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Assume that G is a group. Define $G'=\langle\{ghg^{-1}h^{-1}\mid g,h\in G \}\rangle$. Then $G'$ is a normal subgroup of $G$. Prove If $H$ is a subgroup of $G$ and $G'\subseteq H$, then $H$ is normal subgroup of $G$.

Proof: Let $a \in G$ abd $h \in H$.

Then $aha^{-1}h^{-1} \in G$. Then $aha^{-1}h^{-1} \in H$. Then $aha^{-1}h^{-1}H =H$

Then $aha^{-1}H =H$, since $h^{-1} \in H$, since H is a subgroup.

Then $aha^{-1} \in H$. Therefore, H is normal subgroup of G.

Nadia C
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    What is your definition of $\langle X\rangle$ for sets $X$? You should only really need to show normality. – anon Oct 13 '13 at 23:05
  • You didn't get the definition of a subgroup generated by some elements. – Secret Math Oct 13 '13 at 23:06
  • @rschwieb, That is not the same question. My question is If commutator subgroup is a subset of a group H, H is a subgroup of G. Then H is normal. Remove your post, please. What if G/G' is a subgroup of G/H? That is my question read it carefully before answering. – Nadia C Oct 14 '13 at 00:16
  • Your proof is basically correct, but you need to fix the first two "then" below "Proof": the argument is $$a\in G,,,h\in H\implies [a,h]=aha^{-1}h{-1}\in G'\stackrel{\text{given}}\le H\implies [a,h]\in H\ldots;\text{etc.}$$ After this, you can simplify by writing $$\left(aha^{-1}\right)h\in H\implies aha^{-1}\in H$$ and that's all! – DonAntonio Oct 14 '13 at 03:55
  • @NadiaC Hi: The duplicate answers your first question, and in the post you asked me to delete, I gave a hint for the second question. Here it is again: Deduce that $G/G'$ is abelian, so all of its subgroups are normal. (Using that, you can answer your second question.) In general it's best to focus on a single question in a post, if possible. Good luck! – rschwieb Oct 14 '13 at 17:43

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