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If $G$ is a group and $G'$ is generated by $\{xyx^{-1}y^{-1}|x,y\in G\}$, then $G'\trianglelefteq G$ and $G/G'$ is Abelian.

At first, I thought this is easy because I thought $g[x,y]g^{-1}=[gxg^{-1},gyg^{-1}]$. But it was a mistake. I could not prove that it is also subgroup of G and $G/G'$ is abelian. I need help.

Arnaud D.
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김일희
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  • $g[x,y]g^{-1}=[gxg^{-1},gyg^{-1}]$ is perfectly correct. And you don't need to prove that $G'$ is a subgroup, it is one by definition. As for the fact that $G/G'$ is abelian, try proving that commutators are trivial. – Arnaud D. May 15 '16 at 12:37
  • Thank you for your answer. I proved that $G/G'$ is abelian. But why don' need to show that $G'$ is a subgroup? Then Is the Commutator 'subgroub' of $G$ already a subgroup of $G$ by just definition???? – 김일희 May 15 '16 at 12:46
  • @김일희 No, I think the intention of Arnaud was: for any subset $;S;$ of a group $;G;$ , by definition $;\langle S\rangle;$ is the subgroup generated by elements of $;S;$ , and here $;G'=\langle [g,h];:;g,h\in G\rangle;$ . When you say "subgroup generated by...", you already have a subgroup! – DonAntonio May 15 '16 at 12:48
  • $G'$ is defined as the subgroup of $G$ generated by the commutators, so there is no need to prove it is a subgroup. – Arnaud D. May 15 '16 at 12:49
  • Ohhh I had forgotten what 'generated' means. I was so stupid... haha Everyone Thank you very much. – 김일희 May 15 '16 at 12:57

1 Answers1

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The commutator subgroup of $G$ is the subgroup generated by the set $C$ of elements of the form $[x,y]=xyx^{-1}y^{-1}$.

Suppose you have a subset $S$ with the property that, for every $g\in G$ and every $x\in S$, $gxg^{-1}\in S$. Then the subgroup $H$ generated by $S$ is normal in $G$.

Indeed, an element of the subgroup $H$ is of the form $$ x=x_1^{a_1}x_2^{a_2}\dots x_n^{a_n} $$ where $a_i=\pm1$ and $x_i\in S$, for $i=1,2,\dots n$. Then, if $g\in G$, we clearly have $$ gxg^{-1}=(gx_1g^{-1})^{a_1}(gx_2g^{-1})^{a_2}\dots(gx_ng^{-1})^{a_n} $$ so $gxg^{-1}\in H$, because $gx_ig^{-1}\in S$, for $i=1,2,\dots,n$.

In the case of the commutator subgroup, $$ g[x,y]g^{-1}=[gxg^{-1},gyg^{-1}]\in C $$ so, by the remark above, $G'=\langle C\rangle$ is normal in $G$.

egreg
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