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 $y$.

That is, if and only if $x y x^{-1} y^{-1} \in N$ for all $x, y \in G$.


Proof 1

Let $x, y \in G$.

First we establish the following:

\(\ds \paren {x N} \paren {y N}\) \(=\) \(\ds \paren {y N} \paren {x N}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {x N} \paren {y N}\) \(=\) \(\ds \paren {y N} \paren {x N} N\) Quotient Group is Group: $N$ is the identity of $G / N$
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {x N}^{-1} \paren {y N}^{-1} \paren {x N} \paren {y N}\) \(=\) \(\ds N\) applying $\paren {x N}^{-1} \paren {y N}^{-1}$ to both sides
\(\ds \leadstoandfrom \ \ \) \(\ds \sqbrk {x N, y N}\) \(=\) \(\ds N\) Definition of Commutator of Group Elements
\(\ds \leadstoandfrom \ \ \) \(\ds \sqbrk {x, y} N\) \(=\) \(\ds N\) Commutator of Quotient Group Elements
\(\text {(1)}: \quad\) \(\ds \leadstoandfrom \ \ \) \(\ds \sqbrk {x, y}\) \(\in\) \(\ds N\) Commutator of Quotient Group Elements

$\Box$


Sufficient Condition

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:

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

$\Box$


Necessary Condition

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.

$\blacksquare$


Proof 2

Let $G / N$ be abelian.

Then:

\(\ds \forall a N, b N \in G / N: \, \) \(\ds a N b N\) \(=\) \(\ds b N a N\)
\(\ds \leadsto \ \ \) \(\ds a b N\) \(=\) \(\ds b a N\)
\(\ds \leadsto \ \ \) \(\ds a b \paren {b a}^{-1}\) \(\in\) \(\ds N\)
\(\ds \leadsto \ \ \) \(\ds a b a^{-1} b^{-1}\) \(\in\) \(\ds N\)

$\blacksquare$


The argument reverses.

Suppose that $\forall a, b \in G: a b a^{-1} b^{-1} \in N$.

\(\ds \forall a, b \in G: \, \) \(\ds a b a^{-1} b^{-1}\) \(\in\) \(\ds N\)
\(\ds \leadsto \ \ \) \(\ds a b \paren {b a}^{-1}\) \(\in\) \(\ds N\)
\(\ds \leadsto \ \ \) \(\ds a b N\) \(=\) \(\ds b a N\)
\(\ds \leadsto \ \ \) \(\ds a N b N\) \(=\) \(\ds b N a N\)


Hence the result.

$\blacksquare$


Sources

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