Derived subgroup is characteristic

This article gives the statement, and possibly proof, of the fact that for any group, the subgroup obtained by applying a given subgroup-defining function (i.e., derived subgroup) always satisfies a particular subgroup property (i.e., characteristic subgroup)}
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Statement

Suppose $G$ is a group. Denote by $G'=[G,G]$ the derived subgroup of $G$, i.e., the subgroup generated by the commutators of pairs of elements of $G$. $G'$ is a characteristic subgroup of $G$.

Related facts

Stronger facts

• Derived subgroup is verbal combined with verbal implies fully invariant and fully invariant implies characteristic

Proof

Proof idea

The proof idea is that for $x,y\in G$ and $\sigma \in \operatorname {Aut} (G)$:

$\sigma ([x,y])=[\sigma (x),\sigma (y)]$

Thus, the set of elements expressible as commutators is invariant under any automorphism. Hence, the subgroup generated by this set is also invariant under any automorphism.

Proof details

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