Order of a product of two elements in a Frobenius Group

Question

Let $G = K \rtimes H$ be a Frobenius group with Kernel $K$ and complement $H$. I would like to show that $$ o(xy) = o(y), \quad \mbox{for every} \ x \in K \ \mbox{and for every} \ y \in H \setminus \{1\} $$

Since $K \trianglelefteq G$ and $K \cap H =1$, I already proved that $o(y) \mid o(xy)$, but I'm with difficulty to show that $o(xy) \mid o(y)$. Any help is very appreciated it.

score 3 · Accepted Answer · answered Dec 16 '24 at 09:01

3

By a straightforward counting argument, a Frobenius group of degree $n$ has exactly $n-1$ fixed-point-free elements, so these must be the non-trivial elements of $K$.

So all elements of $G$ outside of $K$ fix a point, and so lie in a conjugate of $H$. This applies to the element $xy$, so its order divides the order of $H$.

answered Dec 16 '24 at 09:01

Derek Holt

96,726

But, if $o(xy)$ divides $|H|$, not necessarily, this number divides $o(y)$. How can I conclude? – Rodolfo Ferreira de Oliveira Dec 16 '24 at 10:20
1

$O(xy)$ divides $|H|$ and $(xy)^{o(y)} \in K$, but $|H|$ and $|K|$ are coprime, so $(xy)^{o(y)}=1$. – Derek Holt Dec 16 '24 at 13:42
thank you so much – Rodolfo Ferreira de Oliveira Dec 17 '24 at 01:24

Order of a product of two elements in a Frobenius Group

1 Answers1