What are the irreducible representations of $(\mathbb{Z},+)$? I'm wondering what are the irreducible representations of the group ($\mathbb{Z}$,+).
Knowing that for $\mathbb{Z}_n$ the 1-dimensional representations are the nth roots of unity, I considered taking the limit when $n \to \infty$ but I think I'm ending up with the U(1) group instead of $\mathbb{Z}$ and I have no other idea...
 A: You are probably looking for the complex representations (is that right?).  Firstly I claim any finite-dimensional simple rep is one-dimensional.  Let $G=\langle g\rangle$ be infinite cyclic, this will stand in place of $\mathbb{Z}$.  If $g$ acts on some nonzero finite-dimensional complex vector space $V$ by a linear map $\rho(g)$ then this linear map has an eigenvector, which spans a one-dimensional subrep.  If the rep was irreducible, this subrep must be all of $V$, so $\dim V=1$.
Now note that if $\lambda \in \mathbb{C}^*$ (I mean the non-zero complex numbers) then $g \mapsto \lambda$ induces a homomorphism $\rho_\lambda: G \to \mathsf{GL}_1(\mathbb{C})$, i.e. a representation of $G$.  Clearly any one-dimensional representation arises this way.  Furthermore if $\lambda \neq \mu$ then the representations afforded by $\rho_\lambda$ and $\rho_\mu$ are not isomorphic: not much conjugacy happens in $\mathsf{GL}_1$ !
In conclusion, the simple representations are in natural bijection with $\mathbb{C}^*$: this is called the character group.
A: If $\rho\colon \mathbb Z\to\mathrm{GL}(n,\mathbb C)$ is a complex representation, then because $\rho$ is a homomorphism, it is entirely determined by $\rho(1)$ (since $\rho(n)=\rho(1)^n$).  So if $\rho$ and $\rho'$ are two non-isomorphic representations of $\mathbb Z$, then $\rho(1)$ and $\rho'(1)$ determine two non-conjugate matrices in $\mathrm{GL}(n,\mathbb C)$.  Since two matrices are conjugate over $\mathbb C$ iff they have the same Jordan canonical form, we get that the $n$-dimensional complex representations of $\mathbb Z$ (up to isomorphism) are in one-to-one correspondence with the $n\times n$ Jordan canonical forms.
EDIT: I didn't notice you asked for irreducible representations of $\mathbb Z$.  In this case, I guess see mt_'s answer.
