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I am trying to understand the first paragraph (Section 9.1) of Chapter 9 in Serre's Linear Representations of Finite Groups.

Let me reproduce it here.

Let $G$ be a finite group and let $\chi_1, \ldots, \chi_h$ be its distinct irreducible characters. A class function on $G$ is a character if and only if it is a linear combination of the $\chi_i$'s with non-negative integer coefficients. We will denote by $R^+ (G)$ the set of these functions, and by $R (G)$ the group generated by $R^+ (G)$, i.e., the set of differences of two characters. We have $$ R (G) = \mathbb{Z} \chi_1 \oplus \ldots \oplus \mathbb{Z} \chi_h $$.

My questions:

  • What is $h$? Is it the number of characters of all irreducible representations of all elements in the group? For example, is $h = 5! \times 7$ for $G = S_5$?
  • If $G = S_n$, can $$ R (G) = \mathbb{Z} \chi_1 \oplus \ldots \oplus \mathbb{Z} \chi_h $$ be written as $$ R (G) = \mathbb{Z} \chi_1 + \ldots + \mathbb{Z} \chi_h $$?

Thanks in advance!

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    $\begingroup$ I am not sure what the confusion with $h$ is. It is the number of irreducible characters. I don't see why any elements should be involved. And the answer to the second question is yes, though that would be a weaker statement. $\endgroup$ – Tobias Kildetoft May 21 '17 at 17:08
  • $\begingroup$ @TobiasKildetoft, I might be asking really obvious questions. Is the number of irreducible characters equal to the number of irreducible representations? $\endgroup$ – Omar Shehab May 21 '17 at 17:29
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    $\begingroup$ Indeed, yes. And while it is fine to ask obvious questions, this indicates that you should go back and read the preceding 8 chapters before this one. $\endgroup$ – Tobias Kildetoft May 21 '17 at 17:37
  • $\begingroup$ @TobiasKildetoft, could you please elaborate why replacing $\oplus$ with $+$ would be weaker? $\endgroup$ – Omar Shehab May 29 '17 at 15:39
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In general, if $G$ is a finite group, the number $h$ of irreducible characters of $G$ is equal to the number of conjugacy classes of $G$. This essentially follows from the fact that the dimension of the space of class functions on $G$ is easily seen to be equal to the number of conjugacy classes, combined with the fact that the irreducible characters $\chi_1, \dots, \chi_h$ form an orthonormal basis for this space (you can find these facts in section 2.5 of Serre's book for example).

Also, to answer your second question, it is known that the conjugacy classes of the symmetric group $\mathbb{S}_n$ are in one to one correspondence with the partitions of $n$. Hence by the above fact, the number of irreducible characters of $\mathbb{S}_n$ is equal to $p(n)$, where $p(n)$ is the partition function.

Thus, since $p(5) = 7$, then $h = 7$ for $G = \mathbb{S}_5$.

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