## Derived subgroups are characteristic

Prove that $G^k$ is characteristic in $G$ for all $k$.

We proceed by induction on $k$.

For the base case, if $k = 0$, then $G^0 = G$ is characteristic in $G$.

Suppose now that $G^k$ is characteristic in $G$ and let $\alpha$ be an automorphism of $G$.

We have $\alpha[G^{k+1}] = \alpha[[G,G^k]]$ $= [\alpha[G],\alpha[G^k]]$ $= [G,G^k]$ $= G^{k+1}$. Thus $G^{k+1}$ is characteristic in $G$.