Let $p,q\in\mathbb{N}$, with $p\neq q$, prime numbers and consider the sets:
$$H_{p}=\biggl\{\frac{a}{p^{k}}: a\in\mathbb{Z}, k\in\mathbb{N}\cup\{0\}\biggr\}$$
$$H_{q}=\biggl\{\frac{a}{q^{k}}: a\in\mathbb{Z}, k\in\mathbb{N}\cup\{0\}\biggr\}$$
It is easy to see that $H_{p}$ and $H_{q}$ are both $\mathbb{Z}-$submodules of the $\mathbb{Z}$-module $\mathbb{Q}$. The question is:
Is there an isomorphim of $\mathbb{Z}$-modules between $H_{p}$ and $H_{q}$? Calculate $\operatorname{Hom}_{\mathbb{Z}}(H_{p},H_{q})$
This question comes from Exercise 3 of Chapter I of the book "Estructuras Algebraicas II" written by Enzo Gentile. Here is my aproach: suppose that such an isomorphim exists. Then it follows that $\mathbb{Z}_{p^{\infty}}=\frac{H_{p}}{\mathbb{Z}}$ is isomorphic to $\mathbb{Z}_{q^{\infty}}=\frac{H_{q}}{\mathbb{Z}}$ as a $\mathbb{Z}-$module in contradiction with Exercise 9 of the same chapter:
$$\mathbb{Z}_{p^{\infty}} \ \text{is isomorphic to} \ \mathbb{Z}_{q^{\infty}} \ \text{if and only if} \ p=q$$
Does anyone knows a different proof? What about $\operatorname{Hom}_{\mathbb{Z}}(H_{p},H_{q})$?
Let’s analyze a potential nontrivial homomorphism $\varphi: H_p \to H_q$ as $\mathbb{Z}$-modules. First, set $a = \varphi(1)$ and $b_k = \varphi(1/p^k)$. Thus, we have
$$a = \varphi(1) = \varphi(p^k\cdot1/p^k) = p^k\varphi(1/p^k) = p^kb_k,$$
so $a = p^kb_k$. Since $a, b_k\in H_q$, there are integers $c,d,s,t$ such that $a = c/q^s$ and $b_k = d/q^t$. Thus we have
$$\frac{c}{q^s} = \frac{p^kd}{q^t}, \text{ or that } cq^t = p^kdq^s.$$
However, Euler’s Theorem tells us that $p^k| c$ for all $k$, which is absurd. (If $c=0$ then $a,b_k=0$ and so $\varphi$ is trivial). Thus, no nontrivial homomorphism can exist between $H_p$ and $H_q$ when $p\ne q$.