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By $\mathbb{Z}_p$ of course I mean $p$-adic integers.

I am fairly certain that there is no non-trivial group homomorphism between $\mathbb{Z}_p$ and $\mathbb{Z}_q$ for $p\neq q$ as additive groups but I can't come up with a simple proof.

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    I guess $p,q$ prime right? – 2017-02-06
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    @JulioMaldonadoHenríquez yes, – 2017-02-06

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Note that $q$ is invertible in $\mathbb{Z}_p$. So in particular for any $a$ in $\mathbb{Z}_p$ there is a unique $a'$ such that $qa'=a,$ a unique $a''$ such that $q^2a''=a$, and so on.

However in $\mathbb{Z}_q$, $0$ is the only element infinitely divisible by $q$ in this sense. Therefore everything must get sent to $0$.

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I notice that the solution given by @Nate makes use of the ring structure of $Z_p$ ($q$ invertible, etc.). Since the original question is about a group homomorphism, I think it is worth trying at a "group theoretic only" answer. Among the many definitions of the group $(Z_p, +)$, the one in terms of the projective limit of the $(Z/p^n Z,+)$'s implies immediately the following property :

(P) A non zero subgroup (resp. quotient) of $Z_p$ is of the form $p^n Z_p$ (resp. $Z_p/p^n Z_p \cong Z/p^n Z$ ).

It follows that any non zero group homomorphism $f : Z_p \to Z_q$ is necessarily injective (apply (P) to $Ker f$ in $Z_p$ and to $Im f\cong Z_p/Kerf$ in $Z_q$). Consider then $Z_p$ as a subgroup of $Z_q$ via $f$. By (P) again, the subgroup $pZ_p$ will be of the form $q^n Z_q$, and the quotient $Z/pZ$ will inject into $Z/q^nZ$ : impossible if $p\neq q$.

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    (P) is not true. For instance $\mathbb{Z}$ is a legitimate subgroup of $\mathbb{Z}_p$ which is not in that form. – 2017-02-08
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    Yes, you are right. I forgot to add "open" (a bad habit). But now that I realize that your original question was about just a homomorphism of groups, not a morphism (continuous) of profinite groups, I wonder if there is not a hidden continuity argument in Nate's solution. So I have questions for @Nate: - is your "infinite divisibility by q" argument equivalent to "q is invertible in Z_p but not in Z_q" ? - since you seem to use elements which are in both Z_p and Z_q (as far as I understand) how do you show that they are not in the kernel of your homomorphism ? Thank you in advance. – 2017-02-09
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    I know you didn't ask the question to me but I am sure there is no continuity argument involved. (But as you said this is due to the fact that $q$ is an invertible element of $\mathbb{Z}_p$ so uses the ring structure.) But I'm not sure what your second question really means. (if you mean $q$ as the shared element, this simply refers to same element added $q$ times so it 'carries' over with the homomorphism. – 2017-02-09
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    Ah, o.k. To be sure, let me recap the argument in my own words. For any non zero a in Z_p, for any n>0, because q is invertible we can take a_n = q^{-n}a in Z_p and write a = (q^n)a_n (multiplication in the ring Z_p) = q^n . a_n (addition q^n times) . So far, so good. Now let f be a group homomorphism Z_p ---> Z_q . Then f(a) = q^n . f(a_n) = (q^n)f(a_n) in Z_q . But if f(a) is not null, this implies that the q-adic valuation of f(a_n) is < 0 for n large, and f(a_n) cannot belong to Z_q . Conclusion : f is the zero homomorphism. Isn't it ? PS: How do you type symbols in a comment ? – 2017-02-09
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    You are saying that the fact that $\mathbb{Z}_p$ is not $p^n$ relies on valuation of $\mathbb{Z}_p$. I'm sure this is not true but let me think a bit before giving a proof why $1/p$ is not in $\mathbb{Z}_p$. And to type symbol just use dollar sign as normal? Also use \mathbb{Z} (with dollar sign of course) for $\mathbb{Z}$. – 2017-02-10
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For prime $p$ and $q$ you already have the answer.

I would however adjust Nate's reasoning as follows: there is no need for doing "infinite" division by $q$. Simply, since in $\mathbb Z_p$ you can divide by $q$ then for some $a$ we have $qa = 1$. In case homomorphism $h: \mathbb Z_p\rightarrow \mathbb Z_q$ exists we have $h(a+a+...+a) = h(a)+...+h(a)$ ($q$ summands in each) and therefore $h(1) = qh(a) = 0$, so homomrphism is trivial.

Same proof if $p$ and $q$ are just co-prime.

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    I tried to decipher your proof and it seems that (again!) you are assuming $\mathbb{Z}_p=\mathbb{Z}/p\mathbb{Z}$. In this case just by looking at carnality the statement holds. – 2017-02-07
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    Sorry, I glitched there. In group theory $\mathbb Z_p$ is often used for $\mathbb Z/p\mathbb Z$, thus there exists quite an opportunity for the confusion. – 2017-08-02