# 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.

*blynn@cs.stanford.edu*💡