# Proving a specific group isomorphism

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.

• Perhaps you meant $G = \{(a_1, a_2, \dots) | \; a_i \in \mathbb{Z} \;\forall \; i \in \mathbb{Z}_{>0}\}$ (i.e. sequences with infinite length)? – ToucanNapoleon Nov 12 '16 at 23:07
• Oops! You're right I did. I misread the problem in the textbook; should get my eyes checked. We're going to let $G=\mathbb{Z}^{\oplus\mathbb{N}}$. Will edit now. – Aidan Nov 12 '16 at 23:09
• This is not the correct implication to prove injectivity. For injectivity of $f$, you have to show $f(x) = f(y) \Rightarrow x = y$ (or equivalently $x \neq y \Rightarrow f(x) \neq f(y)$). What you are showing is that $f(x) \neq f(y) \Rightarrow x \neq y$, which holds for all well-defined functions. – ToucanNapoleon Nov 12 '16 at 23:11
• @HarrySmit you're right, apologies. Let me fix this in my scratch work and I'll come back and edit everything thoroughly. – Aidan Nov 12 '16 at 23:13
• No problem at all, it's great to see you trying to figure this out yourself. – ToucanNapoleon Nov 12 '16 at 23:17

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 $$\psi\bigl((a_1,a_2,\dotsc),(b_1,b_2,\dotsc)\bigr)= (a_1,b_1,a_2,b_2,\dotsc)$$
• Oops! I misread the problem in the textbook; should get my eyes checked. We're going to let $G=\mathbb{Z}^{\oplus\mathbb{N}}$. Will edit now. – Aidan Nov 12 '16 at 23:10
• @Aidan I modified the answer. Your $\phi$ is still not injective, of course. – egreg Nov 12 '16 at 23:37