Criterion for a Quotient Group to be Abelian

Theorem 1: Let $G$ be a group and let $H$ be a normal subgroup of $G$. Then $G / H$ is abelian if and only if for all $g_1, g_2 \in G$ we have that $[g_1, g_2] := g_1^{-1}g_2^{-1}g_1g_2\in H$.

The element $[g_1, g_2] = g_1^{-1}g_2^{-1}g_1g_2$ is called the commutator of $g_1$ and $g_2$. Thus Theorem 1 says that if $G$ is a group and $H$ is a normal subgroup of $G$ then $G/H$ is abelian if and only if $H$ contains all commutators of pairs of elements from $G$.

• Proof: $\Rightarrow$ Suppose that $G/H$ is abelian. Let $g_1, g_2 \in G$. Then $g_1H, g_2H \in G/H$, and since $G/H$ is abelian we have that:

(1)

\begin{align} \quad (g_1H)(g_2H) &= (g_2H)(g_1H) \\ \end{align}

• Since $(g_1H)(g_2H) = (g_1g_2H)$ and since $(g_2H)(g_1H) = (g_2g_1H)$, we have from above that:

(2)

\begin{align} \quad g_1g_2H &= g_2g_1H \\ \quad g_2^{-1}g_1g_2H &= g_1H \\ g_1^{-1}g_2^{-1}g_1g_2H &= H \end{align}

• Since $H$ is a group, $e \in H$. So $g_1^{-1}g_2^{-1}g_1g_2 \in H$, which holds true for all $g_1, g_2 \in G$.

• $\Leftarrow$ Suppose that $g_1^{-1}g_2^{-1}g_1g_2 \in H$ for all $g_1, g_2 \in G$. Let $g_1H, g_2H \in G/H$. Then $(g_2g_1)^{-1}g_1g_2 \in H$ for all $g_1, g_2 \in G$. Therefore:

(3)

\begin{align} \quad (g_1g_2H) = (g_2g_1H) \end{align}

• That is:

(4)

\begin{align} \quad (g_1H)(g_2H) = (g_2H)(g_1H) \end{align}

• Since this holds for all $g_1H, g_2H \in G/H$ we conclude that $G/H$ is abelian. $\blacksquare$

Corollary 2: Let $G$ be a group and let $H$ be a normal subgroup of $G$. If $G$ is abelian then $G/H$ is abelian.

• Proof: Suppose that $G$ is abelian. Then $g_1g_2 = g_1g_2$ for all $g_1, g_2 \in G$. So $g_1^{-1}g_2^{-1}g_1g_2 = e \in H$ for all $g_1, g_2 \in G$. So by Theorem 1 above we see that $G/H$ is an abelian group. $\blacksquare$