# Normal Subgroups that Intersect Trivially

Let $H$ and $K$ be normal subgroups of a group $G$ such that they intersect trivially.

Why is it then the case that

$$hk=kh;\;\;\;\; \forall h\in H,\;\forall k\in K?$$

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The equation is equivalent to $h^{-1}k^{-1}hk = 1$. Show that it holds for any $h \in H$ and $k \in K$. –  Mikko Korhonen May 20 '12 at 21:01
Ha. I put this on a homework assignment a few weeks ago. Not to worry, the papers have all been handed in and the solutions posted to the web. –  Gerry Myerson May 21 '12 at 3:47

Note that $hkh^{-1}k^{-1} = (hkh^{-1})k^{-1} \in K$, because $hkh^{-1}\in K$ (by normality of $K$) and $k^{-1}\in K$.

Also, $hkh^{-1}k^{-1} = h(kh^{-1}k^{-1})\in H$, because $kh^{-1}k^{-1}\in H$ (by normality of $H$), and $h\in H$.

So $hkh^{-1}k^{-1} \in H\cap K=\{1\}$. So $hkh^{-1}k^{-1}=1$.

Note that you don't even need $H$ and $K$ to be normal. You just need $H$ and $K$ to normalize each other, that is, $H\subseteq N_G(K)$ and $K\subseteq N_G(H)$.

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Because $hkh^{-1}k^{-1}$ is both in $H$ and $K$.
$H,K\vartriangleleft G\Rightarrow \left[ H,K\right] \leq H\cap K=1\Rightarrow \left[ h,k\right] =1,\forall h\in H$ and $k\in K$