Can there be semidirect products $(\mathbb{Z}_p \times \mathbb{Z}_p) \rtimes \mathbb{Z}_q$ having $p <q$?

I've seen this group for $p>q$ values but not for $p<q$ values, therefore can someone please help me with this question?

$p, q$ are distinct primes.

Thanks a lot in advance.


Yes, but only when $p=2$, $q=3$. I assume you mean the cyclic group of order $p$ for $\Bbb{Z}_p$ based on your tags. I will denote this by $\newcommand\FF{\Bbb{F}}\FF_p$, (meaning the finite field with $p$ elements), or $Z_p$ depending on whether or not the field structure is relevant.

The automorphism group of $\FF_p^2$ is $\newcommand\GL{\operatorname{GL}}\GL_2(\FF_p)$, which has order $(p^2-1)(p^2-p)=p(p-1)^2(p+1)$. Thus if $q> p$ is a prime, we must have $q=p+1$. The only prime $p$ for which $p+1$ is prime as well is $p=2$, and indeed in this case, we have the automorphism of $\FF_2^2$ given by $$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix},$$ which has order 3. Thus we can form a nontrivial semidirect product $(Z_2^2)\rtimes Z_3$


In general, the automorphism group of $\FF_p^n$ is $\GL_n(\FF_p)$, which has order $$\prod_{i=0}^{n-1}(p^n-p^i)=p^{n(n-1)/2}(p-1)^{n-1}\prod_{i=0}^{n-1}\frac{p^{n-i}-1}{p-1}.$$ In general $\frac{p^{n-i}-1}{p-1}=1+p+\cdots + p^{n-i-1}$ will be divisible by primes larger than $p$, and a semidirect product $Z_p^n\rtimes Z_q$ with $q > p$ will exist if and only if $q\mid \frac{p^{n-i}-1}{p-1}$ for some $0\le i < n$.

For example, if $p=5$, $n=3$, then $(5^3-1)/(5-1)=1+5+25=31$, which is prime, so we have the semidirect product $Z_5^3\rtimes Z_{31}$.

  • $\begingroup$ Thanks @jgon yes I meant cyclic group of order p. What will be the possibility of existence of a semidirect product like $\mathbb{Z}_p^{n} \rtimes \mathbb{Z}_q$ when $p<q$ ? Can we think of it also in a similar manner? $\endgroup$ – Buddhini Angelika Dec 19 '18 at 15:53
  • $\begingroup$ @BuddhiniAngelika Edited to address your comment. $\endgroup$ – jgon Dec 19 '18 at 16:47
  • $\begingroup$ Thanks a lot @jgon $\endgroup$ – Buddhini Angelika Dec 19 '18 at 18:40

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