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I am trying to prove:

Suppose $p$ divides $|G|$ and let $\chi \in Irr(G)$ if $\chi$ is faithful and its degree is less than $p$ then any $p$-Sylow subgroup of $G$ is abelian.

I have tried to decompose the restriction of $\chi$ to $P$, a $p$-group, into irreducibles and I guess I should try to prove that this gives me all one dimensional constituents, but I have no idea how to proceed.

Any hints, links to good sources or other help is much appreciated.

Thanks.

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    $\begingroup$ It's restriction to a Sylow $p$-subgroup $P$ is faithful and must be a sum of linear characters, so $P$ is abelian. (Remember that the degree of an irreducible character of $P$ must divide $|P|$.) $\endgroup$ – Derek Holt Apr 8 '16 at 19:18
  • $\begingroup$ That's a really nice quick little argument. Much obliged! (I'll accept if you choose to repost as an answer.) $\endgroup$ – Winston Apr 8 '16 at 20:02
  • $\begingroup$ @DerekHolt Sounds like an answer, not a comment. $\endgroup$ – zibadawa timmy Apr 8 '16 at 20:02
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OK, I'll make it an answer. The restriction $\chi_P$ of $\chi$ to $P \in {\rm Sy}_p(G)$ decomposes into irreducible characters of $P$. Since the degree of any such character divides $|P|$, it must be power of $p$. But $\deg(\chi) < p$, so all of the irreducible constituents of $\chi_P$ have degree $1$. Since $\chi$ and hence $\chi_P$ is faithful, so $P$ must be abelian.

By the way, there is no need to assume that $\chi$ is irreducible, just that it is faithful.

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