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 💡