## Quotient Groups

Let $H$ be a normal subgroup of $G$. Then it can be verified that
the cosets of $G$ relative to $H$ form a group. This group is called
the *quotient group* or *factor group*
of $G$ relative to $H$ and is denoted $G/H$.

It can be verified that
the set of self-conjugate elements of $G$ forms an abelian group $Z$
which is called the *center* of $G$. Note the center consists
of the elements of $G$ that commute with all the elements of $G$. Clearly
the center is always a normal subgroup.

**Theorem:** A group $G$ of order $p^2$ where $p$ is prime is always abelian.

**Proof:** From a previous theorem,
the number of invariant elements is a positive multiple of $p$, so the
center has order $p$ or $p^2$. The latter case implies $G$ is abelian,
so consider the case $|Z| = p$. Then $|G/Z| = p$ so $G/Z$ is cyclic,
thus we may decompose $G$ into the cosets $Z, Z g, ...,Z g^{p-1}$ for
some $g \in G$. The product of any two elements $z_1 g^\lambda , z_2 g^\mu$
is $z_1 z_2 g^{\lambda + \mu} = z_2 g^{\mu} z_1 g^\lambda$, thus $G$ is
abelian and $|Z| = p^2$ in fact.

Define the *commutator* of two elements $g,h$ of a group $G$
by $u = g^{-1}h^{-1}g h$. We have $u = 1$ if and only if $g h = h g$.
In an abelian group, all commutators are equal to the identity. Consider
the set of all commutators $\{u_1,...,u_m\}$
as $g, h$ run through all the elements of $G$. This set is not necessarily
closed under the group operation. We define the
*commutator group* $U$ to be the group generated by this set.
If $U = G$ we say $G$ is a *perfect group*.

**Theorem:** The commutator group $U$ of a group $G$ is normal.
$G/U$ is abelian. $U$ is contained in every normal subgroup that has an
abelian quotient group.

**Proof:** Let $x \in G$. Then
$x^{-1} g^{-1} h^{-1} g h x = a^{-1} b^{-1} a b$
where $a = x^{-1} g x, b = x^{-1} h x$, thus $U$ is normal.

Consider the commutator of two cosets $U x, U y$. We have

since $x^{-1} y^{-1} x y \in U$, hence $G / U$ is abelian.

Lastly if $R$ is any normal subgroup of $G$ with an abelian quotient group, then for any $x,y\in G$ we have $R x^{-1}y^{-1} x y = R$ since all commutators of $G / R$ must be equal to the identity, thus $R$ contains $x^{-1} y^{-1} x y$ hence $R \supset U$.

**Theorem:** If $A,B$ are normal subgroups of $G$ with only the identity
element in common then every element of $A$ commutes with every element of $B$.

**Proof:**
Consider $u = a^{-1}b^{-1}a b = (a^{-1}b^{-1}a)b = a^{-1}(b^{-1}a b)$
where $a \in A, b \in B$. Then since $A, B$ are normal,
$a^{-1}b a \in B$ and $b^{-1} a b \in A$, thus $u \in A \cap B = \{1\}$,
hence $a,b$ commute.