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Let $G$ be a Frobenius group with $H$ Frobenius complement and $K$ Frobenius kernel.

I read that $|H|$ divides $|K| - 1$, but I don't know why this holds, has anyone an idea?

Daphné Janssens
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    Let $H$ act on $K$ (conjugation), $K$ will decompose as $H$-orbits, $1$ of size $1$, the others will have size $|H|$. So the result follows. (I used the fact that $hkh^{-1} \not = k$ for $ h \not = 1$). – Verbe Dec 29 '17 at 21:08

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I guess my answer is equivalent to the one above, but I think about this stuff from primarily a combinatorial rather than algebraic viewpoint. So: $K$ can be written as a permutation group on $|K|$ points. The Frobenius condition transfers to the statement that the Frobenius complement, written on these same points, fixes exactly one of them, and is therefore a fixed-point free permutation group on $|K| - 1$ points, thus $|H|||K| - 1$.

luxerhia
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Ken Lebensold
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  • You are right that it is equivalent to the above comment. The action of $H$ on $K$ by conjugation is the same as the action of $H$ on the $|K|$ points you are discussing. – Ben Blum-Smith Apr 20 '20 at 16:08