Prove $ a: H_1 × H_2 \rightarrow H_1 + H_2 $ is an isomorphism

Question

I want to know that how to prove the following proposition:

Given an abelian group $G$, if $H_1$ and $H_2$ are any subgroups of $G$ such that $H_1 \cap H_2 = \{0\}$, then the map $a$ is an isomorphism $$ a: H_1 × H_2 \rightarrow H_1 + H_2 $$

Actually, I have got that the kernel of map a is $\{0,0\}$. $\forall a \in H_1, \ \forall b \in H_2$, then we have $H_1 × H_2$ is direct product which means $a × b \in H_1 × H_2$, besides, $H_1 + H_2$ is direct sum which means $a + b \in H_1 + H_2$

Answer 1

Let $h\in H_1+H_2$, then $h=h_1+h_2$ for some $h_1\in H_1,h_2\in H_2$. Hence every $h\in H_1+H_2$ can be mapped from $h_1h_2\in H_1 × H_2$, which means $a: H_1 × H_2 \rightarrow H_1 + H_2$ is indeed surjective.