# $(4\mathbb{Z}, +)$ isomorphic to $(5\mathbb{Z}, +)$

The question asks me to determine if $4\mathbb{Z}$ and $5\mathbb{Z}$ (with standard addition) are isomorphic and if so to give the isomorphism.

My attempts: What I am having difficulty with is showing a mapping that preserves the operation. i.e., $\phi(a+b) = \phi(a) + \phi(b)$.

What I have so far is:

$\phi(a+b) = \phi(5a/4) + \phi(5b/4)$.

$\phi(a+b) = 5/4(a+b) = 5/4a + 5/4b = \phi(a) + \phi(b)$.

Therefore, my conclusion is $4\mathbb{Z}$ and $5\mathbb{Z}$ are isomorphic and the isomorphism is $\phi(n) = \frac{5}{4}n$.

Can someone either confirm this or point me in the right direction? I really appreciate everyone willingness to help one another! Thank you!

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Seems fine to me. Since $n$ is divisible by 4, $\frac{5}{4}n\in 5\mathbb{Z}$. You just need to show injectivity and surjectivity. – Joe Johnson 126 Dec 14 '10 at 0:11
Also, there is only one infinite cyclic group. Each group you listed is an infinite cyclic group. – Sean Tilson Dec 14 '10 at 0:41
@Sean Tilson: "There is only one infinite cyclic group" up to isomorphism. – Arturo Magidin Dec 14 '10 at 2:53
@ Arturo:but of course... (and that isomorphism need not be unique!) (good catch though) @user: you should take my mistake to mean that cook people don't really worry about two things being different when they happen to be isomorphic. Unless they are super cool, then they worry a lot about the choices of isomorphisms. – Sean Tilson Dec 14 '10 at 5:16

Two comments: your argument only shows that $\phi$ is a homomorphism; you haven't shown it is an isomorphism (one-to-one and onto). This is a major thing: you cannot conclude it is an isomorphism just form knowing it is a homomorphism. You need to show it is invertible, or show it is one-to-one and show it is onto.
Second comment is more of a nit-pick: you should not say that $\phi$ is the isomorphism; there may be more than one isomorphism (in fact, there is; you can also take $\psi(4n) = -5n$). You should say (once you establish it, of course), that $\phi$ is an isomorphism.
To add to Arturo's answer: both groups are cyclic. Any homomorphism from a cyclic group is uniquely determined by what it does to a generator. Now, to be onto a cyclic group, you need to hit a generator of that group. A little bit of thought will tell you that what I just said implies that $\pm\phi$ are the only two isomorphisms there can be. – Alex B. Dec 14 '10 at 4:12
HINT $\rm\ \ \ \mathbb Z\ \cong n\ \mathbb Z\$ via $\rm\ x\to\ n\ x\$ for $\rm\: n\ne 0\:.\$ Thus $\rm\ n\ \mathbb Z\ \cong\ \mathbb Z\ \cong\: n'\ \mathbb Z\$ by transitivity of $\rm\: \cong\:$