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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
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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
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3 Answers

up vote 9 down vote accepted

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 $p^n$.

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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
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@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
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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 $p-1.$

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As already explained by Geoff, the number of conjugacy classes of size $>1$ is divisible by $p-1$. A reference is:

A. Mann: Groups with few class sizes and the centralizer equality subgroup. Israel J. Math. 142(2004), 367-380.

The result (with a proof quite similar to Geoff's) is on page 376 (after Corollary 18).

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Thanks for the reference. –  Amin Mar 19 '13 at 12:23
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