# 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

$(U x^{-1})(U y^{-1})(U x)(U y) = U x^{-1} y^{-1} x y = U$

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.

Ben Lynn blynn@cs.stanford.edu 💡