# If $N \trianglelefteq G$ and $G$ is $\frac{3}{2}$-fold transitive, then $N$ is transitive or semiregular.

If $N \trianglelefteq G$ and $G$ is $\frac{3}{2}$-fold transitive, then $N$ is transitive or semiregular.

I proved that if $N$ is intransitive, then $G$ will be imprimitive and Frobenius but I don't know how to prove the semiregulairty of $N$. Any help is appreciated.

I am going to assume that $G$ is finite, although have not said so!
So you want to prove that a normal subgroup $N$ of a Frobenius group $G$ (acting on a set $X$) is transitive or semiregular.
So suppose $N$ is not semiregular, and hence $N_\alpha \ne 1$ for some $\alpha \in X$. Since $G$ is transitive, for all $\beta \in X$, there exists $g \in G$ such that $g(\alpha)=\beta$, and $gN_\alpha g^{-1} = N_\beta$, so the stabilizers $N_\alpha$ are all conjugate in $G$ and hence all have the same order.
Let $p$ be a prime dividing $|N_\alpha|$ and, for each $\alpha \in X$, choose $P_\alpha \in {\rm Syl}_p(N_\alpha)$. Since $p|(|X|-1)$, $p$ does not divide $|X|$, and hence each $P_\alpha \in {\rm Syl}_p(N)$. Now for $\alpha,\beta \in X$, $P_\alpha$ and $P_\beta$ are conjugate in $N$, so there exists $g \in N$ with $gP_\alpha g^{-1} = P_\beta$. But since $\alpha$ is the unique fixed point of $P_\alpha$ (and similarly for $P_\beta$), we must have $g(\alpha)=\beta$, so $N$ is transitive.
• Why $p \,\big|\, (|X|-1)$? – user Dec 13 '14 at 14:51
• We are assuming that $G$ is a Frobenius group, so $P_\alpha$ is acting semiregularly on $X \setminus \{ \alpha \}$. So its orbits all have the same length, which is a power of $p$ and divides $|X|-1$. – Derek Holt Dec 13 '14 at 15:51