Show that G', the commutator subgroup of G, is normal in G. Prove that any subgroup A with G' $\subseteq$ A $\subseteq$ G is normal in G.

Question

Show that G', the commutator subgroup of G, is normal in G.
Prove that any subgroup A with G' $\subseteq$ A $\subseteq$ G is normal in G.

So the definition of the commutator subgroup is that;

In the group G, let G' be the subgroup generated by the set {$xyx^{-1}y^{-1}$ : x,y in G}.

So to show it is normal in G I must show that ($xyx^{-1}y^{-1})^{-1}G'xyx^{-1}y^{-1}$=G' but I just can't manipulate it to work.

Thanks

score 2 · Answer 1 · answered May 13 '14 at 17:22

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Suppose $g \in A$. You know that $k= hgh^{-1}g^{-1} \in A$, because $k= hgh^{-1}g^{-1} \in G' \subset A$. Hence $hgh^{-1} = kg \in A$.

Note that this proof is shorter than the standard proof that $G'$ is normal in $G$.

answered May 13 '14 at 17:22

Stephen Montgomery-Smith

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would it be possible to explain how I would continue on from $(xyx^{-1}y^{-1})^{-1}G'(xyx^{-1}y^{-1})=G'$ so i can get a better understanding? – ZZS14 May 13 '14 at 17:26
I don't see how this would even help. – Stephen Montgomery-Smith May 13 '14 at 17:29
that's proving it is normal is it not? – ZZS14 May 13 '14 at 17:29
No. To prove $A$ is normal you need to show that if $g \in A$, and $h \in G$, then $hgh^{-1} \in A$. – Stephen Montgomery-Smith May 13 '14 at 17:30
The first part of the question is: Show that G', the commutator subgroup of G, is normal in G. – ZZS14 May 13 '14 at 17:31
but a subset X is normal if $g^{-1}Xg=X$ so why can't I just prove that? – ZZS14 May 13 '14 at 17:32
To do the first part of the question, you need to prove that if $g \in G'$, and $h \in G$, then $hgh^{-1} \in G'$. You can get that from my answer by setting $A = G'$. (I'm sure the person who made the question meant for you to go via the method proposed by @Sam.) – Stephen Montgomery-Smith May 13 '14 at 17:33
A subset is normal if $h^{-1} X h = X$, for all $h \in G$. That's what I did. – Stephen Montgomery-Smith May 13 '14 at 17:34
I don't understand how you have shown that $h^{-1}Xh=X$ – ZZS14 May 13 '14 at 17:39
I showed that $h^{-1} X h \subset X$. To finish off the proof, you note that $h X h^{-1} \subset X$, because $h^{-1} \in G$. Hence $X \subset h^{-1} X h$. – Stephen Montgomery-Smith May 13 '14 at 17:41
i really don't understand how you have shown it thoguh – ZZS14 May 13 '14 at 17:53
I think you need to review the definitions, and check through them carefully. – Stephen Montgomery-Smith May 13 '14 at 18:06
I follow that if g in G and k=$xyx^{-1}y^{-1}$ then k in G since k in G' and G' $\subset$ G. Thus $xyx^{-1}=ky$ in G but then I don't see how you can then say $xyx^{-1} \subset G$ or anything from that point. – ZZS14 May 13 '14 at 18:09
Look up the definition of what it means for $A \subset B$. – Stephen Montgomery-Smith May 13 '14 at 18:13
oh ok, i think I've got it, but what about after that part. – ZZS14 May 13 '14 at 18:14

score 0 · Answer 2 · answered May 13 '14 at 17:20

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A subgroup $N$ of $G$ is normal if $gng^{-1}$ is in $N$ for all $g\in G$ and $n\in N$. You can use this to check normality of the derived subgroup.

answered May 13 '14 at 17:20

EPS

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Show that G', the commutator subgroup of G, is normal in G. Prove that any subgroup A with G' $\subseteq$ A $\subseteq$ G is normal in G.

2 Answers2