(1) If $A$ is a elementary abelian p-group. And $Q=\langle t\rangle$ is a group of order q ($q\neq p$ prime numbers). For which primes p,q does the semidirect product $A\rtimes Q$ exist (so $Q\not\leq C_G(A)$)?
If this semidirect product exists, then Q acts with an universal power automorphism on A. That means:
There exists an integer $r\in\mathbb{Z}$, such that $r\not\equiv 1\mod\ p$ (since $Q\not\leq C_G(A))$ and $r^q\equiv 1\mod\ p$.
(2) If r exists these two properties are easy to check. But why does this r exists? Why is there no other possibility how $Q$ can act on $A$?
(3) And if we got two different semidirect products $A_1\rtimes Q_1$ and $A_2\rtimes Q_2$ where $A_1\cong A_2\cong A$ and $\langle t_1\rangle=Q_1\cong Q\cong Q_2=\langle t_2\rangle$. So wiht (2) we get two differnt universial power automorphisms, s.t. f.a. $a\in A$ holds $a^{t_1}=a^{r_1}$ and $a^{t_2}=a^{r_2}$ with $r_1\neq r_2$. Why does always hold: $A_1\rtimes Q_1\cong A_2\rtimes Q_2$?
Thanks for help.
Ah. I checked it. "$\Leftarrow$" is only an application of Sylows theorem. For "$\Rightarrow$" we have to find a nontrivial semidirect product when $q|p^m-1$. But how can i construct this semidirect product?
If we want to get a power automorphism. Then $Q=\langle t\rangle$ has to act on every $\langle a_i\rangle$ if $A=Dr_i^{n}\langle a_i\rangle$. This is an universal power automorphism if it acts on every $\langle a_i\rangle$ in the same way, right? So we only have to look at one of these $\langle a_i\rangle$. And we get an action of $Q$ on $\langle a_i\rangle$, if and only if there exists $r\in\mathbb{Z}$ wich satisfies the above conditions, right?