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Let $G$ be a group and $H,K \subset G$ be normal subgroups such that $H \cap K = \{e\}$ and $G=HK$. I need to show that the map, which I have denoted $\phi$, $H \times K \longrightarrow G$ given by $(h,k) \mapsto hk$ is a group isomorphism. I have shown that it is a homomorphism: for some $h_1,h_2 \in H, k_1,k_2 \in K, \phi(h_1k_1,h_2k_2)=(h_1k_1)(h_2k_2)=\phi(h_1,k_1)\phi(h_2,k_2)$.

I have also shown that it is injective: take $(h_1,k_1) \in H \times K$ and $(h_2,k_2) \in H \times K$ such that $(h_1,k_1) \neq (h_2,k_2)$. Then we have $\phi(h_1,k_1)=h_1k_1 \neq h_2k_2 =\phi(h_2,k_2)$.

However, I am not sure how to show that $\phi$ is surjective to complete the proof.

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It is surjective because of the way $HK$ has been defined, namely as the set consisting of products $hk$ with $h\in H$ and $k\in K$. – Tobias Kildetoft Oct 5 '12 at 8:27

To show that it's surjective you need to show that the image of the function is all of $G$. But you have that $G = HK$ so you simply need to show that for all $hk \in HK$ there exists something in $H \times K$ which maps to it under phi.

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You did not show that $\phi$ is a homomorphism: you need to show that $\phi(h_1h_2,k_1k_2)=\phi(h_1,k_1).\phi(h_2,k_2)$, that is $h_1h_2k_1k_2= h_1k_1h_2k_2$. This means that you have to show that $h_2k_1=k_1h_2$ voor all $h_2 \in H$ and all $k_1 \in K$. Now look at $h_2^{-1}k_1h_2k_1^{-1} = h_2^{-1}(k_1h_2k_1^{-1}) = (h_2^{-1}k_1h_2)k_1^{-1}$, this is an element which is both in the normal subgroups $H$ and in $K$ (look carefully at the brackets), and hence it must equal the identity element.

To show injectivity: assume $\phi(h,k)=1$, then $hk=1$. Again $h=k^{-1} \in H\cap K = 1$, so $h=1=k$, and $(h,k) = (1,1)$ the identity element.

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