Simple and non-abelian implies perfect

Statement

Suppose $G$ is a simple non-abelian group. Then, $G$ is a perfect group, i.e., $G$ equals its own derived subgroup.

Related facts

• Characteristically simple and non-abelian implies perfect
• Simple and non-abelian implies centerless

Facts used

1. Derived subgroup is normal

Proof

Given: A simple non-abelian group $G$.

To prove: The derived subgroup $[G,G]$ equals $G$.

Proof:

No.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The derived subgroup of $G$ is normal in $G$.	Fact (1)			Fact-direct.
2	The derived subgroup of $G$ is either the trivial subgroup or the whole group $G$.		$G$ is simple.	Step (1)	Step-given direct, combined with the definition of simple.
3	The derived subgroup of $G$ cannot be the trivial subgroup.		$G$ is non-abelian.		By definition, the derived subgroup is trivial if and only if the group is abelian.
4	The derived subgroup of $G$ is the whole group $G$.			Steps (2), (3)	Step-combination direct.