Let $G$ be a group and let $H_1,H_2\unlhd G$ such that $H_1\cap H_2 = \{1\}$ and $H_1H_2=G.$ Prove $G\simeq H_1\times H_2$ [duplicate]
Solution 1:
The idea is to show that if $x\in H_1$ and $y\in H_2$, then $xy=yx$.
This is equivalent to showing that $xy(yx)^{-1}=1$, so to $$ xyx^{-1}y^{-1}=1 $$ Now note that $xyx^{-1}\in H_2$, so $xyx^{-1}y^{-1}\in H_2$ as well. Also $yx^{-1}y^{-1}\in H_1$, so $xyx^{-1}y^{-1}\in H_1$. Both statements follows from the subgroups being normal.
Since $H_1\cap H_2=\{1\}$ we have the proof.
The rest follows easily: the map $H_1\times H_2\to G$ defined by $(h_1,h_2)\mapsto h_1h_2$ is quite obviously a homomorphism, due to the above property. Prove it is an isomorphism.