# Two Normal Subgroups Intersecting Trivially Commute Each Other

## Problem 196

Let $G$ be a group. Assume that $H$ and $K$ are both normal subgroups of $G$ and $H \cap K=1$. Then for any elements $h \in H$ and $k\in K$, show that $hk=kh$.

## Proof.

It suffices to show that $h^{-1}k^{-1}hk \in H \cap K$.
In fact, if this it true then we have $h^{-1}k^{-1}hk=1$, and thus $hk=kh$.

Since $h\in H$ and $H$ is a normal subgroup of $G$, we see that the conjugate $k^{-1}hk\in H$.
Thus we have
$h^{-1}k^{-1}hk =h^{-1}(k^{-1}hk)\in H. \tag{*}$

Also, since $k^{-1}\in K$ and $K$ is a normal subgroup of $G$, we have the conjugate $h^{-1}k^{-1}h\in K$.
Hence, we see that
$h^{-1}k^{-1}hk =(h^{-1}k^{-1}h)k\in K. \tag{**}$

From (*) and (**), we see that the element $h^{-1}k^{-1}hk$ is in both $H$ and $K$, hence in $H\cap K$ as claimed.

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