Prove that is characteristic in for all .

We begin with a lemma.

Lemma 1: Let be a group, let be characteristic, and let be an automorphism of . Then the mapping given by is an automorphism of . Proof: (Well-defined) Suppose ; then . Since is characteristic, , so that . Thus . (Homomorphism) We have . (Injective) Suppose . Then , so that (since is characteristic) . (Surjective) Let . Since is surjective, for some .

Lemma 2: Let be a group. Suppose such that is characteristic and is characteristic. Then is characteristic. Proof: Let . By Lemma 1, the mapping given by is an automorphism of . Since is characteristic, we have ; that is, if , then for some . Thus , and we have . Thus . Now suppose . Since is characteristic in , we have for some . Then . Since is characteristic in , we have for some . Thus , where . Thus we have . Hence is characteristic.

We proceed by induction on .

For the base case, if , then . Clearly then is characteristic in .

For the inductive step, let and suppose that is characteristic. Consider . By definition, is the subgroup of containing such that . Now by definition, is characteristic in . By Lemma 2, is characteristic in .

By induction, is characteristic in for all .

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