## The higher centers in a group are characteristic

Prove that $Z_k(G)$ is characteristic in $G$ for all $i$.

We begin with a lemma.

Lemma 1: Let $G$ be a group, let $K \leq G$ be characteristic, and let $\varphi$ be an automorphism of $G$. Then the mapping $\psi : G/K \rightarrow G/K$ given by $\psi(aK) = \varphi(a)K$ is an automorphism of $G/K$. Proof: (Well-defined) Suppose $aK = bK$; then $ab^{-1} \in K$. Since $K$ is characteristic, $\varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1} \in K$, so that $\varphi(a)K = \varphi(b)K$. Thus $\psi(aK) = \psi(bK)$. (Homomorphism) We have $\psi((ab)K) = \varphi(ab)K$ $= (\varphi(a)K)(\varphi(b)K)$ $= \psi(aK)(\psi(bK)$. (Injective) Suppose $\psi(aK) = \psi(bK)$. Then $\varphi(ab^{-1}) \in K$, so that (since $K$ is characteristic) $aK = bK$. (Surjective) Let $aK \in G/K$. Since $\varphi$ is surjective, $aK = \varphi(b)K = \psi(bK)$ for some $b \in G$. $\square$

Lemma 2: Let $G$ be a group. Suppose $K \leq H \leq G$ such that $K \leq G$ is characteristic and $H/K \leq G/K$ is characteristic. Then $H \leq G$ is characteristic. Proof: Let $\varphi \in \mathsf{Aut}(G)$. By Lemma 1, the mapping $\psi : G/K \rightarrow G/K$ given by $\psi(aK) = \varphi(a)K$ is an automorphism of $G/K$. Since $H/K$ is characteristic, we have $\psi[H/K] = H/K$; that is, if $h \in H$, then $\psi(hK) = \varphi(h)K = xK$ for some $x \in H$. Thus $\varphi(h)x^{-1} \in K$, and we have $\varphi(h) \in H$. Thus $\varphi[H] \leq H$. Now suppose $h \in H$. Since $H/K$ is characteristic in $G/K$, we have $hK = \psi(xK) = \varphi(x)K$ for some $x \in H$. Then $\varphi(x^{-1})h = k \in K$. Since $K$ is characteristic in $G$, we have $k = \varphi(y)$ for some $y \in K$. Thus $h = \varphi(xy)$, where $xy \in HK = H$. Thus we have $H \leq \varphi[H]$. Hence $H \leq G$ is characteristic. $\square$

We proceed by induction on $k$.

For the base case, if $k=0$, then $Z_k(G) = 1$. Clearly then $Z_k(G)$ is characteristic in $G$.

For the inductive step, let $k \geq 0$ and suppose that $Z_k(G) \leq G$ is characteristic. Consider $Z_{k+1}(G) \leq G$. By definition, $Z_{k+1}(G)$ is the subgroup of $G$ containing $Z_k(G)$ such that $Z_{k+1}(G)/Z_k(G) = Z(G/Z_k(G))$. Now by definition, $Z_{k+1}(G)/Z_k(G)$ is characteristic in $G/Z_k(G)$. By Lemma 2, $Z_{k+1}$ is characteristic in $G$.

By induction, $Z_k(G)$ is characteristic in $G$ for all $k$.