Let $G$ be a finite simple group and $\chi$ be an irreducible character of $G$ of degree $p$ (prime). Prove that any $p$-sylow subgroup of $G$ is of size $p$.

I have no idea how to solve the problem. There is also a small hint to the question.

Hint: if $P$ is a non abelian Sylow-$p$ subgroup then $Z(P) \subseteq Z(\chi)$.

$Z(\chi)=\{\,g\in G : |\chi(g)|=|\chi(1)|\,\}$

I didn't get how to use the hint to my advantage.

I started with the argument that $G$ cannot be a $p$-group since a simple $p$-group has to be abelian in which case it will be abelian of prime order and hence all irreducible characters are of degree $1$. Hence $|G|=p^a. m$ (with $m>1$). But I don't get how to proceed further.


migrated from mathoverflow.net Apr 2 '17 at 22:06

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  • $\begingroup$ Is this from Isaacs's book? $\endgroup$ – Yemon Choi Apr 2 '17 at 22:03
  • $\begingroup$ Yeah that's right!! $\endgroup$ – Riju Apr 2 '17 at 22:05
  • 2
    $\begingroup$ Hint to start you off: Set $K = \rm{ker} \chi.$ Then from the book's hint we see ( if $P$ is non-Abelian) that for any non-identity $x \in Z(P)$, we have $xK \in Z(G/K)$. But $G$ is simple, so $K = 1$. Hence $x \in Z(G),$ a contradiction as $G$ is non-Abelian simple. Thus $P$ is Abelian. Now you should have enough information in the book to know that $\chi(x) = 0$ whenever $1 \neq x \in P.$ $\endgroup$ – Geoff Robinson Apr 2 '17 at 23:59
  • $\begingroup$ Another hint: Theorem 3.8 in Isaacs's book is relevant (also for proving the hint). $\endgroup$ – ladisch Apr 6 '17 at 9:03
  • $\begingroup$ @Geoff Robinson I have proved that $\chi(x)$=0 for all x$\in$P and x$\neq$1. But then I don't get how is it giving the result? $\endgroup$ – Riju Apr 6 '17 at 19:23

Since $G$ is simple, we must have $Z(\chi)=1$, as $Z(\chi)$ is always a normal subgroup. So by Theorem 3.8 in Isaacs's book, $\chi$ vanishes on $Z(P)\setminus \{1\}$, and $\chi(1)=p$ by assumption. It follows that $Z(P)$ has order $p$. Let $1\neq x\in Z(P)$, and $D\colon G\to \operatorname{GL}(p,\mathbb{C})$ a representation affording $\chi$. Then $D(x)$ is a matrix of order $p$ and has trace $\chi(x)=0$. It follows that the eigenvalues of $D(x)$ are exactly the different $p$-th roots of unity (including $1$), each with multiplicity $1$. Then the centralizer of $D(x)$ in $\operatorname{GL}(p,\mathbb{C})$ is abelian, and thus $P$ is abelian. Thus $P=Z(P)$ has order $p$.

  • $\begingroup$ how did you deduce that Z(P) has order p if $\chi(x)$=0 for all x$\in$Z(P) and x $\neq$ 1and $\chi(1)$=p $\endgroup$ – Riju Apr 6 '17 at 22:42
  • 2
    $\begingroup$ @Riju: the restriction $\chi_Z$ of $\chi$ to $Z=Z(P)$ is a character. Thus the inner product $[\chi_Z,1_Z] = \chi(1)/|Z|$ is an integer. (Indeed, we see that $\chi_Z$ must be the regular character of $Z$.) $\endgroup$ – ladisch Apr 6 '17 at 22:47

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