Let $G$ be a finite group, and let $k$ be a field, which should be algebraically closed, I think. How to describe all homomorphisms $G\rightarrow k^*$ (i.e. one-dimensional representations: $k^*=\mathrm{GL}_1$) or just find the number of them?

I've got the following:

First, we assume $G$ is abelian. Since each finitely generated abelian group is a product of cyclic groups ($C_{k}$ or $\mathbb{Z}$), $G=C_{k_1}\times C_{k_2}\times\cdots \times C_{k_l}$. Let $n=|G|=k_1 k_2\cdots k_l$. So, for each homomorphism $\phi: G\rightarrow k^*$ we have $\phi(G)\subset Z(t^n-1)=H$ (I mean zeros of $t^n-1\in k[t]$).

Since each finite subgroup of $k^*$ is cyclic (let $H=C_m$) we are to investigate homomorphisms $$C_{k_1}\times C_{k_2}\times\cdots\times C_{k_l}\rightarrow C_m.$$ Is it necessarily the case that $m=n$? I know that in $\mathbb{Z}/p\mathbb{Z}$, we have $Z(x^p-1)=\{1\}$, so $m\ne n$. But how about algebraically closed fields?

However that may be, are there any other ways to describe such homomorphisms? To describe homomorphisms $C_{k_1}\times C_{k_2}\times\cdots\times C_{k_l}\rightarrow C_m$?

In conclusion, let me note that the case of nonabelian $G$ reduces to abelian one.

  • $\begingroup$ I've made some minor phrasing, grammar, and MathJax edits, hopefully making your question easier to understand; if you don't like any of the edits, please feel free to undo them. However, I would ask that you not use "$\mathbb{C}_n$" to mean "the cyclic group of order $n$", especially when you are elsewhere talking about algebraically closed fields; it is very confusing with the standard use of $\mathbb{C}$ for the complex numbers. The notations $C_n$ or $Z_n$ are much more common. $\endgroup$ – Zev Chonoles May 19 '13 at 4:02

Every homomorphism $C_{k_1}\times \cdots \times C_{k_l}\to k^\times$ factors as a product of homomorphism $C_{k_{\large i}}\to k^\times$, since the original is determined by where it sends the coordinates' generators. Each such coordinate homomorphism may be chosen independently. In particular the image of the full homomorphism is at most the $\ell={\rm lcm}(k_1,\cdots,k_l)$th roots of unity in $k^\times$ (which exist because $k$ is alg. closed), and it is indeed possible to obtain this as a surjective image.

To see that $\mu_\ell$, the $\ell$th roots of unity in $k$, are an "upper bound" for the image, note that everything in the image must be $\ell$-torsion since everything in $G$ is $\ell$-torsion itself. To see that this bound can be reached, it is simply a matter to see that the subgroups of $C_{\large \ell}$ of orders $k_i$ ($1\le i\le l$) are a generating set for the entire cyclic group $C_{\large\ell}$ (nice elementary number theory exercise).

The full collection of homomorphisms into $k^\times$ is itself a group under pointwise multiplication, called the dual group, and it is (noncanonically) isomorphic to the original finite abelian group.

If the original group $G$ is nonabelian then the homomorphisms factor through the abelianization $G^{\rm ab}:=G/[G,G]$, as you allude to, and there is no nice dual group for the full nonabelian group.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy