Abelian Quotient Group
 It has been suggested that this page or section be merged into Quotient Group is Abelian iff All Commutators in Divisor. (Discuss) |
Theorem
Let $G$ be a group.
Let $H$ be a normal subgroup of $G$.
Let $G / H$ denote the quotient group of $G$ by $H$.
Then $G / H$ is abelian if and only if $H$ contains every element of $G$ of the form $a b a^{-1} b^{-1}$ where $a, b \in G$.
Proof
Let $G / H$ be abelian.
Then:
| \(\, \displaystyle \forall a H, b H \in G / H: \, \) | \(\displaystyle a H b H\) | \(=\) | \(\displaystyle b H a H\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle a b H\) | \(=\) | \(\displaystyle b a H\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle a b \paren {b a}^{-1}\) | \(\in\) | \(\displaystyle H\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle a b a^{-1} b^{-1}\) | \(\in\) | \(\displaystyle H\) | $\quad$ | $\quad$ |
The argument reverses.
Suppose that $\forall a, b \in G: a b a^{-1} b^{-1} \in H$.
| \(\, \displaystyle \forall a, b \in G: \, \) | \(\displaystyle a b a^{-1} b^{-1}\) | \(\in\) | \(\displaystyle H\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle a b \paren {b a}^{-1}\) | \(\in\) | \(\displaystyle H\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle a b H\) | \(=\) | \(\displaystyle b a H\) | $\quad$ | $\quad$ | ||||||||
| \(\displaystyle \leadsto \ \ \) | \(\displaystyle a H b H\) | \(=\) | \(\displaystyle b H a H\) | $\quad$ | $\quad$ |
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \zeta$
