Semidirect product $\mathbb{F}_q \rtimes \mathbb{F}_q'$

Let $\mathbb{F}_q$ be an additive group of finite field and $\mathbb{F}_q' \simeq \mathbb{Z}_{q-1}$ be a multiplicative group of finite field.

I want to build a semidirect product $\mathbb{F}_q \rtimes \mathbb{F}_q'$. Every semidirect product is determined by the choise of homomoprhism $\mathbb{F}_q' \rightarrow Aut(\mathbb{F}_q)$. Such homomorphism is determined by its value on a generator of $\mathbb{F}_q'$.

We know that $\mathbb{F}_q \simeq \mathbb{Z}_{p}^d$, where $q = p^d$. Note that $Aut(\mathbb{Z}_{p}^d) \simeq GL_d(\mathbb{Z}_{p})$. Every image of a generator of $\mathbb{F}_q'$ has order, dividing $q-1$.

Thus a homomorphism $\mathbb{F}_q' \rightarrow Aut(\mathbb{F}_q)$ is determined by a choice of a matrix $A \in GL_d(\mathbb{Z}_{p})$ such that $A^{q-1} = E$. But I don't know how to get an explicit function law of such homomorphism.

• What do you mean by a function law? – Tobias Kildetoft Oct 31 '16 at 9:49
• It's indefinitely notion. For example, $f(x) = x^2$. – user272816 Oct 31 '16 at 9:55
• There are several semidirect products here (corresponding to choice of $A$). IMHO a most natural semidirect product is the one where an element $z\in \Bbb{F}_q'$ acts via the (additive or, actually $\Bbb{F}_p$-linear) mapping $z:x\mapsto zx$ for all $x\in\Bbb{F}_q$. You can equally well use mappings like $x\mapsto z^kx$ for some exponent $k\in\Bbb{Z}$. But, to find the corresponding $d\times d$ matrices $A$, you need to specify a basis of $\Bbb{F}_q$ over $\Bbb{F}_p$. – Jyrki Lahtonen Oct 31 '16 at 9:57
• I would much rather represent elements of this ($k=1$) semidirect product as affine transformations, and use an augmented matrix. Here we only need to augment a $1\times1$ matrix to a $2\times2$ matrix. Dietrich seems to have done exactly that. – Jyrki Lahtonen Oct 31 '16 at 9:57
• You are looking for a description of the conjugacy classes in ${\rm GL}(d,p)$ of cyclic subgroups whose order divides $q-1$. They are generated by semisimple elements, so some kind of description is possible, but it would be complicated. – Derek Holt Oct 31 '16 at 11:33

1 Answer

This is, in a special case, the finite affine group ${\rm Aff}(q)$, which has "function law" matrix multiplication, see here for the general case. So we have the group $${\rm Aff}(q)=\{\begin{pmatrix} x & y \cr 0 & 1 \end{pmatrix}\mid x\in \mathbb{F}_q^{\times}, y\in \mathbb{F}_q\},$$ of order $q(q-1)$, where the homomorphism is given by $\gamma\colon \mathbb{F}_q^{\times} \rightarrow Aut(\mathbb{F}_q)$, $\gamma_s(v)=sv$.

Edit: In general there will be other possibilities, but we still can represent the semidirect product as affine transformations.

• Is it clear that there are no other possible choices of homomorphism to the group of matrices? – Tobias Kildetoft Oct 31 '16 at 9:53
• Is it true that all semidirect products are isomorphic to affine group over finite field? – user272816 Oct 31 '16 at 9:58
• No, in general there will be other possibilities, including the direct product. But I don't really understand what the question is asking! – Derek Holt Oct 31 '16 at 9:59
• I want to get all possible groups, including the direct product. – user272816 Oct 31 '16 at 10:05
• Then follow Jyrki's answer in the comments. See also your previous questions here and here. – Dietrich Burde Oct 31 '16 at 10:14