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Question

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

Solution

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

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

so $\phi([a,b])\in [G,G]$ and therefore $G^1$ is a characteristic subgroup. Assume now that for $i>1$ we have that $G^k$ is a characteristic subgroup of $G$ for all $k. Then for $\phi\in Aut(G)$, $a\in G$ and $b\in G^{i-1}$ we have again that

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

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

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