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.

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