ is a $p\dash$group.
Factor $\abs{G}$ into primes and suppose $\exists q$ such that $q \divides \abs{G}$ but $q \neq p$.

By Cauchy, we can then get a subgroup $\generators{c}$ such that $\abs{\generators{c}} \divides q$, but then $\abs{G} \neq p^n$.

$\qed$