# Non-Abelian Simple Group is Equal to its Commutator Subgroup

## Problem 149

Let $G$ be a non-abelian simple group. Let $D(G)=[G,G]$ be the commutator subgroup of $G$. Show that $G=D(G)$.

Contents

## Definitions/Hint.

We first recall relevant definitions.

• A group is called simple if its normal subgroups are either the trivial subgroup or the group itself.
• The commutator subgroup $D(G)=[G,G]$ is a subgroup of $G$ generated by all commutators $[a,b]=a^{-1}b^{-1}ab$ for $a,b\in G$.

The commutator subgroup $D(G)=[G,G]$ is a normal subgroup of $G$.
For a proof, see: A condition that a commutator group is a normal subgroup.

## Proof.

Note that the commutator subgroup $D(G)$ is a normal subgroup.
Since $G$ is simple, any normal subgroup of $G$ is either the trivial group $\{e\}$ or $G$ itself. Thus we have either $D(G)=\{e\}$ or $D(G)=G$.
If $D(G)=\{e\}$, then for any two elements $a,b \in G$ the commutator $[a,b]\in D(G)=\{e\}$.

Thus we have
$a^{-1}b^{-1}ab=[a,b]=e.$ Therefore we have $ab=ba$ for any $a,b\in G$. This means that the group $G$ is abelian, which contradicts with the assumption that $G$ is non-abelian.
Therefore, we must have $D(G)=G$ as required.

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##### Two Quotients Groups are Abelian then Intersection Quotient is Abelian

Let $K, N$ be normal subgroups of a group $G$. Suppose that the quotient groups $G/K$ and $G/N$ are both...

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