$G_{1}, G_{2} \triangleleft G$, $G_{1}G_{2} = G$, $G_{1} \cap G_{2} = \{ e \}$ implies $G_{1} \times G_{2} \cong G$?

Question

I'm looking for a proof of the following statement:

Let $G$ be a group, and let $G_{1}$, $G_{2}$ be subgroups of $G$, such that:

1. $G_{1}$, $G_{2} \triangleleft G$;

2. $G_{1} \cap G_{2} = \{ e \}$

3. $G_{1}G_{2} = G$.

Show that $G \cong G_{1} \times G_{2}$.

I know the variant of the theorem with $g_{1}g_{2} = g_{2}g_{1}$ as a condition, for all $g_{1} \in G_{1}$, $g_{2} \in G_{2}$, instead of $G_{1}$, $G_{2} \triangleleft G$.

I know from conditions 2. and 3. that every $g \in G$ can be represented uniquely as a product $g = g_{1}g_{2}$, $g_{1} \in G_{1}$, $g_{2} \in G_{2}$, so I know that $g \mapsto (g_{1}, g_{2})$ is a bijection, but I can't prove that it's a homomorphism. Is this the right way to go, or do I need to figure out another isomorphism?

Answer 1

Maybe it's better to consider the map $\varphi :G_1\times G_2\to G$ such that $\varphi((g_1,g_2))=g_1 g_2$ and prove that is an homomorfism.

To prove that is an homomorfism consider that $g\cdot h=h\cdot g\forall g\in G_1\forall h\in G_2$.

! In fact $\varphi((g_1, h_1)(g_2, h_2))=\varphi((g_1 g_2, h_1 h_2))=g_1 g_2 h_1 h_2$. But $[g_2, h_1]=g_2^{-1} h_2^{-1} g_1 h_2$ but note that $G_1, G_2$ are normal in $G$ then $g_2^{-1} h_2^{-1} g_1\in G_2$ and $h_2^{-1} g_1 h_2\in G_1$ then $[g_2, h_1]\in G_1\cap G_2=1$. Then $g_2^{-1} h_2^{-1} g_1 h_2=1$ then $g_1 h_2= h_2 g_1$. Finally you have that $\varphi((g_1, h_1)(g_2, h_2))=g_1 g_2 h_1 h_2=g_1 h_1 g_2 h_2=\varphi((g_1, h_1))\varphi((g_2, h_2)$