# Deduce that the number of inequivalent degree $1$ complex representations of $G$ are equal to $|G|$.

Describe all the one-dimensional complex representations of a finite abelian group $G$. Deduce that the number of inequivalent degree $1$ complex representations of $G$ are equal to $|G|$.

attempt: By the fundamental theorem of abelian groups , $G$ can be expressed as a direct product of cyclic groups $G = C_1 \times ... \times C_n$,

where $|C_i| = n_i$. And recall every irreducible complex representation of $G$ is 1 dimensional (i.e a homomorphism between $G → \mathbb{C}^{\times}$.

Then for each generator $x_i \in C_i$ , then by the representation $\phi : G → \mathbb{C}^{\times}$, then $\phi(x_i)^{n_i} = 1$ ,so $n_i$ is a root of $1$. So we map each $x_i$ to a $n_i$ root of $1$ and extend this to all powers of $x_i$, the number of distinct homomorphism of $G$ into $\mathbb{C}^{\times}$ defined by this process equals $|G|$.

Can someone please help me? I need to know if this is fine, and if not please can someone please provide feedback or better approach. Thank you !

• looks fine to me. If you've already proved that the sums of the squares of the degrees of the irr. reps. of an arbitrary finite group is the order of the group, then you can use that for an easier proof, but I assume you haven't done that yet or else you would've thought of it! Feb 18, 2016 at 8:35