# Tuesday August 27th

Let $G$ be a finite group and $p$ a prime.
TFAE:

- $\abs{H} = p^n$ for some $n$
-
 Every element of $H$ has order $p^\alpha$ for some $\alpha$.

If either of these are true, we say $H$ is a *$p\dash$group*.

Let $H$ be a $p\dash$group, last time we proved that if $p \divides [G : H]$ then $N_G(H) \neq H$.

## Sylow Theorems

Let $G$ be a finite group and suppose $\abs{G} = p^n m$ where $(m, n) = 1$. Then

### Sylow 1

> Idea: take a prime factorization of $\abs{G}$, then there are subgroups of order $p^i$ for *every* prime power appearing, up to the maximal power.

1. $G$ contains a subgroup of order $p^i$ for every $1 \leq i \leq n$.
2. Every subgroup $H$ of order $p^i$ where $i < n$ is a normal subgroup in a subgroup of order $p^{i+1}$.

*Proof:*
By induction on $i$. For $i=1$, we know this by Cauchy's theorem.
If we show (2), that shows (1) as a consequence.

So suppose this holds for $i < n$. Let $H \leq G$ where $\abs{H} = p^i$, we now want a subgroup of order $p^{i+1}$. Since $p\divides [G: H]$, by the previous theorem, $H < N_G(H)$ is a proper subgroup (?).

Now consider the canonical projection $N_G(H) \to N_G(H) /H$. Since
$$
p \divides [N_G(H) : H] = \abs{N_G(H)/ H}
,$$
by Cauchy there is a subgroup of order $p$ in this quotient.
Call it $K$.
Then $\pi\inv(K) \leq N_G(H)$.

*Exercise*: Show that $\abs{\phi\inv(K)} = p^{i+1}$.

It now follows that $H \normal \phi\inv(K)$.
$\qed$

**Definition**:
For $G$ a finite group and $\abs{G} = p^n m$ where $p$ does not divide $m$.

Then a subgroup of order $p^n$ is called a **Sylow $p\dash$subgroup**.

> Note: by Sylow 1, these exist.

### Sylow 2

If $P_1, P_2$ are Sylow $p\dash$subgroups of $G$, then $P_1$ and $P_2$ are conjugate.

*Proof:*
Let $\mathcal L$ be the left cosets of $P_1$, i.e. $\mathcal L = G/P_1$.

Let $P_2$ act on $\mathcal L$ by
$$
p_2 \actson (g + P_1) \definedas (p_2g) + P_1
.$$

By a previous theorem about orbits and fixed points, we have
$$
\abs{\mathcal L_{P_2}} = \abs{\mathcal L} \mod p.
$$

Since $p$ does not divide $\abs{\mathcal L}$, we have $p$ does not divide $\abs{\mathcal L_{P_2}}$.
So $\mathcal L_{P_2}$ is nonempty.

So there exists a coset $xP_1$ such that $xP_1 \in \mathcal L_{P_2}$, and thus
$$
yxP_1 = xP_1 \text{ for all } y\in P_2
.$$

Then $x\inv y x P_1 = P_1$ for all $y\in P_2$, and so $x\inv P_2 x = P_1$.
So $P_1$ and $P_2$ are conjugate.

$\qed$

### Sylow 3

Let $G$ be a finite group, and $p\divides \abs G$.
Let $r_p$ be the number of Sylow $p\dash$subgroups of $G$.

Then

- $r_p \cong 1 \mod p$.

- $r_p \divides \abs G$.

- $r_p = [G : N_G(P)]$

*Proof:*

Let $X = \mathcal S$ be the set of Sylow $p\dash$subgroups, and let $P \in X$ be a fixed Sylow $p\dash$subgroup.

Let $P \actson \mathcal S$ by conjugation, so for $\overline P \in \mathcal S$ let $x \actson \overline P = x \overline P x\inv$.

By a previous theorem, we have
$$
\abs{\mathcal S} = \mathcal{S}_P \mod p
$$

What are the fixed points $\mathcal{S}_P$?
$$
\mathcal{S}_P = \theset{T \in \mathcal S \suchthat xTx\inv  = T \quad \forall x\in P}.
$$

Let $\mathcal T \in \mathcal{S}_P$, so $xTx\inv = T$ for all $x\in P$.

Then $P \leq N_G(T)$, so both $P$ and $T$ are Sylow $p\dash$ subgroups in $N_G(H)$ as well as $G$.

So there exists a $f\in N_G(T)$ such that $T = gPg\inv$.
But the point is that in the normalizer, there is only **one** Sylow $p\dash$ subgroup.

But then $T$ is the unique largest normal subgroup of $N_G(T)$, which forces $T = P$.

Then $\mathcal{S}_P = \theset{P}$, and using the formula, we have $r_p \cong 1 \mod p$.

Now modify this slightly by letting $G$ act on $\mathcal S$ (instead of just $P$) by conjugation.

Since all Sylows are conjugate, by Sylow (1) there is only one orbit, so $\mathcal S = GP$ for $P \in \mathcal S$. But then
$$
r_p = \abs{\mathcal S} = \abs{GP} = [G: G_p] \divides \abs{G}.
$$

Note that this gives a precise formula for $r_p$, although the theorem is just an upper bound of sorts, and $G_p = N_G(P)$.

## Applications of Sylow Theorems

Of interest historically: classifying finite *simple* groups, where a group $G$ is *simple* If $N \normal G$ and $N \neq \theset{e}$, then $N=G$.

*Example:*
Let $G = \ZZ_p$, any subgroup would need to have order dividing $p$, so $G$ must be simple.

*Example:*
$G = A_n$ for $n\geq 5$ (see Galois theory)

One major application is proving that groups of a certain order are *not* simple.

*Applications:*

**Proposition:**
Let $\abs G = p^n q$ with $p > q$.
Then $G$ is not simple.

*Proof:*

> Strategy: Find a proper normal nontrivial subgroup using Sylow theory.
Can either show $r_p = 1$, or produce normal subgroups by intersecting distinct Sylow p-subgroups.

Consider $r_p$, then $r_p = p^\alpha q^\beta$ for some $\alpha, \beta$. But since $r_p \cong 1 \mod p$, $p$ does not divide $r_p$, we must have $r_p = 1, q$.

But since $q < p$ and $q\neq 1 \mod p$, this forces $r_p = 1$.

So let $P$ be a sylow $p\dash$subgroup, then $P < G$. Then $gPg\inv$ is also a sylow, but there's only 1 of them, so $P$ is normal.

$\qed$

**Proposition**:
Let $\abs{G} = 45$, then $G$ is not simple.

*Proof*:
Exercise.

$\qed$

**Proposition**:
Let $\abs{G} = p^n$, then $G$ is not simple if $n > 1$.

*Proof:*
By Sylow (1), there is a normal subgroup of order $p^{n-1}$ in $G$.

$\qed$

**Proposition:**
Let $\abs{G} = 48$, then $G$ is not simple.

*Proof:*

Note $48 = 2^4 3$, so consider $r_2$, the number of Sylow 2-subgroups. Then $r_2 \cong 1 \mod 2$ and $r_2 \divides 48$. So $r_2 = 1, 3$. If $r_2 = 1$, we're done, otherwise suppose $r_2 = 3$.

Let $H \neq K$ be Sylow 2-subgroups, so $\abs H = \abs K = 2^4 = 16$. Now consider $H \intersect K$, which is a subgroup of $G$. How big is it?

Since $H\neq K, \abs{H \intersect K} < 16$. The order has to divides 16, so we in fact have $\abs{H \intersect K} \leq 8$. Suppose it is less than 4, towards a contradiction. Then
$$
\abs{HK} = \frac{\abs H \abs K}{\abs{H \intersect K}} \geq \frac{(16)(16)}{4} = 64 > \abs{G} = 48.
$$

So we can only have $\abs{H \intersect K} = 8$. Since this is an index 2 subgroup in both $H$ and $K$, it is in fact normal. But then
$$
H, K \subseteq N_G(H \intersect K) \definedas X
.$$

But then $\abs X$ must be a multiple of 16 *and* divide 48, so it's either 16 or 28. But $\abs X > 16$, because $H \subseteq X$ and $K \subseteq X$. So then
$$
N_G(H \intersect K) = G \text{ and so } H \intersect K \normal G
.$$

$\qed$