(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?


If $|A| = p^n$, then a semidirect product $A \rtimes Q$ with $Q \not\le C_G(A)$ exists if and only if $q$ divides $p^m-1$ for some $m \le n$.

It is certainly not always true that there exists $r$ such that $a^t = a^r$ for all $a \in A$. For example that is false in $A_4$, which is a semidirect product of groups of order $2^2$ and 3.

Were you supposed to assume that perhaps? If you assume that to be true in (3), and $|A_1|=|A_2|$, $|Q_1|=|Q_2|$ with $r_1,r_2 \not\equiv 1$ mod $p$, then it is indeed correct that $A_1 \rtimes Q_1 \cong A_2 \rtimes Q_2$, because you can define an isomorphism by mapping $t_1$ to a suitable power of $t_2$. But it would not be true without the assumption that $Q$ acts on $A$ with a universal power automorphism in both groups.

  • $\begingroup$ Thanks. I already noticed, that there could be other possibilities to get an action from $Q$ on A. I missunderstood the definition of P(n,p) groups, so I thougt that there has to be a univeral powerautomorphism. Thank you very much for your answer! $\endgroup$
    – Peter
    Aug 31 '12 at 8:22
  • $\begingroup$ Where can i find a proof, for the your first statement, that if $|A|=p^n$ the nontrivial semdidirect product exists if and only if $q|p^m-1$ for some $m\leq n$? $\endgroup$
    – Peter
    Aug 31 '12 at 8:27
  • $\begingroup$ And when do I get such a universal power automorphism? Which conditions on $A$ and $Q$ have to be satisfied? $\endgroup$
    – Peter
    Aug 31 '12 at 8:39
  • 1
    $\begingroup$ @Peter if $A$ is elementary abelian of order $p^n$, its automorphism group is isomorphic to $GL_n(\mathbb{F}_p)$. Its order is $(p^n-1)(p^n-p)(p^n-p^2)\cdots (p^n-p^{n-1})$. If $Q$ is to act via a non-trivial automorphism on $A$, $q$ must divide the order of the automorphism group. This proves Derek's first sentence (and more or less explicitly constructs the semi-direct product, given a suitable $q$). $\endgroup$
    – Alex B.
    Aug 31 '12 at 12:56

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