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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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    Can you find the generators of $U_{16}$? Do you know that the irreducible representations of an abelian group are all homomorphisms into the nonzero complex numbers? Given the generators, can you find all such homomorphisms, by thinking about where the generators have to go?2012-11-03
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    so U16 is generated by {3,5,6,7,9,11,13,15} Is that correct?2012-11-07
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    sorry.. U 16 = { 3,5,7,9,11,13,15 }2012-11-09
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    U16 = {1,3,5,7,9,11,13,15} = C2×C42012-11-09
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    U16=<3>×<15>....<3>={1,3,9,11}...<15>={1,15}2012-11-10
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    Good. Now, how about the other questions I asked in my comment of 3 November?2012-11-11
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    Thanks for helping me.. but i didn't understand the other questions?.. Do you mean the linear character of U 16(ʎ:U 16 -->C*)such that ʎ(x)ʎ(y)=ʎ(xy) for all x,y in U 16 ?2012-11-11
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    You're looking for the character table. That means you're looking for the irreducible characters, right? And if $G$ is abelian, then its irreducible characters are the linear characters, the maps $\chi:G\to{\bf C}^*$ with $\chi(xy)=\chi(x)\chi(y)$ --- in other words, the group homomorphisms from $G$ to ${\bf C}^*$. Now that you know a set of generators for $U_{16}$, can you work out all the homomorphisms from $U_{16}$ to ${\bf C}^*$?2012-11-11
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    ʎ(1)=1, ʎ(15)=-1 ,ʎ(3)=w ,ʎ(9)=w^2 ,ʎ(11)=w^3 where w=e^(2πi⁄4)2012-11-11
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    Yes, that's *one* homomorphism from $U_{16}$ to ${\bf C}^*$. Can you work out all of them?2012-11-11
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    there is 8 characters______ʎ0(x)=1 , for all x in U162012-11-12
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    ʎ1(1)=1, ʎ1(15)=-1, ʎ1(3)=w, ʎ1(9)=w^2, ʎ1(11)=w^32012-11-12
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    ʎ2(1)=1, ʎ2 (15)= 1, ʎ2 (3)=w^2, ʎ2 (9)=w^3, ʎ2 (11)=w,2012-11-12
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    ʎ3(1)=1, ʎ3 (15)= -1, ʎ3 (3)=w^3, ʎ3 (9)=w, ʎ3 (11)=w^2,2012-11-12
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    ʎ4(1)=1, ʎ4(15)=1, ʎ4(3)=-1, ʎ4(9)=1, ʎ4(11)=-1______and so on ..and by finding ʎ(xy) for every x,y in U16 I finish the character Table .. Is that right?2012-11-12
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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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    so U16={3,5,6,7,9,11,13,15} Is that correct?2012-11-06
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    U16 = {1,3,5,7,9,11,13,15} = C2×C42012-11-09
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    Alexander, I believe OP is using $U_{16}$ to mean the group of units modulo 16, which is a group of 8 elements. Finding 16 homomorphisms is going to be rather a challenge.2012-11-13
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    Evidently so. I usually see $U_{16}$ as the group of complex $16$th roots of unity, isomorphic to $\mathbb{Z}_{16}$.2012-11-13
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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