0
$\begingroup$

Let $G$ be a finite group and $N\unlhd G$. For $A\subseteq G$, let $k_G(A)$ be the number of conjugacy classes of G which intersects A non-trivially. Let $\frac{G}{N}$ be a Frobenius group with kernel $\frac{K}{N}$ which is an elementary abelian $p$-group and cyclic complement isomorphic to $\frac{G}{K}$. Suppose that for any coset $xK$, we have a conjugacy class of $G$ and $\frac{G}{K}$ is isomorphic to $\mathbb{Z}_5$ such that $(|K|, 5)=1$ and $k_G(G-K)=5$. Can we conclude that $G$ is a Frobenius group with kernel $K$ and complement $\mathbb{Z}_5$? Is $K$ abelian?

We know that $G$ is the semiderect product of $K$ and $\mathbb{Z}_5$ by the Schur–Zassenhaus theorem.

  • 0
    I don't understand your statement "Suppose that for any coset $xK$ we have a conjugacy class of $G$". You haven't siad anything about that conjugacy class, so the statement does not make much sense. Also, if $G$ was a Frobenius group with kernel $K$, then you would have $k_G(G \setminus K) =4$, not $5$. – 2017-02-12
  • 0
    We mean that for every coset $xK$ of $\frac{G}{K}$, we have $xK=cl_G(x)$ (for $x\in G-K$). Now if we suppose that finite group $G$ has at most 5 conjugacy classes whose sizes are multiples of p, for every prime $p$ and $k_G(G-K)=5$, then what can we say about the structure of $G$? – 2017-02-13
  • 0
    There are only four cosets $xK$ with $x \in G \setminus K$. So if $xK = {\rm cl}_G(x)$ for each of these cosets, then $k_G(G \setminus K) =4$. – 2017-02-13

0 Answers 0