Consider the Pr$\ddot{\text{u}}$fer $p$-group $\mathbb{Z}(p^\infty)$ (where $p$ is a prime number). Define $f:\mathbb{Z}(p^\infty)\to \mathbb{Z}(p^\infty)$ by $f(x)=xz$ for all $x\in \mathbb{Z}(p^\infty)$ and $z\in \mathbb{Z}$. Then prove that $f$ is an isomorphism only if $z$ is not divisible by $p$.
Proof. For any $x\in \mathbb{Z}(p^\infty), x=a/p^n+\mathbb{Z}$ with $a/p^n\in \mathbb{Q}, a\in \mathbb{Z},n\in \mathbb{N}$. If $z$ is divisible by $p$, then $f(x)=xz=x(tp)=(a/p^n+\mathbb{Z})tp=b/p^m+\mathbb{Z}$. Hence for any $y=a/p^m+\mathbb{Z}\in \mathbb{Z}(p^\infty)$ there exists $x=a/p^n+\mathbb{Z}\in \mathbb{Z}(p^\infty)$ such that $f(x)=y$. This gives $f$ surjective.
My problem is in
(1) proving injectivity of $f$
(2) How to prove that $f$ is not injective if $z$ is not divisible by $p$.
For (1): $\ker(f)=\{a/p^n+\mathbb{Z}\in\mathbb{Z}(p^\infty): (a/p^n+\mathbb{Z})tp=\mathbb{Z}\}=\{a/p^m+\mathbb{Z} =\mathbb{Z} \}$. Which is possible if $a/p^m=0$. This seems not to make sense for me.
Consider the commutative diagram$\require{AMScd}$: \begin{CD} \Bbb Z/p^n\Bbb Z@>\theta_n>>\Bbb Z(p^\infty)\\ @Vf_nVV@VVfV\\ \Bbb Z/p^n\Bbb Z@>>\theta_n>\Bbb Z(p^\infty) \end{CD} where $\theta_n$ is the canonical homomorphism into the direct limit, which is injective. If $f$ is injective, then each $f_n$ is injective as well, hence $p\nmid z$ (otherwise $p^{n-1}\in\ker f_n$). Conversely, if $p\nmid z$, then each $f_n$ is injective, hence: \begin{align} \ker f &=\bigcup_{n\in\Bbb N}\theta_n[\ker(f\circ\theta_n)]\\ &=\bigcup_{n\in\Bbb N}\theta_n[\ker(\theta_n\circ f_n)]\\ &=\{0\} \end{align}