Let $G=\mathbb{Z}^{\oplus \mathbb{N}}$. That is: $G=\{(a_1,a_2,\ldots)\mid a_i\in\mathbb{Z}, \forall i\in\mathbb{Z}^{>0}\}$. Prove $G\times G\cong G$.
What I've done so far:
Let $\phi:G\times G\to G$ such that $\big((a_1,a_2,a_3,\ldots),(b_1,b_2,b_3,\ldots)\big)\mapsto(a_1,b_1,a_2,b_2,a_3,b_3,\ldots)$.
First I show that $\phi$ is a homomorphism.
Take $(a,b),(c,d)\in G\times G$. I must show $\phi\big((a,b)(c,d)\big)=\phi\big((a,b)\big)\phi\big((c,d)\big)$.
Well, \begin{equation*} \begin{split}\phi\big((a,b)(c,d)\big)&=(a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2,\ldots)\\ & =(a_1,b_1,a_2,b_2,\ldots)(c_1,d_1,c_2,d_2,\ldots)\\ &=\phi\big((a,b)\big)\phi\big((c,d)\big). \end{split} \end{equation*} So $\phi$ is a homomorphism.
Now I must show $\phi$ is bijective. First I show $\phi$ is injective.
Of course, $\phi$ is not injective: indeed $$ \phi\bigl((1,0,\dotsc),(-1,0,\dotsc)\bigr)=(0,0,\dotsc) $$
An isomorphism can be defined by $$ \psi\bigl((a_1,a_2,\dotsc),(b_1,b_2,\dotsc)\bigr)= (a_1,b_1,a_2,b_2,\dotsc) $$