2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Why would anyone downvote my question without explaining? $\endgroup$ – Meitar Jan 11 '15 at 0:01
3
$\begingroup$

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 $K, \phi_g|_K(H) = H.$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.