Endomorphism ring of Prüfer group It is claimed in Wikipedia that the ring of endomorphisms of the Prüfer $p$-group is given by the p-adic integers, but I want to know how to prove it
So far I have if $\phi \in End_{\mathbb{Z}}(\mathbb{Z}_{p^\infty}, \mathbb{Z}_{p^\infty})$ then $\phi (\langle \frac{1}{p^k} \rangle ) = \langle \frac{1}{p^l} \rangle $ because the image of a submodule is a submodule, then $l \leq k$ because the number of elements of $\phi (\langle \frac{1}{p^k} \rangle )$ doesn't exceed $p^k$.
So ,if we consider $\phi ( \frac{1}{p}  )$ is equal to $0$ 
or $\frac{a}{p}$ with $(a,p) =1$ .  If it is $\frac{a}{p}$, $\phi $ is determined because   $\phi ( \frac{1}{p^k }  )= \frac{b}{p^k}$ with $a \equiv b$ mod $p$, otherwise we do the same with $\phi ( \frac{1}{p^2 }) $ , it cannot be $\frac{a}{p^2}$ because it contradicts that $\phi ( \frac{1}{p}) =0 $ so is $0$ or $\frac{a}{p}$ and we continue the process.
With this I think that all the endomorphisms are determined but that's all.
Thank you in advance.
 A: Since this hasn't been answered so far (and if you hadn't done it already on your own), I am gonna try to delineate a much more general result towards this direction.
The key word you're looking is called Matlis' Theorem.
I presume you're familiar with some categorical notions like inverse (projective) limit, direct (inductive) limit (if not is a good chance to try :)). In every book of modern abstract algebra you can find the aforementioned explicitly written with many examples. For the moment you can think of them as a kind of generalized intersections and unions respectively, where simultaneously we preserve the structure of the underlying material (modules, groups, rings...). In fact those ideas behind the limits generalize the ideas of products and coproducts, pullbacks and pushouts, equalizers and coequalizers and other related notions. For the moment we are working out in a category of modules, so keep in mind that the above limits do exist indeed (that is, categories of modules are complete and co-complete, i.e. bicomplete).
We are seeking of an isomorphism of rings

$$ End_{\mathbb{Z}}(\mathbb{Z}_{p^{\infty}}) \cong \hat{\mathbb{Z}}_p.$$

Where the right-hand side denotes the $p$-adic integers, or maybe better in our case the inverse limit of the inverse system $\{\mathbb{Z}/{(p)}^n,\psi_{m,n}\}_{n \in \mathbb{N}}$, which is called the $p$-adic completion of $\mathbb{Z}$. The definitions and theory behind the aforementioned can be found in almost any book of abstract algebra nowadays as well. To this end, we're gonna need some tools and comments, which I introduce below, and a good reference to find their proofs is Rotman's book "An Introduction to Homological Algebra".

Proposition:
If $\{M_{i},\psi_{i,j}\}_{i \in \mathbb{N}}$ is an inverse system of left $R$-modules, then there is a natural isomorphism
$$ Hom_R(A, \varprojlim M_i) \cong \varprojlim Hom_R(A, M_i),$$
for every left $R$-module $A$.

Similarly one can define the dual notion of inverse systems, and deal with direct systems instead and an analogous proposition exists.

Proposition:
If $\{M_{i},\psi_{i,j}\}_{i \in \mathbb{N}}$ is a direct system of left
$R$-modules, then there is an isomorphism
$$ Hom_R(\varinjlim M_i, A) \cong \varprojlim Hom_R(M_i, A).$$
for every left $R$-module $A$.

Now think of the case where we have as our material the family of cyclic groups $\{ \mathbb{Z}/(p)^n \}_{n \in \mathbb{N}}$, for some fixed $p$ prime. But once we look on it with the homomorphisms
$$ \psi_{m,n}: \mathbb{Z}/(p)^m \rightarrow \mathbb{Z}/(p)^n , \,\, m \geq n, $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, a+(p)^m \mapsto a+(p)^n,$$
which give the above inverse system  $\{\mathbb{Z}/{(p)}^n,\psi_{m,n}\}_{n \in \mathbb{N}}$, with the corresponding inverse limit being $p$-adic integers $\hat{\mathbb{Z}}_{p}$ as I said earlier, and by just think the homomorphisms $$\gamma_{n+1,n} : \mathbb{Z}/p^n\mathbb{Z} \rightarrow \mathbb{Z}/p^{n+1}\mathbb{Z},$$
induced by multiplication by $p$, we form the direct system $\{\mathbb{Z}/{(p)}^n,\gamma_{m,n}\}_{n \in \mathbb{N}}$ whose direct limit is the Prüfer group, that is $\varinjlim \mathbb{Z}/{(p)^n} \cong \mathbb{Z}_{p^{\infty}}$.
Now, after the prologue, Matlis proved himself that there is the following isomorphism (proof can be found in Lam's book Lectures on Modules and Rings)

Theorem:
Given a local commutative Noetherian ring $(R, \mathfrak{m})$, the endomorphism ring $End_R(E)$, is isomorphic to $\hat{R}$, the $\mathfrak{m}$-adic completion of $R$.

Where by $E$ we denote the injective hull of the quotient $R/ \mathfrak{m}$, as  $R$-module and usually is referred as the standard module.
Let's see now, how does this apply to our case. Firstly, we have the following fact

The Prüfer group $\mathbb{Z}_{p^{\infty}}$, is the injective hull of the cyclic groups $\mathbb{Z}_{p^{n}}$, for every $n \geq 1$.

The fact that is injective module (here as abelian group), follows directly, since it is direct summand of the divisible group $\mathbb{Q}/\mathbb{Z}$ (see structure theorem of divisible abelian groups), hence is itself divisible which implies injectivity since $\mathbb{Z}$ is a principal ideal domain. To see that is the smallest such an injective module we exploit its presentation
$$\mathbb{Z}_{p^{\infty}}= <a_0,a_1,a_2,...| \,\, pa_0=0, \, pa_{n+1}=a_n, \,\, n\geq 1>,$$
indeed, if $D$ is another abelian divisible group such that $\mathbb{Z}_p\subseteq D$, we can construct $x_1,x_2,\ldots\in D$ such that $px_1=\overline 1$, and $px_{n+1}=x_n$, hence it is not hard to see that the subgroup generated by $\bar 1$ and the $x_i\,'s$ is isomorphic to $\mathbb{Z}_{p^{\infty}}$.
Having all the above in mind we can replace in Matlis theorem our data. So we have
$$\,\,\,\, End_\mathbb{Z}(\mathbb{Z}_{p^{\infty}})=Hom_\mathbb{Z}(\mathbb{Z}_{p^{\infty}},\mathbb{Z}_{p^{\infty}}) \cong Hom_\mathbb{Z}( \varinjlim \mathbb{Z}/{(p)^n}, \mathbb{Z}_{p^{\infty}}) \cong \varprojlim Hom_\mathbb{Z}(\mathbb{Z}/{(p)^n},\mathbb{Z}_{p^{\infty}}) \stackrel{(*)}{\cong}  \varprojlim \mathbb{Z}/{(p)^n} \cong \hat{\mathbb{Z}}_p.$$
Note that $(*)$ is the only subtle point, since if you check Matlis' Theorem you'll see that is proven for local rings and apparenly $\mathbb{Z}$ isn't such a ring. In that case we can apply the theorem after localizing our ring over a prime ideal. In our case we just simply have the ideal $(p)$ for the fixed prime which gives the local ring $\mathbb{Z}_{(p)}$ . So the isomorphism follows.
(My answer is extended since I've no idea what you know or not, because many things are introductory if you know them already, you can just skip them.)
Cheers!
