Quotient Group is Abelian iff All Commutators in Divisor/Proof 1
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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
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.
$\blacksquare$