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For a finite group G the number of irreducible representations over an algebraically closed field F is at most the number of conjugacy classes whose sizes are coprime to the characteristic of F.

What about over fields that aren't algebraically closed?

The motivation for this question is essentially knowing when all of the irreducible representations of a group over a particular field (in particular a finite one) have been found.

As an example, $A_4$ has four conjugacy classes and has four irreducible representations over fields such as $\mathbb{C}$ or $\mathbb{F}_7$ but only has three irreducible representations over $\mathbb{F}_5$. In the first two cases the original theorem holds so we know we have found all irreducible representations but for the third case, how do we know that there aren't any other possible irreducible representations?

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You'll want to read about $K$-conjugacy classes which are exactly what you are asking about and described in this question.

For example, the $\mathbb{F}_5$-conjugacy classes of $A_4$ are precisely the sets of elements with the same order. Usually the elements of order 3 split into two, but they are related by $x$ in one class means $x^{-1}$ in the other. However, $x^5$ and $x$ are always in the same $\mathbb{F}_5$-class, and $x^5 = x^{-1}$ when $x$ has order 3.

  • Pazderski, Gerhard. "On the number of irreducible representations of a finite group." Arch. Math. (Basel) 44 (1985), no. 2, 119–125. MR780258 DOI:10.1007/BF01194075
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  • $\begingroup$ Oh wow. This seems a bit intense but no doubt what I was after. Thank you! $\endgroup$
    – john
    Jun 1, 2013 at 14:44
  • $\begingroup$ No problem. For finite fields, it is not so bad. You get to use two kinds of conjugation $x^g = g^{-1} x g$ and $x^q = x \cdot x \cdots x$. Both $x^g$ and $x^q$ are $\mathbb{F}_q$-conjugate to $x$, and those are the only kinds of conjugacy you need to consider. $\endgroup$
    – Jack Schmidt
    Jun 1, 2013 at 14:50
  • $\begingroup$ Oh. That doesn't seem too bad at all! So then the number of irreducible representations is less than or equal to the number of $\mathbb{F}_q$-conjugacy classes with orders coprime to $q$? $\endgroup$
    – john
    Jun 1, 2013 at 14:54
  • $\begingroup$ Yes. Even better: exactly equal, I believe. $\endgroup$
    – Jack Schmidt
    Jun 1, 2013 at 14:57
  • $\begingroup$ (If you do character theory: Modular character tables are square, and the Galois group acts nicely on both its rows and columns.) $\endgroup$
    – Jack Schmidt
    Jun 1, 2013 at 14:59

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