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.

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