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Let $H \triangleleft G$ and $K \triangleleft G$ such that $H \cap K = {e}$. Prove that $hk = kh, \forall h \in H, \forall k \in K $.

So here's what I have:

Since $H \triangleleft G$ and $K \triangleleft G$, $\forall h\in H, \forall k\in K$,

$khk^{-1} = h'$ for some $h' \in H$ and $hkh^{-1} = k'$ for some $k' \in K$.

So, $khk^{-1} = khk^{-1}h^{-1}h \implies khk^{-1}h^{-1} = h'h^{-1} \in H$ and $hkh^{-1} = k^{-1}khkh^{-1} \implies khkh^{-1} = kk' \in K$.

From there, I thought that starting with $h'h^{-1}kk' = kk'h'h^{-1}$ and trying to prove $hk = kh$ would be the right idea, but I'm stuck.

Perhaps I went to wrong direction somewhere (maybe even in the beginning) or just didn't think of the right steps that follow what I have.

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    It is easier than you expect it. Show that $k^{-1}hkh^{-1}$ is in $H\cap K$. – 2017-02-26
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    I do have a habit of making it more complicated. Where does that aspect come into play? Do I take a different approach or from what I have work towards that? – 2017-02-26
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    Okay, I definitely see how that is much easier. How would I show that $k^{-1}hkh^{-1} \in H \cap K$? – 2017-02-26
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    Yeah, a different approach. To show that it belongs to $H$, decompose the product into two factors each of which is in $H$. (Use the normality hypothesis.) Do the same to show it is in $K$. – 2017-02-26
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    Yes, I see it now. Thank you very much. – 2017-02-26
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    You're welcome. – 2017-02-26

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