Realization of the metacyclic group of order $21$

Question

I would like to understand the nonabelian groups of order $pq$ (with $q \mid p-1$) 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!

Answer 1

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.

Answer 2

To explicitly build up a (nontrivial) semidirect product $\mathbb Z_3\stackrel{\varphi}{\ltimes}\mathbb Z_{7}$, we must send $0$ to the identity map of $\mathbb Z_{7}$, say $Id_{\mathbb Z_{7}}$, and $1$ and $2$ to two distinct automorphisms of $\mathbb Z_{7}$ of order $3$$^\dagger$, say $\varphi_1$ and $\varphi_2$, such that $Id_{\mathbb Z_{7}}(=$ $\varphi_0=$ $\varphi_{1+2})=$ $\varphi_1\varphi_2$. Since $\operatorname{Aut}(\mathbb Z_{7})$ acts transitively on the set $X$ of the generators of $\mathbb Z_{7}$, the only option is: \begin{alignat}{1} &\varphi_1(0)=0,\space\space\space{\varphi_1}_{|X} = (1,2,4)(3,6,5) \\ &\varphi_2(0)=0,\space\space\space{\varphi_2}_{|X} = (1,4,2)(3,5,6) \\ \tag1 \end{alignat} This can be easily generalized to every semidirect product $\mathbb Z_3\stackrel{\varphi}{\ltimes}\mathbb Z_{p}$, where $p$ is a prime such that $3\mid p-1$$^{\dagger\dagger}$. Since $(\operatorname{Aut}(\mathbb Z_p)\cong)\mathbb Z_p^\times$ is cyclic of order $p-1$, it has exactly one subgroup of order $3$, namely exactly two elements of order $3$, say $x$ and $x^2$. Therefore, $(1)$ generalizes to: \begin{alignat}{1} &\varphi_1(0)=0,\space\space\space{\varphi_1}_{|X} = (1,x,x^2)(i_2,i_2x,i_2x^2)\dots(i_k,i_kx,i_kx^2) \\ &\varphi_2(0)=0,\space\space\space{\varphi_2}_{|X} = (1,x^2,x)(i_2,i_2x^2,i_2x)\dots(i_k,i_kx^2,i_kx) \\ \tag2 \end{alignat} where $k:=\frac{p-1}{3}$ and, iteratively, $i_1:=1$ and $i_{j+1}:=\min\{X\setminus\bigcup_{l=1}^j\{i_l,i_lx,i_lx^2\}\}$ for $j=1,\dots,k-1$.

Edit. This can be further generalized to every semidirect product $\mathbb Z_q\stackrel{\varphi}{\ltimes}\mathbb Z_{p}$, where $q,p$ are primes such that $q\mid p-1$$^{\dagger\dagger\dagger}$. Since $(\operatorname{Aut}(\mathbb Z_p)\cong)\mathbb Z_p^\times$ is cyclic of order $p-1$, it has exactly one subgroup of order $q$, namely exactly $q-1$ elements of order $q$, say $x^i$ ($i=1,\dots,q-1$). Therefore, $(2)$ generalizes to: \begin{alignat}{1} &\varphi_i(0)=0,\space\space\space{\varphi_i}_{|X} = \prod_{r=1}^k\alpha^{(i)}_r \\ \tag3 \end{alignat} for $i=1,\dots,q-1$, where: $$k:=\frac{p-1}{q}$$ and $$\alpha^{(i)}_r:=(j_r,j_rx^{\sigma^i(q-1)},j_rx^{\sigma^i(1)},j_rx^{\sigma^i(2)},\dots,j_rx^{\sigma^i(q-2)})$$ being $\sigma:=(1,2,\dots,q-1)$ and, iteratively:

• $j_1:=1$
• $j_{m+1}:=\min\{X\setminus\bigcup_{l=1}^m\{\alpha^{(i)}_l\}\}$, for $m=1,\dots,k-1$ (here $\{\alpha^{(i)}_l\}$ is the set made of the entries of the cycle $\alpha^{(i)}_l$).

---

$^\dagger$Under a homomorphism, the order of the image of an element divides the order of the element.

$^{\dagger\dagger}$Hence it accounts for the whole sequence of your "interesting group orders", but $55$.

$^{\dagger\dagger\dagger}$Which, incidentally, accounts for the leftover case $55$ of your orders' sequence.

Answer 3

"What is it the symmetry group of?" While some of the links suggest this might be a possible approach, I don't understand them very well; nor did I ever know that this can always be done.

However, to "realize" the group (similar to @kan't), the fact that the Sylow-7 is normal implies rather easily that we have a semi-direct product: $$\Bbb Z_7\rtimes \Bbb Z_3.$$

This can be written unambiguously because there are two automorphisms to choose from, both giving the same group (by a theorem, since they both have image the one $\Bbb Z_3\le\Bbb Z_6$).

It's meta cyclic because the quotient by $\Bbb Z_7$ is $\Bbb Z_3.$

Answer 4

Supplementing the other fine answers with a quick explanation as to why we cannot realize this group $G$ as a collection of symmetries of a geometrical object of real dimension $\le3$.

Let $a\in G$ be an element of order seven. The semidirect product structure implies that we can find an element $b\in G$ of order three such that $bab^{-1}=a^2$. It follows that $b^2ab^{-2}=a^4$, so the elements $a,a^2,a^4$ are each others conjugates in $G$. If $\rho:G\to GL(V)$ is a faithful, complex representation of $G$, then $\rho(a)$ must have an eigenvalue $\lambda\neq1$ that can only be a seventh root of unity, call it $\zeta$. The conjugacy relations imply that $\rho(a^2)$ as well as $\rho(a^4)$ are similar to $\rho(a)$. Therefore $\zeta^2$ and $\zeta^4$ must also be eigenvalues of $\rho(a)$.

If $V$ were a real vector spaces of dimension $3$, then the characteristic polynomial of $\rho(a)$ would be a cubic with real coefficients, implying that $\rho(a)$ should also have a real eigenvalue (that can only be equal to $1$). Altogether we have shown that $\rho(a)$ would have no less than four distinct eigenvalues in this case. That's a contradiction.

---

There is a complex faithful representation of $G$ determined by $$ \rho(a)=\left(\begin{array}{ccc}\zeta&0&0\\ 0&\zeta^4&0\\ 0&0&\zeta^2\end{array}\right) $$ and $$ \rho(b)=\left(\begin{array}{ccc}0&0&1\\ 1&0&0\\ 0&1&0\end{array}\right). $$ See this old thread, possibly several others also, for a description of the full character table of this group.

---

Observe the difference with the dihedral group $$D_7=\langle a,b\mid a^7=1=b^2, bab^{-1}=a^{-1}\rangle.$$ For a representation $\rho:D_7\to GL(V)$ we can only infer that $\rho(a)$ has the eigenvalues $\zeta$ and $\zeta^{-1}=\overline{\zeta}$. Those are exactly the eigenvalues of a rotation of $\Bbb{R}^2$ by the angle $2\pi/7$, leading to the description of $D_7$ as the group of symmetries of a regulat heptagon. In other words, the orbits of the acting group $K$ in $H\rtimes K$ on the set of eigenvalues of elements of $H$ played a key role.