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I would like to understand the nonabelian groups of order $pq$ (with $q | p-1$) better. For $q=2$ this is the dihedral group which I am comfortable with.

For each $pq$ I know that there is exactly one of these groups. It is a semidirect product. Its Sylow structure is $n_q = p$ and $n_p = 1$. I don't know much about them.

I calculated the following interesting group orders 21, 39, 55, 57, 93. And I will ask about 21.

What is the nonabelian group of order 21 the symmetry of?

I have been researching this and not found a good answer. I don't think it is the symmetry of rotations of a polyhedra or any twisting puzzle. I have seen that the fano plane has 7 lines and 3 points on each line but I don't know if it can be used. Are these groups acting naturally on a code of design of some type? Or is there a better way to understand them at a deeper level? thanks!

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    There is a theorem that says that every finite group is the automorphism group of a graph. – Gerry Myerson Aug 18 '20 at 06:30
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    https://mathoverflow.net/questions/993/is-every-finite-group-a-group-of-symmetries may interest you. And also https://math.stackexchange.com/questions/3254492/can-all-groups-be-thought-of-as-the-symmetries-of-a-geometrical-object – Gerry Myerson Aug 18 '20 at 06:31

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Over every field $F$ there's a group of affine transformations

$$x \mapsto ax + b, a \in F^{\times}, b \in F$$

acting on the affine line $\mathbb{A}^1(F)$ (which as a set is just $F$). Equivalently this is a group of $2 \times 2$ matrices

$$\left[ \begin{array}{cc} a & b \\ 0 & a \end{array} \right].$$

Over a finite field $F = \mathbb{F}_q$ we get a family of nonabelian (except when $q = 2$) groups of order $q(q - 1)$ which are semidirect products constructed from the action of $\mathbb{F}_q^{\times}$ on $\mathbb{F}_q$ by multiplication. Furthermore we can consider subgroups of this group by restricting $a$ to a subgroup of $F^{\times}$. All of the groups you're interested in can be constructed this way.

The specific group you're interested in occurs when $q = 7$ and $a$ is restricted to lie in the subgroup $(\mathbb{F}_7^{\times})^2$ of square elements of $\mathbb{F}_7^{\times}$. It's a Frobenius group and according to that page it also acts on the Fano plane.

Qiaochu Yuan
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    The action on the Fano plane for $q=7$ is a sort of small numbers coincidence, isn't it? In this case $\text{PSL}_2(\mathbb{F}_7)$ just happens to be isomorphic to $\text{GL}_3(\mathbb{F}_2)$. – ancient mathematician Aug 18 '20 at 08:21