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Can each of the following be a character of a finite group $G$?

$(i) \ \ (2,0,\frac{1}{2},\frac{1}{2},0)$

$(ii) \ \ (3,-1,0,4,0)$

$(iii) \ \ (2,2,2,2)$

$(iv) \ \ (1,0,-1,0)$

I think that the first one cannot be a group as there are $1/2$'s and this contradicts some theorem about characters being algebraic integers. I can see no reason why the others cant be characters of finite groups. Is this correct?

How would we give an example of a finite group with the given character? Why should I be considering?

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    $\begingroup$ It depends a bit whether the first number is (as is conventional) the degree - i.e. the character associated with the identity element. What do you know (a) about the degree of a character; (b) the inner product of a character with itself (and definition of this); (c) the relationship between a character and the conjugacy classes of the group? These should help you to analyse the situation in more detail and more accurately. $\endgroup$ – Mark Bennet May 12 '15 at 16:17
  • $\begingroup$ The first is always $\chi(1)$ $\endgroup$ – Permian May 12 '15 at 16:18
  • $\begingroup$ Remember that algebraic integers are simply roots to polynomial equations with integer coefficients (this includes complex numbers, like $i$). So, everything in sight here is an algebraic integer. One fact that will be helpful is how the size of character values relate to the degree of the character. $\endgroup$ – pjs36 May 12 '15 at 16:20
  • $\begingroup$ This is a test of what you know about characters and whether you can relate theory to real situations. It is well worth doing more work on this yourself, because if I give you an answer now it will consolidate my knowledge, not yours. You mention algebraic integers as one thing you vaguely know - what else do you know which might apply? And can you see one of the possibilities which is an obvious candidate, and find a group for that one? $\endgroup$ – Mark Bennet May 12 '15 at 17:01

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