Criterion for a Quotient Group to be Abelian

# 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. $\blacksquare$