# Characters of p-groups

Berkovich mentioned the following result of Mann in his book on p-groups:

The number of nonlinear irreducible characters of given degree in a p-group is divided by p-1.

Do you know any reference for this statement? Also, I would like to ask you if a similar assertion holds for conjugacy classes.

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What is a nonlinear character? – Qiaochu Yuan Mar 17 '13 at 23:51
Qiaochu: this is the group theorists' terminology for characters of degree greater than 1, i.e. which are not homomorphims into the complex numbers – Yemon Choi Mar 18 '13 at 6:15

Here is a more elementary argument than Geoff's. The number of linear characters of $P$ divides the order of $P$ and is hence a power $p^r$ of $p$. All the other characters have degrees $d_i$ of the form $p^{a_i}$. Hence $$p^n = p^r +\sum_i p^{2a_i}.$$ I claim that if $p^n$ is written as a sum of powers of $p$, then the number $N$ of summands satisfies $N\equiv 1$ (mod $p-1$), and the proof follows. One way to prove the claim is to consider the smallest summand $p^k$. Considering the sum mod $p^k$ shows that the number $N_k$ of summands equal to $p^k$ is a multiple of $p$. Hence we can replace them with $N_k/p$ summands equal to $p^{k+1}$ without affecting the number of summands mod $p-1$. Now continue this argument until reaching $p^n$.
It's easier than that: letting $N$ be the number of terms in that sum on the right, just look at this equation mod $p-1$. Since $p \equiv 1 \bmod p-1$, the equation becomes $1 \equiv 1 + N \bmod p-1$, so $N \equiv 0 \bmod p-1$. – KConrad Mar 18 '13 at 7:07
@Richard,@KConrad: And Richard's answer and/or KCs comment makes the situation with the conjugacy class question transparent too: using the modified class equation gives $1 \equiv 1 + (k(P) -|Z(P)|)$ (mod $p-1$), where $k(P)$ is the number of conjugacy classes of $P$, so the number of conjugacy classes of $p$ of size greater than $1$ is diisible by P-1$. Simpler than my approach! – Geoff Robinson Mar 18 '13 at 7:57 Thank you very much for your answers and comments. – Amin Mar 19 '13 at 12:22 In fact, if$M$is the number of nonlinear irreducible characters, then$M\equiv 0$(mod$p^2-1$) if$n\equiv r$(mod 2);$M\equiv p-1$(mod$p^2-1$) if$n$is odd and$r$is even; and$M\equiv -p+1$(mod$p^2-1$) if$n$is even and$r$is odd. – Richard Stanley Mar 19 '13 at 13:00 For every non-trivial irreducible character$\chi$of a finite$p$-group$P,$we may choose an element$z in P$such that$\chi(z) = \chi(1) \omega$for some primitive$p$-th root of unity$\omega$(this is an easy consequence of Schur's Lemma and the fact that the image of$P$in the associated representation has non-trivial center). Now let$\zeta$be a primitive$p^{e}$-th root of unity, where$P$has exponent$p^{e}.$Then${\rm Gal}(\mathbb{Q}[\zeta]/\mathbb{Q})$acts on the irreducible characters of$P,$and it is clear that$\chi$is in an orbit of length divisible by$p-1.$Since$\chi$was an arbitrary non-trivial irreducible character, and since all irreducible characters in the same orbit have the same degree it follows both that the total number of non-trivial irreducible characters, and the total number of irreducible non-linear characters of$P$are multiples of$p-1$(in fact, the number of irreducible characters of$p$of any fixed degree$p^{d} >1$is a multiple of$p-1.$And yes, a similar statement holds for conjugacy classes. If$C$is a non-trivial conjugacy class of$P$, say containing an element$x,$then the length of the conjugacy class only depends on$\langle x \rangle$. If$x$has order$p^{e},$then$\langle x \rangle$has$p^{e-1}(p-1)$generators, but if$y$is another generator, and$y^{p^{e-1}} \neq x^{p^{e-1}},$then$x$and$y$are not conjugate within$p.$Since$\langle x \rangle$contains$p-1$elements of order$p,$it follows that the number of conjugacy classses of non-trivial elements of$P$which have length$[P:C_{P}(x)]$is divisible by$p-1.$In particular, the number of conjugacy classes of length greater than$1$is divisible by$p-1.$- As already explained by Geoff, the number of conjugacy classes of size$>1$is divisible by$p-1\$. A reference is: