Group Generated by Commutators of Two Normal Subgroups is a Normal Subgroup

Normal Subgroups Problems and Solutions in Group Theory

Problem 129

Let $G$ be a group and $H$ and $K$ be subgroups of $G$.
For $h \in H$, and $k \in K$, we define the commutator $[h, k]:=hkh^{-1}k^{-1}$.
Let $[H,K]$ be a subgroup of $G$ generated by all such commutators.

Show that if $H$ and $K$ are normal subgroups of $G$, then the subgroup $[H, K]$ is normal in $G$.

 
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Proof.

We first prove that a conjugate of each generator is in $[H,K]$.
Let $h \in H, k \in K$. For any $g \in G$, we have by inserting $g^{-1}g=e$ inbetweens
\begin{align*}
g[h,k]g^{-1}&=ghkh^{-1}k^{-1}=(ghg^{-1})(gkg^{-1})(gh^{-1}g^{-1})(gk^{-1}g^{-1})\\
& = (ghg^{-1})(gkg^{-1})(ghg^{-1})^{-1}(gkg^{-1})^{-1}.
\end{align*}


Now note that $ghg^{-1} \in H$ since $H$ is normal in $G$, and $gkg^{-1} \in K$ since $K$ is normal in $G$. Thus $g[h,k]g^{-1} \in [H, K]$.
By taking the inverse of the above equality, we also see that $g[k,h]g^{-1} \in [H, K]$. Thus the conjugate of the inverse $[h,k]^{-1}=[k,h]$ is in $[H, K]$.


Next, note that any element $x \in [H,K]$ is a product of generators or their inverses. So let us write
\[x=[h_1, k_1]^{\pm 1}[h_2, k_2]^{\pm 1}\cdots [h_n, k_n]^{\pm 1},\] where $h_i \in H, k_i\in K$ for $i=1,\dots, n$.
Then for any $g \in G$, we have
\begin{align*}
gxg^{-1}=(g[h_1, k_1]^{\pm 1}g^{-1})(g[h_2, k_2]^{\pm 1}g^{-1})\cdots(g [h_n, k_n]^{\pm 1} g^{-1}).
\end{align*}

We saw that the conjugate of a generator, or its inverse, by $g \in G$ is in $[H,K]$.
Thus $gxg^{-1}$ is also in $[H, K]$.
This proves that the group $[H,K]$ is a normal subgroup of $G$.

Related Question.

Another problem about a commutator group is
A condition that a commutator group is a normal subgroup


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  1. 10/01/2016

    […] You might also be interested in the problem: A condition that a commutator group is a normal subgroup […]

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