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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.

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  • $\begingroup$ 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)? $\endgroup$ – ToucanNapoleon Nov 12 '16 at 23:07
  • $\begingroup$ 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. $\endgroup$ – Aidan Nov 12 '16 at 23:09
  • $\begingroup$ 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. $\endgroup$ – ToucanNapoleon Nov 12 '16 at 23:11
  • $\begingroup$ @HarrySmit you're right, apologies. Let me fix this in my scratch work and I'll come back and edit everything thoroughly. $\endgroup$ – Aidan Nov 12 '16 at 23:13
  • $\begingroup$ No problem at all, it's great to see you trying to figure this out yourself. $\endgroup$ – ToucanNapoleon Nov 12 '16 at 23:17
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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 $$ \psi\bigl((a_1,a_2,\dotsc),(b_1,b_2,\dotsc)\bigr)= (a_1,b_1,a_2,b_2,\dotsc) $$

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  • $\begingroup$ 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. $\endgroup$ – Aidan Nov 12 '16 at 23:10
  • $\begingroup$ @Aidan I modified the answer. Your $\phi$ is still not injective, of course. $\endgroup$ – egreg Nov 12 '16 at 23:37
  • $\begingroup$ Thanks, sorry. I'm trying to do a couple of things at once here and of course my current map won't permit an isomorphism, finite or otherwise. I'll do some scribbles using your approach and see if I can break through. $\endgroup$ – Aidan Nov 12 '16 at 23:40
  • $\begingroup$ I'm stuck on showing a homomorphism using the map you provided, so I edited my progress showing what I've tried so far. Pretty confident it's wrong, but I'm not seeing how to separate the terms in the tuple. $\endgroup$ – Aidan Nov 12 '16 at 23:59
  • $\begingroup$ @Aidan Changing the question is not the correct way to go, as it makes the answer not understandable. Please, roll back. $\endgroup$ – egreg Nov 13 '16 at 0:04

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