# August 30th

Recall the Sylow theorems:

- $p$ groups exist for *every* $p^i$ dividing $\abs{G}$, and $H(p) \normal H(p^2) \normal \cdots H(p^n)$.

- All Sylow $p\dash$subgroups are conjugate.

- Numerical constraints

  - $r_p \cong 1 \mod p$,

  - $r_p \divides \abs{G}$ and $r_p \divides m$,

## Internal Direct Products

Suppose $H, K \leq G$, and consider the smallest subgroup containing both $H$ and $K$.
Denote this $H \vee K$.

If either $H$ or $K$ is normal in $G$, then we have $H\vee K = HK$.

There is a "recipe" for proving you have a direct product of groups:

**Theorem (Recognizing Direct Products)**:
Let $G$ be a group, $H \normal G$ and $K\normal G$, and

1. $H\vee K = HK = G$,

2. $H \intersect K = \theset{e}$.

Then $G \cong H \times K$.

*Proof:*
We first want to show that $hk = kh ~\forall k\in K, h\in H$.
We then have
$$
hkh\inv k\inv = (hkh\inv)k\inv \in K = h(kh\inv k\inv) \in H \implies hkh\inv k\inv \in H\intersect K = \theset{e}.
$$

So define

\begin{align*}
\phi: H \cross K \to G \\
(h, k) \mapsto hk
,\end{align*}

*Exercise:* check that this is a homomorphism, it is surjective, and injective.

$\qed$

*Applications:*

**Theorem:**
Every group of order $p^2$ is abelian.

*Proof:*
If $G$ is cyclic, then it is abelian and $G \cong \ZZ_{p^2}$.
So suppose otherwise. By Cauchy, there is an element of order $p$ in $G$.
So let $H = \generators{a}$, for which we have $\abs{H} = p$.

Then $H \normal G$ by Sylow 1, since it's normal in $H(p^2)$, which would have to equal $G$.

Now consider $b\not\in H$. By Lagrange, we must have $o(b) = 1, p$, and since $e\in H$, we must have $o(b) = p$.
This uses fact that $G$ is not cyclic.

Now let $K = \generators{b}$.
Then $\abs{K} = p$, and $K \normal G$ by the same argument.

$\qed$

**Theorem:**
Let $\abs{G} = pq$ where $q\neq 1 \mod p$ and $p < q$.
Then $G$ is cyclic (and thus abelian).

*Proof:*
Use Sylow 1.
Let $P$ be a sylow $p\dash$subgroup.
We want to show that $P \normal G$ to apply our direct product lemma, so it suffices to show $r_p = 1$.

We know $r_p = 1 \mod p$ and $r_p \divides \abs{G} = pq$, and so $r_p = 1,q$. It can't be $q$ because $p < q$.

Now let $Q$ be a sylow $q\dash$subgroup.
Then $r_q \cong 1 \mod 1$ and $r_q \divides pq$, so $r_q = 1, q$.
But since $p< q$, we must have $r_q = 1$. So $Q \normal G$ as well.

We now have $P \intersect Q = \emptyset$ (why?) and
$$
\abs{PQ} = \frac{\abs P \abs Q}{ \abs{P \intersect Q} } = \abs P \abs Q = pq,
$$

and so $G = PQ$, and $G \cong \ZZ_p \cross \ZZ_q \cong \ZZ_{pq}$.

$\qed$

*Example:*
Every group of order $15 = 5^1 3^1$ is cyclic.

## Determination of groups of a given order

| Order of G 	| Number of Groups 	| List of Distinct Groups                    	|
|------------	|------------------	|--------------------------------------------	|
| 1          	| 1                	| $\theset{e}$                               	|
| 2          	| 1                	| $\ZZ_2$                                    	|
| 3          	| 1                	| $\ZZ_3$                                    	|
| 4          	| 2                	| $\ZZ_4, \ZZ_2^2$                           	|
| 5          	| 1                	| $\ZZ_5$                                    	|
| 6          	| 2                	| $\ZZ_6, S_3$ (*)                           	|
| 7          	| 1                	| $\ZZ_7$                                    	|
| 8          	| 5                	| $\ZZ_8,\ZZ_4 \cross \ZZ_2, \ZZ_2^3, D_8,Q$ 	|
| 9          	| 2                	| $\ZZ_9, \ZZ_3^2$                           	|
| 10         	| 2                	| $\ZZ_{10}, D_5$                            	|
| 11         	| 1                	| $\ZZ_{11}$                                 	|

We still need to justify 6, 8, and 10.

## Free Groups

Define an *alphabet* $A = \theset{a_1, a_2, \cdots a_n}$,
and let a *syllable* be of the form $a_i^m$ for some $m$.
A *word* is any expression of the form $\prod_{n_i} a_{n_i}^{m_i}$.

We have two operations,

- Concatenation, i.e. $(a_1 a_2) \star (a_3^2 a_5) = a_1 a_2 a_3^2 a_5$.

- Contraction, i.e. $(a_1 a_2^2) \star (a_2\inv a_5) = a_1 a_2^2 a_2\inv a_5 = a_1 a_2 a_5$.

If we've contracted a word as much as possible, we say it is *reduced*.

We let $F[A]$ be the set of reduced words and define a binary operation

\begin{align*}
f: F[A] \cross F[A] &\to F[A] \\
(w_1, w_2) &\mapsto w_1 w_2 \text{ (reduced) }
.\end{align*}

**Theorem:**
$(A, f)$ is a group.

*Proof:*
Exercise.

$\qed$

**Definition:**
$F[A]$ is called the **free group generated by $A$**.
A group $G$ is called *free* on a subset $A \subseteq G$ iff $G \cong F[A]$.

*Examples:*

1. $A = \theset{x} \implies F[A] = \theset{x^n \suchthat n \in \ZZ} \cong \ZZ$.

2. $A = \theset{x, y} \implies F[A] = \ZZ \ast \ZZ$ (not defined yet!).

Note that there are not relations, i.e. $xyxyxy$ is *reduced*.
To abelianize, we'd need to introduce the relation $xy = yx$.

*Properties:*

1. If $G$ is free on $A$ and free on $B$ then we must have $\abs A = \abs B$.

2. Any (nontrivial) subgroup of a free group is free.

    > (See Fraleigh or Hungerford for possible Algebraic proofs!)

**Theorem:**
Let $G$ be generated by some (possibly infinite) subset $A = \theset{A_i\mid i\in I}$ and $G'$ be generated by some $A_i' \subseteq A_i$.

Then

a. There is at most one homomorphism $a_i \to a_i'$.

b. If $G \cong F[A]$, there is exactly *one* homomorphism.

**Corollary:**
Every group $G'$ is a homomorphic image of a free group.

*Proof:*
Let $A$ be the generators of $G'$ and $G = F[A]$, then define

\begin{align*}
\phi: F[A] &\to G' \\
a_i &\mapsto a_i
.\end{align*}

This is onto exactly because $G' = \generators{a_i}$, and using the theorem above we're done.

$\qed$

## Generators and Relations

Let $G$ be a group and $A \subseteq G$ be a generating subset so $G = \generators{a \mid a\in A}$.
There exists a $\phi: F[A] \surjects G$, and by the first isomorphism theorem, we have $F[A] / \ker \phi \cong G$.

Let $R = \ker \phi$, these provide the *relations*.

*Examples:*

Let $G = \ZZ_3 = \generators{[1]_3}$. Let $x = [1]_3$, then define $\phi: F[\theset{x}] \surjects \ZZ_3$.

Then since $[1] + [1] + [1] = [0] \mod 3$, we have $\ker \phi = \generators{x^3}$.

Let $G = \ZZ \oplus \ZZ$, then $G \cong \generators{x,y \mid [x,y] = 1}$.

> We'll use this for groups of order 6 -- there will be only one presentation that is nonabelian, and we'll exhibit such a group.