Let $G$ be an inner direct product of subgroups $H$ and $K$. Prove that the function $\phi : G \rightarrow H \times K$ such that $\phi (g) = (h,k)$ for $g = hk$, where $h \in H$ and $k \in K$, is injective and surjective.
Here's what I have:
Injectivity: Let $h_1, h_2 \in H$ and $k_1, k_2 \in K$ such that $h_1k_1 = h_2k_2$. WTS: $g_1 = g_2$. If $\phi (g_1) = \phi (g_2) \Longleftrightarrow (h_1, k_1) = (h_2, k_2) \Longleftrightarrow h_1k_1 = h_2k_2 \Longleftrightarrow g_1 = g_2$. so $\phi$ is injective.
Surjectivity: Consider some $g' \in G$. Want to show that $\phi (g) = g'$. And I'm stuck.
For the injectivity, I know I have a tendency to overcomplicate proofs, this may be the case. For the surjectivity, I don't know what to take $g$ as in order to get some $g'$.
A group $G$ is the internal direct product of $H$ and $K$ if $G=HK$, $H\cap K=1$, and $H,K$ are both normal subgroups of $G$.
Let $g\in G$.
Since $G=HK$, $g=hk$ for some $h\in H$ and $k\in K$.
Hence $(h,k)\in H\times K$ where $\phi(g)=\phi(hk)=(h,k)$.
By the way, you can try to prove also that $\phi$ is a well-defined homomorphism so that we can conclude that $G$ is isomorphic to $H\times K$.