# September 9th

## Series of Groups

Recall that a *simple* group has no nontrivial normal subgroups.

*Example:*

\begin{align*}
\ZZ_6 \normal \generators{[3]} \normal \generators{[0]} \\
\ZZ_6 / \generators{[3]} &= \ZZ_3 \\
\generators{[3]} / \generators{[0]} &= \ZZ_2
.\end{align*}

**Definition:**
A *normal series* (or an *invariant series*) of a group $G$ is a finite sequence $H_i \leq G$ such that $H_i \normal H_{i+1}$ and $H_n = G$, so we obtain
$$
H_1 \normal H_2 \normal \cdots \normal H_n = G
.$$

**Definition:**
A normal series $\theset{K_i}$ is a **refinement** of $\theset{H_i}$ if $K_i \leq H_i$ for each $i$.

**Definition:**
We say two normal series of the same group $G$ are *isomorphic* if there is a bijection from
$$
\theset{H_i / H_{i+1}} \iff \theset{K_j / K_{j+1}}
$$

**Theorem (Schreier):**
Any two normal series of $G$ has isomorphic refinements.

**Definition:**
A normal series of $G$ is a **composition series** iff all of the successive quotients $H_i / H_{i+1}$ are **simple**.

Note that every finite group has a composition series, because any group is a maximal normal subgroup of itself.

**Theorem (Jordan-Holder):**
Any two composition series of a group $G$ are isomorphic.

*Proof:*
Apply Schreier's refinement theorem.

$\qed$

*Example:*
Consider $S_n \normal A_n \normal \theset e$.
This is a composition series, with quotients $Z_2, A_n$, which are both simple.

**Definition:**
A group $G$ is **solvable** iff it has a composition series in which all of the successive quotients are **abelian**.

*Examples:*

- Any abelian group is solvable.

- $S_n$ is not solvable for $n\geq 5$, since $A_n$ is not abelian for $n\geq 5$.

**Recall Feit-Thompson:**
Any nonabelian simple group is of *even* order.

**Consequence:**
Every group of *odd* order is solvable.

## The Commutator Subgroup

Let $G$ be a group, and let $[G, G] \leq G$ be the subgroup of $G$ generated by elements $aba\inv b\inv$, i.e. every element is a *product* of commutators. So $[G, G]$ is called *the commutator subgroup*.

**Theorem:**
Let $G$ be a group, then

1. $[G,G] \leq G$

2. $[G,G]$ is a normal subgroup

3. $G/[G, G]$ is abelian.

4. $[G,G]$ is the smallest normal subgroup such that the quotient is abelian,

    > I.e., $H \normal G$ and if $G/N$ is abelian $\implies [G, G] \leq N$.

*Proof of 1:*

$[G, G]$ is a subgroup:

- Closure is clear from definition as generators.
- The identity is $e = e e\inv e e\inv$.
- So it suffices to show that $(aba\inv b\inv)\inv \in [G, G]$, but this is given by $bab\inv a\inv$ which is of the correct form.

$\qed$

*Proof of 2:*

$[G, G]$ is normal.

Let $x_i \in [G, G]$, then we want to show $g \prod x_i g\inv \in [G, G]$, but this reduces to just showing $gx g\inv \in [G, G]$ for a single $x\in [G, G]$.

Then,

\begin{align*}
g(aba\inv b\inv ) g\inv &= (g\inv aba\inv) e (b\inv g) \\
&= (g\inv aba\inv)(gb\inv b g\inv)(b\inv g) \\
&= [(g\inv a)b (g\inv a)\inv b\inv] [bg\inv b\inv g] \\
&\in [G, G]
.\end{align*}

$\qed$

*Proof of 3:*

$G/[G, G]$ is abelian.

Let $H = [G, G]$.
We have $aH bH = (ab) H$ and $bH aH = (ba)H$.

But $abH = baH$ because $(ba)\inv(ab) = a\inv b\inv a b \in [G, G]$.

$\qed$

*Proof of 4:*

$H \normal G$ and if $G/N$ is abelian $\implies [G, G] \leq N$.

Suppose $G/N$ is abelian.
Let $aba\inv b\inv \in [G, G]$.

Then $abN = baN$, so $aba\inv b\inv \in N$ and thus $[G, G] \subseteq N$.

$\qed$

## Free Abelian Groups

*Example:*
$\ZZ \cross \ZZ$.

Take $e_1 = (1, 0), e_2 = (0, 1)$.
Then $(x,y) \in \ZZ^2$ can be written $x(1,0) + y(0, 1)$, so $\theset{e_i}$ behaves like a basis for a vector space.

**Definition:**
A group $G$ is *free abelian* if there is a subset $X\subseteq G$ such that every $g \in G$ can be represented as
$$
g = \sum_{i=1}^r n_i x_i,\quad x_i \in X,~n_i \in \ZZ.
$$

Equivalently, $X$ generates $G$, so $G = \generators{X}$, and if $\sum n_i x_i = 0 \implies n_i = 0~~\forall i$.

If this is the case, we say $X$ is a **basis** for $G$.

*Examples:*

- $\ZZ^n$ is free abelian

- $\ZZ_n$ is not free abelian, since $n [1] = 0$ and $n\neq 0$.

    > In general, you can replace $\ZZ_n$ by any finite group and replace $n$ with the order of the group.

**Theorem:**
If $G$ is free abelian on $X$ where $\abs X = r$, then $G \cong \ZZ^r$.

**Theorem:**
If $X = \theset{x_i}_{i=1}^r$, then a basis for $\ZZ^r$ is given by
$$
\theset{(1, 0, 0, \cdots), (0, 1, 0, \cdots), \cdots, (0, \cdots, 0, 1)} \definedas \theset{e_1, e_2, \cdots, e_r}
$$

*Proof:*
Use the map $\phi: G \to \ZZ^r$ where $x_i \mapsto e_i$, and check that this is an isomorphism of groups.

**Theorem:**
Let $G$ be free abelian with two bases $X, X'$, then $\abs X = \abs X'$.

**Definition:**
Let $G$ be free abelian, then if $X$ is a basis then $\abs X$ is called the *rank* of of $G$.