# Formula for the number of inequivalent k-dimensional complex representations of an abelian group

Let G be an abelian group, $|G|=n$. Prove that the number of inequivalent k-dimentional complex representions of G is equal to the coefficient of $t^k$ in series $(1-t)^{-n}$. Find this coefficient.

I begin. the coefficient of $t^k$ in series $(1-t)^{-n}$ is equal to $\frac{n(n-1)\cdots(n-k+1)}{k!}$.

It is well known facts are 1)any irreducible complex representation of anabelian group G is one-dimensional; 2)the number of irreducible isomorphism classes of representations of a ﬁnite abelian group equals the order of the group. Use Maschke's Theorem.k-dimentional complex representions of G can be represented as direct sum of irreducible representation. It is a scalar matrix. its elements $a_{ii}$ is an one-dimensional irreducible complex representation of the group G. $a_{ii}$ can take n values. But then the number k-dimentional complex representions of G is equal to $n^k.$ And the number of inequivalent k-dimentional complex representions of G is equal to $\frac{n^k}{k!}.$ I undenstand that I am wrong, because generally speaking this number is not integer generally speaking.

where I am mistaken?

• By the way the coefficient of $t^k$ is another one (the exponent of $(1-t)$ is negative), and I think this exercise is intended to be solved without caring of calculating it explicitly.. Dec 27, 2013 at 17:58
The generating function is equal to $$\prod_{j=1}^n \sum_{i=0}^\infty x^i.$$ The coefficient of $x^k$ equals, therefore, to the number of ways of writing $k$ as $k=i_1 + i_2 + \dotsc + i_n.$ Now, $i_1$ is the number of copies of the first irrep, $i_2$ the number of copies of the second irrep, etc.