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 $a + b \in H_1 + H_2$

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

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

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