Let $G$ finite group and $\sigma \in \text{Aut}(G)$, suppose that at most two prime numbers divide $o(\sigma)$. Show that $\left \langle \sigma \right \rangle$ has a regular orbit on $G$.
Suppose $o(\sigma)=p^\alpha q^\beta$ with $\alpha, \beta$ non-negative numbers, $\alpha+\beta>0$. Decompose (as a partition) $G$ in orbits under the action of $\left \langle \sigma \right \rangle$: \begin{gather} G=x_1^{\left \langle \sigma \right \rangle}\cup \dots \cup x_n^{\left \langle \sigma \right \rangle} \end{gather} Write $\lambda_i=x_i^{\left \langle \sigma \right \rangle}$ and $m=\mathrm{lcm}\{o(\lambda_1), \dots ,o(\lambda_n)\}$ and suppose $m<p^\alpha q^\beta$.Without loss of generality we can assume $m\le p^{\alpha -1} q^\beta$. Looking at each orbit, $\left \langle \sigma \right \rangle$ acts with a non trivial kernel that contains a subgroup of order $p$; being $\left \langle \sigma \right \rangle$ abelian, there is a unique subgroup of order $p$, i.e. $\left \langle \sigma^{p^{\alpha -1} q^\beta} \right \rangle$. This means that this subgroup of order $p$ acts trivially on $G$ and this is impossible since $\left \langle \sigma \right \rangle \le Aut(G)$. Then $m=p^\alpha q^\beta$ and there are two orbits $\lambda_i, \lambda_j$ such that $p^\alpha$ divedes $o(\lambda_i)$ and $q^\beta$ divides $o(\lambda_j)$. By direct check $\left \langle \sigma \right \rangle$ acts faithfully on $\Lambda=\lambda_i \cup \lambda_j$, and since it is abelian (cyclic) the action is regular.
I think that this proof works (correct if I'm wrong). The only problem is that this exercise is presented after the section of semidirect products but this notion is not used. Maybe it can be done studying $\Omega_\pi(G \rtimes \left \langle \sigma \right \rangle)$ for $\pi=\{p,q\}$, but I can't find a proof in this way. Does someone know a proof using the semidirect products?
By direct check $\langle\sigma\rangle$ acts faithfully on $\Lambda=\lambda_i\cup\lambda_j$, and since it is abelian (cyclic) the action is regular.
This line is incorrect. A cyclic group can act faithfully on a set without acting regularly. If the action is not transitive, it cannot be regular, and may not even be semiregular.
Here is a way to solve the problem. Let $\Sigma=\langle \sigma\rangle$. Assume that $o(\sigma)=p^{\alpha}q^{\beta}$ with $\alpha\geq \beta$ and $\alpha>0$. If $\beta = 0$, that is if $o(\sigma)=p^{\alpha}$, then let $\tau=\rho=\sigma^{p^{\alpha-1}}$. If $\beta>0$, then let $\tau = \sigma^{p^{\alpha-1}q^{\beta}}$ and $\rho = \sigma^{p^{\alpha}q^{\beta-1}}$. In either case the elements $\tau$ and $\rho$ generate the only minimal subgroups of $\Sigma$.
Since both $\tau$ and $\rho$ are nonidentity automorphisms of $G$, the subgroups $C_G(\tau), C_G(\rho)$ are proper subgroups of $G$. $G$ cannot be the union of any two proper subgroups, so there is an element $g\in G-(C_G(\tau)\cup C_G(\rho))$. I claim that the orbit $\Sigma g$ is regular.
By definition, $\Sigma$ acts transitively on $\Sigma g$. To see that $\Sigma$ acts regularly on $\Sigma g$, it suffices (since $\Sigma$ is abelian) to show that $C_{\Sigma}(g) = \{1\}$. This equality holds because, by the choice of $g$, we have $\tau, \rho\notin C_{\Sigma}(g)$. Since any nonidentity subgroup of $\Sigma$ contains either $\tau$ or $\rho$, this forces $C_{\Sigma}(g)=\{1\}$. \\\