Let $W$ be a quasi-primitive faithful and irreducible $H$-module with $H$ and $W$ of odd order. Suppose that the Fitting subgroup $F(H)$ is cyclic (so $H \le \Gamma(W)$ the semilinear group). Then each element of $W \setminus \{0\}$ generates a regular orbit under the action of $F(H)$.

I'm struggling with this, any idea? Of course $H$ must be solvable by Feit-Thompson Theorem. I don't figure out how to prove a subgroup of prime order in $F(H)$ that centralizes one element $x \in W\setminus \{0\}$ must centrlize every other non zero element, if this is the way to preceed.

Edit: quasi-primitive means that whenever $N$ is a normal subgroup of $H$, the restriction $W_N$ is homogeneous.

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    $\begingroup$ Doesn't that follow immediately from $W_{F(H)}$ being homogeneous? $\endgroup$ – Derek Holt Jun 18 '18 at 20:09
  • $\begingroup$ Thank you Derek, I found my own way while dining. $\endgroup$ – Lorban Jun 18 '18 at 20:54

Here's my solution. Call $\mathbb{F}$ the base field of $V$. The restriction $V_{F(G)}$ is homogeneus and we can write $V_{F(G)}=fU$, so $U$ a faithful irreducible $F(H)$-module. But $F(H)$ is cyclic and so $\dim_{\mathbb{F}}(U)=1$, this means that $U \lesssim \mathbb{F}^{\times}$ and $F(G)$ acts as scalar multiplication on each irreducible constituent. So, if $v \in \mathbb{F}$ and $F(G)=<g>$ then $x^g=xc$ for a $c \in \mathbb{F}$. Being $V_{F(G)}$ homogeoneus $c$ is constant over all $F(G)$-irreducible constituent and then $F(G)$ acts as scalar multiplication on $V$ by a constant $c$ that has order $|F(G)|$ in $\mathbb{F}$. From this follows that every non zero element for a regular $F(G)$-orbit.

Once did it, I realize that this problem maybe isn't worth to be published to SE. However I cannot find a quicker way to prove it, as suggested by Derek Holt in a comment.

  • $\begingroup$ I just figured out that a cyclic group acting irreducibly on a finite vector space doesn't make it one-dimentional. Some semisempleness is required. Anyway it is well known a a nilpotent group acting irreducibility on a finite vector space must have coprime order with the characteristic. So we may fix choosing $\mathbb{K}$ a splitting field for $F(H)$ and apply mashke's theorem to $U\otimes \mathbb{K}$ and then go back to the previous base field someway. If I have time, I'll fix finely. $\endgroup$ – Lorban Jun 24 '18 at 17:25
  • $\begingroup$ *acting irreducibly and faithfully $\endgroup$ – Lorban Jun 24 '18 at 17:36

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