## Lower centers are characteristic

Prove that $G^{(k)}$ is characteristic in $G$ for all $k$.

We proceed by induction on $k$.

For the base case, let $k = 0$. Then $G^{(0)} = G$ is characteristic in $G$.

For the inductive step, suppose that for some $k \geq 0$, $G^{(k)}$ is characteristic in $G$. Let $\alpha$ be an automorphism of $G$. Then $\alpha[G^{(k+1)}] = \alpha[[G^{(k)}; G^{(k)}]]$ $= [\alpha[G^{(k)}; \alpha[G^{(k)}]]]$ $= [G^{(k)}; G^{(k)}]$ $= G^{(k+1)}$. Thus $G^{(k+1)}$ is characteristic in $G$.