Quotient Group is Abelian iff All Commutators in Divisor

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Theorem

Let $G$ be a group.

Let $N$ be a normal subgroup of $G$.

Let $G / N$ be the quotient group of $G$ by $N$.


Then the quotient group $G / N$ is abelian if and only if:

$\forall x, y \in G: \sqbrk {x, y} \in N$

where $\sqbrk {x, y}$ denotes the commutator of $x$ and $y$.


Proof

Let $x, y \in G$.

Then:

\(\text {(1)}: \quad\) \(\ds \paren {x N} \paren {y N}\) \(=\) \(\ds \paren {y N} \paren {x N}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \sqbrk {x N, y N}\) \(=\) \(\ds N\) Definition of Commutator of Group Elements
\(\ds \leadstoandfrom \ \ \) \(\ds \sqbrk {x, y} N\) \(=\) \(\ds N\) Commutator of Quotient Group Elements


Let $G / N$ be abelian.

Then by definition:

$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$

and it follows from $(1)$ that:

$\forall x, y \in G: \sqbrk {x, y} \in N$


Conversely, let:

$\forall x, y \in G: \sqbrk {x, y} \in N$

Again it follows from $(1)$ that:

$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$

That is, that $G / N$ is abelian.

$\blacksquare$


Sources