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We know that if $G$ be a finite group and $F$ be an algebraically closed field whose characteristic does not divide the order of $G$, then the number of inequivalent irreducible $F$-representations of $G$ equals the class number of $G$.

Now, if we suppose that the field $F$ is not necessarily an algebraically closed field but its characteristic does not divide the order of $G$, is it true that the number of inequivalent irreducible $F$-representations of $G$ cannot exceed the class number of $G$?

If the answer is yes, how can we prove that?

I know that each irreducible representation of $G$ over $F$ is completely reducible by Maschke's Theorem because the characteristic of $F$ does not divide the order of $G$ and $G$ is finite. Also, each representation of $G$ over $F$ can be considered as a representation of $G$ over $\overline{F}$, where $\overline{F}$ is an algebraic closure of $F$. But I have no idea about how they can help.

I will be so grateful for any answers and comments.

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2 Answers 2

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Yes, the number of irreducible representations cannot exceed the class number (in the case of caprice [autocorrect error. Should be ``coprime''] characteristic): Consider the case of an $F$-irreducible module $M$ that becomes reducible over $\bar F$. Then $M$ has an $\bar F$-submodule $N$, and $M$ is simply the direct sum of Galois-conjugates of $N$ (the Galois-sum is an $F$-invariant submodule). That shows that the number of $F$-irreducible representations never exceeds the number of $\bar F$-irreducible representations.

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    $\begingroup$ "caprice" is a cute autocorrect for "coprime"? $\endgroup$
    – Jeremy Rickard
    Nov 16, 2015 at 18:25
  • $\begingroup$ Nice answer! Thank you so much. Just to ensure more, do you mean that if $M_1$ and $M_2$ are two inequivalent irreducible $FG$-modules then $M_1=N_1\bigoplus L_1$ and $M_2=N_2\bigoplus L_2$ as $\overline{F}G$-modules, which $N_1$ is not isomorphic to $N_2$ as an $\overline{F}G$-module because $M_1$ is not isomorphic to $M_2$ as an $FG$-module? Would you please explain this part a little more? $\endgroup$
    – A-213
    Nov 16, 2015 at 20:29
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    $\begingroup$ @JeremyRickard Yes, indeed. I should fix it, but its just sooooo cute... $\endgroup$
    – ahulpke
    Nov 17, 2015 at 1:39
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    $\begingroup$ @A-213 In your example, if $N_1$ was isomorphic to $N_2$, you could pair up all Galois images similarly and thus construct an isomorphism between the Galois closures which must be $M_1$ and $M_2$ (as they are simple modules). $\endgroup$
    – ahulpke
    Nov 17, 2015 at 3:07
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    $\begingroup$ @A-213 The only references I'm aware of would be general texts about Galois theory, alas I'm not aware of a text specific to the applications in representation theory. $\endgroup$
    – ahulpke
    Nov 18, 2015 at 16:29
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By Maschke's theorem and the Artin-Wedderburn theorem, the group algebra $k[G]$ decomposes as a finite product $\prod_i M_{n_i}(D_i)$ of matrix algebras over finite-dimensional division algebras $D_i$ over $k$. Here the product runs over all simple modules of $k[G]$.

The center of this product is $\prod_i Z(D_i)$, and hence the number of simple modules is at most the dimension of the center. The center of $k[G]$ has a natural basis given by sums over the conjugacy classes of $G$, and in particular it always has dimension the number of conjugacy classes of $G$.

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  • $\begingroup$ Thank you so much. I think that your answer is for the case that $k$ is algebraically closed. Am I right? In my question, the characteristic of the field does not divide the order of $G$ and it is not necessarily algebraically closed. $\endgroup$
    – A-213
    Nov 16, 2015 at 23:58
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    $\begingroup$ @A-213: no, I don't assume this (this is why I need the division algebras $D_i$). What makes you say that? $\endgroup$
    – Qiaochu Yuan
    Nov 16, 2015 at 23:58
  • $\begingroup$ I got the point. Thank you so much. $\endgroup$
    – A-213
    Nov 17, 2015 at 0:03

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