# Regular orbits on primitive module

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.

• Doesn't that follow immediately from $W_{F(H)}$ being homogeneous? – Derek Holt Jun 18 '18 at 20:09
• Thank you Derek, I found my own way while dining. – 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 $$F(G)=$$ 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.
• 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. – Lorban Jun 24 '18 at 17:25