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It is known that the intersection of kernels of all ireducible representations of a finite group in characteristic zero is the trivial group.

In characteristic $p>0$ I have understood that this intersection îs $O_p(G)$, the largest normal $p$ - subgroup of $G$. In other words, all irreducible representations of G are coming from the quotient $G/O_p(G)$.

I have some difficulties in proving this similar result in characteristic $p$ dividing the order of $G$. Could someone provide me with a reference for this result? Or shortly explain the argument to me. My intuition is that this is not so difficult.

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  • $\begingroup$ Over an algebraically closed field, this is a consequence of orthogonality relations of Brauer which involve the Brauer irreducible characters and the Brauer characters of the projective indecomposable modules. When the field is nnot algebraically closed, there are various alternative approaches. Back to the algebraically closed case, the crux of the matter is that no non-identity element $y$ of order prime to $p$ can be in the kernel of every irreducible characeristic $p$ representation, so by Cauchy's Theorem, the intersection of those kernels is a $p$-group ( which contains $O_{p}(G)).$ $\endgroup$
    – Geoff Robinson
    Commented Jan 12, 2017 at 0:06
  • $\begingroup$ See mathoverflow.net/questions/69039/… $\endgroup$
    – Benjamin Steinberg
    Commented Jan 12, 2017 at 1:53
  • $\begingroup$ The question here is different from the linked question. The linked question asked whether the intersection of the kernels of the simple module contained $O_{p}(G).$ As I read it, this question asks to prove that the intersection of those kernels is no larger than $O_{p}(G)$ ( which is true, as my comment above indicates). $\endgroup$
    – Geoff Robinson
    Commented Jan 12, 2017 at 5:05
  • $\begingroup$ @GeoffRobinson, my answer proves the equality. Probably yours as well. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 12, 2017 at 11:47
  • $\begingroup$ If g is in the kernel of all irrrps, g-1 is in the radical so taking a large p-power shows g is a p-power order element. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 12, 2017 at 11:50

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