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

Why is it then the case that

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

Martin Sleziak
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rk101
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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
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    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 03:47

3 Answers3

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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 $K\subseteq N_G(H)$.

Arturo Magidin
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Because $hkh^{-1}k^{-1}$ is both in $H$ and $K$.

Egbert
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    Wow! you and @Arturo answered "exactly" at the same time measured with a least count of 1 second. –  May 20 '12 at 21:03
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$H,K\vartriangleleft G\Rightarrow \left[ H,K\right] \leq H\cap K=1\Rightarrow \left[ h,k\right] =1,\forall h\in H$ and $k\in K$

User2040
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  • This post has been long dead. But anyways, this is really assuming what you are trying to prove. The proof that $[H,K] \leq H \cap K$ is about as hard as the proof of the OP's question. Might as well prove this and then draw the conclusion from there. – misogrumpy Aug 10 '18 at 15:37