complex irreps is in bijective correspondence with sequences Let $\{a_n\}$ be a sequences of positive integers such that
$$0 \leq a_n\leq p^n - 1,$$
$$a_n \equiv a_{n +1} \bmod p^n \quad \text{for all $n$}$$
Prove that the complex irreps of the group $ U_{p^{\infty}}$  is in bijective correspondence with  sequences $(a_n)$ 
I know what is the group $ U_{p^{\infty}}.$  I think that it is infinite abelian group. But I can't build its complex irreps. I don't know any what about their order.
I am not sure how to start it.Thanks a lot in advance for any help!
 A: Schur's lemma tells us the endomorphism ring of a simple module is a division ring; in particular since the only division algebra over $\Bbb C$ is $\Bbb C$ itself, the (module) endomorphisms of an irreducible representation must be the scalar multiples of the identity map. Therefore if $A$ is an abelian group and $V$ is an irreducible representation of $A$, for each $a\in A$, the action of $a$ is itself $A$-equivariant hence must be a scalar multiple of the identity; it is easy to prove then that $V$ is necessarily one dimensional. If $A$ is abelian, its irreducible representations are identified with dual group $A^\vee$ of homomorphisms $A\to\Bbb C^\times$ (since $\Bbb C^\times\cong{\rm GL}(\Bbb C^1)$).
You want to show the dual group of the Prufer $p$-group $\Bbb Z(p^\infty)$ is the group of $p$-adic integers $\Bbb Z_p$.
We can consider $\Bbb Z_p=\{(a_0,a_1,a_2,\cdots):a_r\equiv a_{r+1}\bmod p^r\}\subseteq\prod_{r=0}^\infty \Bbb Z/p^r\Bbb Z$ (this is the usual route to concretely constructing an inverse limit). We can also say $\Bbb Z(p^\infty)=\Bbb Z[p^{-1}]/\Bbb Z$ (additively).
Suppose $\alpha\in\hom(\Bbb Z(p^\infty),\Bbb C^\times)$. Show that $\alpha(m/p^r)$ must be a $p^r$th root of unity, $m\in\Bbb Z$. Therefore we can always consider $\alpha(1/p^r)=e^{2\pi i\varphi(r)/p^r}$. What is the relation between $\alpha(p^{-r})$ and $\alpha(p^{-(r+1)})$, in particular between $\varphi(r)$ and $\varphi(r+1)$? Can you now construct a coherent sequence of residues mod powers of $p$ out of $\alpha$ which sits inside $\Bbb Z_p$?
Note this goes further than just bijective correspondence; it is an isomorphism of topological groups. This can be checked as well (since $\Bbb Z_p$ and $\Bbb Z(p^\infty)^\vee$ carry group operations and topology.)
