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Question

Prove that GiG^{i} is a characteristic subgroup of G for all i.

Solution

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Let ϕAut(G)\phi\in Aut(G) and a,bGa,b\in G. Then

ϕ([a,b])=ϕ(a1b1ab)=ϕ(a)1ϕ(b)1ϕ(a)ϕ(b)=[ϕ(a),ϕ(b)]\phi([a,b])=\phi(a^{-1}b^{-1}ab)=\phi(a)^{-1}\phi(b)^{-1}\phi(a)\phi(b)=[\phi(a),\phi(b)]

so ϕ([a,b])[G,G]\phi([a,b])\in [G,G] and therefore G1G^1 is a characteristic subgroup. Assume now that for i>1i>1 we have that GkG^k is a characteristic subgroup of GG for all k<ik<i. Then for ϕAut(G)\phi\in Aut(G), aGa\in G and bGi1b\in G^{i-1} we have again that

ϕ([a,b])=ϕ(a1b1ab)=ϕ(a)1ϕ(b)1ϕ(a)ϕ(b)=[ϕ(a),ϕ(b)]\phi([a,b])=\phi(a^{-1}b^{-1}ab)=\phi(a)^{-1}\phi(b)^{-1}\phi(a)\phi(b)=[\phi(a),\phi(b)]

so by the induction hypothesis ϕ([a,b])Gi\phi([a,b])\in G^i and therefore GiG^i is a characteristic subgroup.

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