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I'm studying elementary level of algebra and I'm trying to prove that the commutator subgroup of a nonabelian simple group is the original group itself. It is trivially false if the group is abelian, but I can't prove it when the group is nonabelian.

The definition of the commutator subgroup of $G$ is $\{a^{-1}b^{-1}ab \mid a, b \in G\}$.

Shaun
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Sphere
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    Hint: the commutator subgroup is a normal subgroup. What is the definition of "simple" group? – Nick Dec 31 '20 at 03:41
  • Thanks. It was much more trivial than I thought. – Sphere Dec 31 '20 at 03:53
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    In fact, the commutator subgroup of $G$ is the subgroup generated by ${a^{-1}b^{-1}ab : a, b \in G}$, that is, if $g \in G$ is an element of the commutator subgroup of $G$ then there exists $k \in \mathbb Z^+$ and $a_1,\dots,a_k,b_1,\dots,b_k \in G$ such that $g = (a_1^{-1}b_1^{-1}a_1b_1) \cdots (a_k^{-1}b_k^{-1}a_kb_k)$. – azif00 Dec 31 '20 at 03:53

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