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I'm supposed to prove that if a group $K$ has normal subgroups $G$ and $H$ with $G \cup H = 1$ and $G \vee H = K$ then there is an isomorphism $\Theta: G \times H \cong K$ defined by $\Theta (g, h) = gh$ for all $g \in G$ and $h \in H$ and so elsewhere I've got this theorem that says that if a group $K$ has subgroups $G$ and $H$ with $G \cup H = 1$ and $G \vee H = K$ then there is an isomorphism $\Theta: G \times H \cong K$ defined by $\Theta (g, h) = gh$ for all $g \in G$ and $h \in H$ as long as $gh = hg$ for $g \in G, h \in H$ so I'm guessing here that what I have to do is show that if $G$ and $H$ are normal subgroups of $K$ then $gh = hg$ for $g \in G, h \in H$. This is not apparent, and it's kind of killing me. Any help y'all could offer would be great. Thanks!

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I suppose you meant $\,G\cap H = 1\,$ in the first line. What is $\,G\, V\, H\,$ for you? Union of these two subgroups or the subgroup generated by them? –  DonAntonio Sep 29 '12 at 15:20
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up vote 1 down vote accepted

Consider the commutator $[g,h]:=ghg^{-1}h^{-1}$. Since both $ghg^{-1}$ and $h^{-1}$ is in $H$, $[g,h]\in H$. Similarly for $G$, thus, $[g,h]\in H\cap G=1$.

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Aww that's so cute. Thanks! –  Toby Carter Sep 29 '12 at 14:19
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