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Show that if $G$ is a group with two normal subgroups $H$ and $K$ such that $G=HK$ and $H\cap K=\{e\}$ then the map $(h,k)\rightarrow hk$ is an isomorphism of groups from $H\times K$ to $G$.

I am not sure whether or not I should be using the homomorphism theorem or not. (If $j:G\rightarrow H$ is a homomorphism then $G/\text{Ker}(j)$ is isomorphic to $\text{Im}(j)$)

Thanks

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No you cannot use that theorem here (but in other situations you will use it often, yes). Note that you have a bijection by $G=HK$ and $H\cap K=1$. The "hard" part is that this actually is a homomorphism. For this note that elements of $H$ commute with elements of $K$, using that these are normal subgroups (note $hkh^{-1}k^{-1}\in H\cap K=1$ for all $h\in H$ and $k\in K$, you have $\in$ by normality.)

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  • $\begingroup$ So do I show that the function is injective, surjective and is a homomorphism and this is proven? I have shown it is injective but cant seem to show surjectivity. Thanks $\endgroup$ – cf12418 Jan 3 '15 at 0:41
  • $\begingroup$ @cf12418: exactly. $H\cap K=1$ implies (here) injectivity, and $G=HK$ surjectivity (how is $HK$ defined?!). $\endgroup$ – Mister Benjamin Dover Jan 3 '15 at 11:15

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