Please check my incomplete proof and give me a hint for make it complete
Cosinder the mapping between G to G' and the cardinality. Because it is one to one(by isomophism property)the cardinality of G and G' $|G|\leq |G'|$
Then,it is onto,the cardinality of G and G' is $|G|\geq |G'|$ . Then the cardianlity $|G|=|G'|$ and because it's one to one and onto it's bijective.
Because it is bijective the it has inverse and the inverse mapping from $G'\rightarrow G$ also bijective too.
I'm not sure I can imply the homomorphism as $\phi (ab)=\phi (ba)=\phi (a)(b)=\phi (b)(a)$ and $\phi (b+a)=\phi (a+b)=\phi (b)+\phi (a)=\phi (a)+\phi (b)$
because I can't prove that is commutative. anyone can give me a hint ?
If you are considering the (algebraic) homomorphism, it is no need to say anything about the cardinality of the domain to the codomain. And it is no need to deduce anything about the abelian, just take $u=\phi^{-1}(a)$ and $v=\phi^{-1}(b)$, then $\phi(uv)=\phi(u)\phi(v)=ab$, so $\phi^{-1}(ab)=uv=\phi^{-1}(a)\phi^{-1}(b)$ is all you need.