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

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 $$a + b \in H_1 + H_2$$

• What have you tried? What is the image of $a$? What is its kernel? – Viktor Vaughn Apr 2 '20 at 4:00
• I think the kernel of map a is $\{0,0\}$. – Zhenyu Wu Apr 2 '20 at 4:02
• If you can prove that, that means $a$ is injective. Can you show that it is surjective? – Viktor Vaughn Apr 2 '20 at 4:04
• Actually, I wonder if the map satisfies surjective. Could you prove that? Thanks a lot! – Zhenyu Wu Apr 2 '20 at 4:09
• What is the definition of $H_1 + H_2$? What does an element of $H_1 + H_2$ look like? – Viktor Vaughn Apr 2 '20 at 4:10

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.

• Your answer is so helpful for me, thanks a lot! – Zhenyu Wu Apr 2 '20 at 4:23