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Let $G$ be a group, $H$, a normal subgroup of $G$, prove that the commutator subgroup $H'$ of $H$ is a normal subgroup in $G$.

My ideas:

I want to prove it like this: $gH'g^{-1} = H'$ $\forall g \in G$ but I don't know how to continue, because $g$ may not lie in $H$ and $ghg^{-1}h^{-1}$ it's not a commutator.

Shaun
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    This is a special case of a more general statement: if $H$ is normal in $G$ and $C< H$ is a characteristic subgroup of $H$, then $C$ is normal in $G$. (A subgroup of $H$ is called characteristic if it is preserved by all automorphisms of $H$.) – Moishe Kohan Feb 08 '22 at 18:31
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    In fact, commutator subgroups are verbal, which implies they are fully invariant, which implies they are characteristic. See here. – Arturo Magidin Feb 08 '22 at 19:16

1 Answers1

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Hint: $g[h_1,h_2]g^{-1}=[gh_1g^{-1}, gh_2g^{-1}]$. Why is this a commutator of $H$?

Mark
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