# Question regarding possiblity for existence of a particular semidirect product

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

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

$$p, q$$ are distinct primes.

Thanks a lot in advance.

## 1 Answer

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 $$(Z_2^2)\rtimes Z_3$$

Edit:

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 $$Z_5^3\rtimes Z_{31}$$.

• 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? – Buddhini Angelika Dec 19 '18 at 15:53
• @BuddhiniAngelika Edited to address your comment. – jgon Dec 19 '18 at 16:47
• Thanks a lot @jgon – Buddhini Angelika Dec 19 '18 at 18:40