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Find the character table of $U_{16}$.

Could you give me a hint or a start? Thank you.

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    Sort of. $\chi(9)$ can't be $w$ or $w^3$, since $\chi(9)=\chi(3^2)=(\chi(3))^2$. Also, $w^2=-1$ (indeed, I'm not sure why you are using the symbol $w$ for what everyone else calls $i$).2012-11-13

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The number of irreducible representations of a group is the number of conjugacy classes of that group. In an abelian group each element is its own conjugacy class, so there are $|G|$ irreducible representations for an abelian group $G$. Now, the number of degree one complex representations of a group $G$ is $[G:G']$, which of course is $|G|$ in an abelian group $G$. So all irreducible complex representations are degree one. Note that we could also have seen this from the formula $\sum_{i=1}^{|G|}d_i=|G|$ where $d_i$ denotes the degree of each irreducible representation.

So, your problem is reduced to finding all $16$ homomorphisms from $U_{16}\rightarrow \mathbb{C}^*$. I bet you can do this.

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    I think, Gerry is right. Most of the questions at this site involving $U_{16}$ mean $(\mathbb{Z}/16\mathbb{Z})^{\times}$.2014-11-12