Let $G$ be a group and let $N$ be a normal subgroup of $G.$ Let $N'$ denote the commutator of $N.$ Prove that $N'$ is a normal subgroup of $G.$

What I do know is that the commutator subgroup is characteristics. What I am not sure about is whether or not $N'$ is characteristic in $N$ or in $G$. If it is characteristic in $G$ then it is invariant under the conjugation automorphism $gN'g^{-1}$ for every $g \in N'$ and I am done. Otherwise it is characteristic in $N$ only, which means I am stuck. I would appreciate your help.

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    $\begingroup$ Why would anyone downvote my question without explaining? $\endgroup$ – Meitar Jan 11 '15 at 0:01

Hint: Let $G$ be a group and $H \leq K$ be two subgroups of $G.$ If $H$ is characteristic in $K$ and $K$ is normal in $G$ then $H$ is normal in $G.$

EDIT: Let $g \in G$ and consider the automorphism $\phi_g:G \to G, x \mapsto gxg^{-1}.$ Since $K$ is normal, $gKg^{-1} = K.$ So $\phi_g|_K$ is an automorphism of $K.$ Since $H$ is a characteristic subgroup of $K, \phi_g|_K(H) = H.$

  • $\begingroup$ Thank you. :) Here's how I tried to prove it using your hint. let gN'g^-1 for an arbitrary g in G. N is normal and N' is a subgroup of G, therefore gN'g^-1 is in N. Now gN'g^-1 is considered an automorphism, and since N' is characteristic gN'g^-1=N for all g in G. $\endgroup$ – Meitar Jan 11 '15 at 0:15
  • $\begingroup$ @Meitar: your idea is correct, but what you wrote is not completely right. you have to write it more clearly. I have added a proof. See if it helps. $\endgroup$ – Krish Jan 11 '15 at 0:24

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