Random Permutations

Happy new year!

Group theory warm-up exercise for the year: what are all the groups of order 2024?

Last year we did 2023, which was rather boring because all groups of order ${2023}$ are abelian. This year the possibilities are much more numerous and the calculation is somewhat involved.

There are some easy initial reductions using Sylow’s theorem. Let ${G}$ be a group of order ${2024 = 2^3 \cdot 11 \cdot 23}$ . Sylow’s theorem implies that there must be a unique subgroup ${P}$ of order ${23}$ . Similarly, in the quotient ${G/P}$ of order ${2^3 \cdot 11}$ there must be a unique subgroup of order ${11}$ . Therefore ${G}$ has a unique subgroup ${N}$ of order ${11 \cdot 23}$ . Let ${Q}$ be a Sylow ${2}$ -subgroup of ${G}$ . Then ${G = NQ \cong N \rtimes_\alpha Q}$ for some homomorphism ${\alpha : Q \to \mathrm{Aut}(N)}$ .

We can easily list the possibilities for ${N}$ and ${Q}$ . There are just two possibilities for ${N}$ : the cyclic group ${C_{253}}$ and the unique nonabelian semidirect product ${C_{23} \rtimes C_{11}}$ . There are five possibilities for ${Q}$ : ${C_8}$ , ${C_4 \times C_2}$ , ${C_2^3}$ , the dihedral group ${D_4}$ , and the quaternion group ${Q_8}$ .

Next, for each choice of ${N}$ and ${Q}$ , we must consider all possibilities for ${\alpha : Q \to \mathrm{Aut}(N)}$ . However, in order to avoid duplication, we must determine when two homomorphisms ${\alpha, \beta : Q \to \mathrm{Aut}(N)}$ induce isomorphic semidirect products.

A quick word about notation. Following standard practice in group theory, we denote group actions on the right. If ${x}$ and ${y}$ are elements of a common group ${G}$ , we denote by ${x^y}$ the conjugate ${y^{-1} x y}$ . We also denote by ${x^\alpha}$ the image of ${x}$ under a given homomorphism ${\alpha}$ . We can then say that the semidirect product ${N \rtimes_\alpha Q}$ is generated by copies of ${N}$ and ${Q}$ subject to the natural-looking relation

$\displaystyle n^q = n^{q^\alpha}.$

This relation asserts that the conjugation action of ${Q}$ on ${N}$ is given by ${\alpha}$ .

Suppose ${f}$ is an isomorphism from ${G_\alpha = N \rtimes_\alpha Q}$ to ${G_\beta = N \rtimes_\beta Q}$ such that ${f(N) = N}$ . Then ${f}$ induces an automorphism ${f_1}$ of ${N}$ as well as an automorphism ${f_2}$ of ${G/N \cong Q}$ . These automorphisms are not arbitrary: they satisfy a certain compatibility relation. Namely, for all ${n \in N}$ and ${q \in Q}$ we have

$\displaystyle (n^{q^\alpha})^{f_1} = (n^q)^f = (n^f)^{q^f} = (n^{f_1})^{(q^{f_2})^\beta}.$

In other words, for all ${q \in Q}$ we have, in ${\mathrm{Aut}(N)}$ ,

$\displaystyle f_1^{-1} q^\alpha f_1 = (q^{f_2})^\beta.$

Written a third way, we have, as elements of ${\mathrm{Hom}(Q, \mathrm{Aut}(N))}$ ,

$\displaystyle \alpha f_1^\iota = f_2 \beta,$

where ${f_1^\iota}$ denotes the inner automorphism of ${\mathrm{Aut}(N)}$ induced by ${f_1}$ . (Alternatively, if somewhat more traditionally we denote function composition in a right-to-left manner, the compatibility relation is ${f_1 \cdot \alpha \cdot f_1^{-1} = \beta \circ f_2}$ .)

Let us denote by ${I_{\alpha,\beta}}$ the set of all isomorphisms ${G_\alpha \to G_\beta}$ such that ${f(N) = N}$ , and by ${C_{\alpha, \beta}}$ the set of all compatible isomorphism pairs ${(f_1, f_2) \in \mathrm{Aut}(N) \times \mathrm{Aut}(Q)}$ , i.e.,

$\displaystyle C_{\alpha,\beta} = \{(f_1, f_2) \in \mathrm{Aut}(N) \times \mathrm{Aut}(Q) : \alpha f_1^\iota = f_2 \beta\}.$

The import of the previous paragraph is that there is a natural map ${I_{\alpha,\beta} \to C_{\alpha,\beta}}$ . Moreover this map is surjective (though typically not injective), because if ${(f_1, f_2)}$ is a compatible isomorphism pair then the map ${f :G_\alpha \to G_\beta}$ defined by ${(nq)^f = n^{f_1} q^{f_2}}$ is an isomorphism (easy exercise). In particular, there is an isomorphism ${f}$ from ${N \rtimes_\alpha Q}$ to ${N \rtimes_\beta Q}$ such that ${f(N) = N}$ if and only if ${C_{\alpha,\beta} \ne \emptyset}$ , i.e., if and only if there is a pair of compatible automorphisms ${(f_1, f_2) \in \mathrm{Aut}(N) \times \mathrm{Aut}(Q)}$ .

We can phrase this conclusion another way. Observe that ${\mathrm{Aut}(N) \times \mathrm{Aut}(Q)}$ acts naturally on ${\mathrm{Hom}(Q, \mathrm{Aut}(N))}$ . The ${\mathrm{Aut}(Q)}$ factor acts by precomposition, while the ${\mathrm{Aut}(N)}$ factor acts by conjugation. Isomorphism classes of split extensions ${N \rtimes Q}$ are in bijection with orbits of ${\mathrm{Aut}(N) \times \mathrm{Aut}(Q)}$ in ${\mathrm{Hom}(Q, \mathrm{Aut}(N))}$ . (Here we consider ${N \rtimes_\alpha Q}$ and ${N \rtimes_\beta Q}$ to be isomorphic as extensions if and only if there is an isomorphism ${f : N \rtimes_\alpha Q \to N \rtimes_\beta Q}$ such that ${f(N) = N}$ . In general this is more restrictive than mere isomorphism as groups.)

Now consider the special case in which ${\alpha = \beta}$ . In this case ${I_{\alpha,\alpha}}$ is a subgroup of ${\mathrm{Aut}(N \rtimes_\alpha Q)}$ and ${C_{\alpha,\alpha}}$ is a subgroup of ${\mathrm{Aut}(N) \times \mathrm{Aut}(Q)}$ , and the natural map ${I_{\alpha,\alpha} \to C_{\alpha,\alpha}}$ is a homomorphism. Let ${K}$ be the kernel. Then ${K}$ consists of all isomorphism ${f : G \to G}$ that restrict to the identity on ${N}$ and induce the identity on ${G/N \cong Q}$ . This implies that there is a map ${h : Q \to N}$ such that ${q^f = q q^h}$ for all ${q \in Q}$ . Since ${f}$ restricts to the identity on ${N}$ , we have

$\displaystyle n^q = (n^q)^f = n^{q^f} = n^{q q^h},$

which implies that ${h}$ takes values in the centre ${Z(N)}$ of ${N}$ . Moreover, ${q^f = qq^h}$ is a homomorphism ${Q \to G}$ , so we must have the relation

$\displaystyle (q_1q_2)^h = (q_1^h)^{q_2^\alpha} (q_2^h).$

This relation means that ${h}$ is a crossed homomorphism from ${Q}$ to ${Z(N)}$ . The group of crossed homomorphisms ${G \to M}$ is usually denoted ${Z^1_\alpha(G, M)}$ . Thus we have a short exact sequence

$\displaystyle Z^1_\alpha(Q, Z(N)) \to I_{\alpha,\alpha} \to C_{\alpha,\alpha}.$

In particular, if ${N}$ is a characteristic subgroup of ${N \rtimes_\alpha Q}$ , ${I_{\alpha,\alpha}}$ is the full automorphism group, so we have a short exact sequence

$\displaystyle Z^1_\alpha(Q, Z(N)) \to \mathrm{Aut}(N \rtimes_\alpha Q) \to C_{\alpha,\alpha}.$

In fact ${C_{\alpha,\alpha}}$ can be identified with the subgroup of automorphisms of ${N \rtimes_\alpha Q}$ preserving ${Q}$ , so this sequence splits and we find that

$\displaystyle \mathrm{Aut}(N \rtimes_\alpha Q) \cong Z_\alpha^1(Q, Z(N)) \rtimes C_{\alpha,\alpha}.$

Let us put all this into practice. Recall that we have two possibilities for ${N}$ : ${N \cong C_{253}}$ or ${N \cong C_{23} \rtimes C_{11}}$ . Let us consider the case ${N \cong C_{253}}$ first. We have

$\displaystyle \mathrm{Aut}(C_{253}) \cong \mathrm{Aut}(C_{23}) \times \mathrm{Aut}(C_{11}) \cong C_{22} \times C_{10} \cong C_2^2 \times C_5 \times C_{11}.$

For each choice of ${Q}$ among ${C_8, C_4 \times C_2, C_2^3, D_4, Q_8}$ we must tabulate the homomorphisms ${Q \to \mathrm{Aut}(C_{253})}$ up to the action of ${\mathrm{Aut}(Q)}$ by precomposition and the conjugation action of ${\mathrm{Aut}(C_{253})}$ . Since ${\mathrm{Aut}(C_{253})}$ is abelian, the latter action is trivial. Any such homomorphism must take values in the Sylow ${2}$ -subgroup ${C_2^2}$ , so we are reduced to tabulating homomorphisms ${Q \to C_2^2}$ up to automorphisms of ${Q}$ .

Case ${Q = C_8}$ : There are four. Among these are groups with the structures ${C_{2024}}$ , ${C_{23} \times (C_{11} \rtimes C_8)}$ , ${C_{11} \times (C_{23} \rtimes C_8)}$ . The last has the structure ${C_{253} \rtimes C_8}$ , and the action of ${C_8}$ on ${C_{253}}$ is fixed-point-free.

Case ${Q = C_4 \times C_2}$ : Write ${Q = \langle x, y \mid x^4 = y^2 = [x,y] = 1\rangle}$ . Any homomorphism ${Q \to C_2^2}$ must kill ${x^2}$ , so factors through ${Q / \langle x^2 \rangle \cong \langle x, y \mid x^2 = y^2 = [x,y] = 1\rangle \cong C_2^2}$ . Naively are ${4^2}$ homomorphisms ${Q/\langle x^2\rangle \to C_2^2}$ , but ${\mathrm{Aut}(Q)}$ acts by swapping ${x}$ and ${xy}$ (while ${y}$ is fixed) and the number of orbits of homomorphisms ${Q \to C_2^2}$ is ${1 \cdot 4 + 3 \cdot 2 = 10}$ . These groups include ${C_{253} \times C_4 \times C_2}$ , ${(C_{253} \rtimes C_4) \times C_2}$ , ${(C_{11} \rtimes C_4) \times C_{23} \times C_2}$ , ${(C_{23} \rtimes C_4) \times C_{11} \times C_2}$ , ${D_{253} \times C_4}$ , ${D_{11} \times C_{23} \times C_4}$ , ${D_{23} \times C_{11} \times C_4}$ , and ${3}$ others.

Case ${Q = C_2^3}$ : There are ${4^3}$ homomorphisms ${C_2^3 \to C_2^2}$ , but only ${5}$ up to the action of ${\mathrm{Aut}(Q) \cong \mathrm{GL}_3(2)}$ . They are in bijection with subgroups of ${C_2^2}$ . The corresponding groups of order ${2024}$ are ${C_{23} \times C_{11} \times C_2^3}$ , ${D_{11} \times C_{23} \times C_2^2}$ , ${D_{23} \times C_{11} \times C_2^2}$ , ${D_{253} \times C_2^2}$ , and ${D_{11} \times D_{23} \times C_2}$ .

Case ${Q = D_4}$ : Any homomorphism ${D_4 \to C_2^2}$ factors through ${D_4 / D_4' \cong C_2^2}$ . Just as in the case of ${C_4 \times C_2}$ there are ${10}$ homomorphisms ${D_4 \to C_2^2}$ up to the action of ${\mathrm{Aut}(D_4)}$ . The groups of order ${2024}$ include ${C_{253} \times D_4}$ and ${9}$ in which ${D_4}$ acts nontrivially on ${C_{253}}$ by conjugation.

Case ${Q = Q_8}$ : Any homomorphism ${Q_8 \to C_2^2}$ factors through ${Q_8 / Q_8' \cong C_2^2}$ . In this case ${\mathrm{Aut}(Q_8)}$ acts on ${C_2^2}$ as ${S_3}$ , so as in the case of ${C_2^3}$ there are just ${5}$ homomorphisms ${Q_8 \to C_2^2}$ up to automorphisms of ${Q_8}$ . Thus there are ${5}$ semidirect products of the form ${C_{253} \rtimes Q_8}$ .

Now consider the case in which ${N \cong C_{23} \rtimes_\alpha C_{11}}$ is nonabelian. Since ${C_{23}}$ is a characteristic subgroup of ${N}$ (being the only subgroup of order ${23}$ ), we have

$\displaystyle \mathrm{Aut}(C_{23} \rtimes C_{11}) \cong Z^1_\alpha(C_{11}, C_{23}) \rtimes C_{\alpha,\alpha},$

where ${C_{\alpha,\alpha}}$ is the subgroup of ${\mathrm{Aut}(C_{23}) \times \mathrm{Aut}(C_{11})}$ consisting of compatible pairs. Consider the compatibility relation ${\alpha f_1^\iota = f_2 \alpha}$ . Since ${\mathrm{Aut}(C_{23}) \cong C_{22}}$ is abelian, ${f_1^\iota}$ is trivial, so the relation reduces to ${\alpha = f_2 \alpha}$ . Since ${\alpha}$ is injective, this implies that ${f_2}$ is trivial. Therefore ${C_{\alpha,\alpha} = \mathrm{Aut}(C_{23}) \cong C_{22}}$ . Meanwhile one checks that ${Z^1_\alpha(C_{11}, C_{23}) \cong C_{23}}$ . Therefore ${\mathrm{Aut}(C_{23} \rtimes C_{11}) \cong C_{23} \rtimes C_{22}}$ . Now if ${Q}$ has order ${8}$ then any homomorphism ${Q \to C_{23} \rtimes C_{22}}$ takes values in a Sylow ${2}$ -subgroup isomorphic to ${C_2}$ , which are all conjugate. Tabulating homomorphisms ${Q \to \mathrm{Aut}(N)}$ up to the natural action of ${\mathrm{Aut}(N) \times \mathrm{Aut}(Q)}$ is therefore equivalent to tabulating homomorphisms ${Q \to C_2}$ up to automorphisms of ${Q}$ , which just amounts to tabulating automorphism classes of subgroups of ${Q}$ of index at most ${2}$ . This case is therefore somewhat easier than the previous one.

Case ${Q = C_8}$ : There are just two. The corresponding groups of order ${2024}$ have the forms ${(C_{23} \rtimes C_{11}) \times C_8}$ and ${C_{23} \rtimes (C_{11} \times C_8)}$ .

Case ${Q = C_4 \times C_2}$ : There are three, corresponding to subgroups isomorphic to ${Q}$ , ${C_4}$ , and ${C_2 \times C_2}$ . The corresponding groups of order ${2024}$ include ${(C_{23} \rtimes C_{11}) \times C_4 \times C_2}$ , ${(C_{23} \rtimes C_{22}) \times C_4}$ , ${(C_{23} \rtimes C_{44}) \times C_2}$ .

Case ${Q = C_2^3}$ : There are two, since all index- ${2}$ subgroups are essentially the same. The corresponding groups of order ${2024}$ are ${(C_{23} \rtimes C_{11}) \times C_2^3}$ and ${(C_{23} \rtimes C_22) \times C_2^2}$ .

Case ${Q = D_4}$ : There are three, since there is a unique subgroup isomorphic to ${C_4}$ while the two subgroups isomorphic to ${C_2 \times C_2}$ are equivalent by an automorphism. There are three corresponding groups with structure ${(C_{23} \rtimes C_{11}) \rtimes D_4}$ .

Case ${Q = Q_8}$ : There are two, since the three subgroups of ${Q_8}$ isomorphic to ${C_4}$ are equivalent by automorphisms. Thus there are just two groups with the structure ${(C_{23} \rtimes C_{11}) \rtimes Q_8}$ .

Thus altogether there are ${34+12 = 46}$ groups of order ${2024}$ .

Random Permutations

Groups of order 2024

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