Any subgroup that contains the subgroup generated by all commutators is normal.
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Hint for normality: for all $g \in G$ and $h \in H$, $g^{1}hg=[g,h^{1}]h$.
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Michael Shumate
First year graduate student at the University of Arkansas. I'm not sure what direction I want to go with research, but so far I have found geometric group theory and algebraic geometry interesting.
Updated on December 06, 2022Comments

Michael Shumate less than a minute
First, I was asked to show that 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.
This was not too difficult to show.
The second part of the question said if $G$ is a group and $H\supseteq G'$, where $G'$ is as in the last part, then $H\trianglelefteq G$ and $G/H$ is Abelian.
I'm not sure of the best way to approach this problem. Any help would be appreciated.

Captain Lama over 6 yearsHint : Normal subgroups of $G$ containing $G'$ correspond to normal subgroups in the quotient $G/G'$.

Michael Shumate over 6 yearsAh. That definitely helps. Thanks for the hint.
