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.

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