The fifty groups of order $72$

A group of order \(9\) is abelian: its center is nontrivial by the class equation; if its center had order \(3\), its quotient by the center would be cyclic, which forces the whole group to be abelian. Indeed, writing every element as \(g^i z\) with \(z\) central shows that any two elements commute. Thus the normal Sylow subgroup \(N\) is \(C_9\) or \(C_3^2\). A Sylow \(2\)-subgroup \(H\) has order \(8\), with \(N\cap H=1\) and \(NH=G\), so \(G=N\rtimes H\). The five groups of order \(8\) are exactly those in the table [@DF04].

The subgroup \(N\) is characteristic because it is the unique Sylow \(3\)-subgroup. Conjugation on the abelian group \(N\) defines an action of \(G/N\) independent of the chosen lifts. Consequently two actions give isomorphic groups exactly when one can be changed to the other by an automorphism of \(H\) and conjugation by an automorphism of \(N\). Necessity follows by restricting an isomorphism to \(N\) and passing to the quotient; sufficiency follows by applying these two automorphisms to the two coordinates of the semidirect product. Different kernel types or quotient types cannot be isomorphic.

For \(N=C_9\), its automorphism group is \((\mathbb Z/9\mathbb Z)^\times\), of order \(6\), whose only elements of \(2\)-power order are \(+\) and \(-\). Thus every action takes values in this two-element subgroup. For \(C_8\) there are two homomorphisms. For \(C_4\times C_2\), its automorphisms have the form \[ r\mapsto r^\varepsilon s^e,\quad s\mapsto r^{2f}s, \qquad \varepsilon\in\{1,3\},\quad e,f\in\{0,1\}. \] On the two signs \((R,S)\) this replaces \(R\) by \(RS^e\) and leaves \(S\) fixed, giving exactly its three listed orbits. For \(C_2^3\), the automorphism group is transitive on nonzero linear functionals to \(C_2\), giving two orbits. For \(D_8\), the automorphisms are \(r\mapsto r^{\pm1}\), \(s\mapsto r^k s\); thus \(R\) is fixed and \(S\) can be replaced by \(R^kS\), giving the three displayed pairs. For \(Q_8\), every sign homomorphism factors through \(Q_8/\langle r^2\rangle\cong C_2^2\). Permuting its three cyclic subgroups of order \(4\) induces all automorphisms of this quotient, giving two orbits. This proves completeness and distinctness in the \(C_9\) column.

For \(C_3^2\), the automorphism group is \(\operatorname{GL}_2(\mathbb F_3)\). Step 2 verifies that the third column lists each \(\operatorname{Aut}(H)\times\operatorname{GL}_2(\mathbb F_3)\)-orbit of homomorphisms \(H\to\operatorname{GL}_2(\mathbb F_3)\) exactly once. Together with the preceding isomorphism criterion, this proves all assertions about the \(42\) groups.

Represent a matrix by its four row-major entries. There are exactly the invertible matrices in the finite set \(\{0,1,2\}^4\) to check. A tuple determines a homomorphism from a presented group exactly when it satisfies all the defining relations. The function valid below tests those relations; candidates therefore contains every homomorphism.

The calculation checks that the listed orbits are disjoint and that their union equals candidates. All arithmetic is exact modulo \(3\).

from itertools import product, combinations

I = (1, 0, 0, 1)
Z = (2, 0, 0, 2)
D = (1, 0, 0, 2)
A = (0, 2, 1, 0)
B = (1, 1, 1, 2)
C = (0, 1, 1, 1)

def mul(x, y):
    a, b, c, d = x
    e, f, g, h = y
    return ((a*e+b*g) % 3, (a*f+b*h) % 3,
            (c*e+d*g) % 3, (c*f+d*h) % 3)

def power(x, n):
    result = I
    for _ in range(n):
        result = mul(result, x)
    return result

def inverse(x):
    a, b, c, d = x
    u = pow((a*d-b*c) % 3, -1, 3)
    return ((u*d) % 3, (-u*b) % 3, (-u*c) % 3, (u*a) % 3)

GL = [x for x in product(range(3), repeat=4)
      if (x[0]*x[3]-x[1]*x[2]) % 3 != 0]
table = {
    "C8": [(I,), (Z,), (D,), (A,), (C,)],
    "C4xC2": [(I,I), (Z,I), (I,Z), (D,I),
              (I,D), (D,Z), (Z,D), (A,I)],
    "C2^3": [(I,I,I), (Z,I,I), (D,I,I), (Z,D,I)],
    "D8": [(I,I), (I,Z), (I,D), (Z,I),
           (Z,D), (D,I), (D,Z), (A,D)],
    "Q8": [(I,I), (Z,I), (D,I), (Z,D), (A,B)],
}

vectors = [v for v in product(range(2), repeat=3) if any(v)]

def xor(x, y):
    return tuple((a+b) % 2 for a, b in zip(x, y))

bases = [(u,v,w) for u,v,w in product(vectors, repeat=3)
         if len({(0,0,0), u, v, w, xor(u,v), xor(u,w),
                 xor(v,w), xor(xor(u,v),w)}) == 8]

def valid(name, t):
    if name == "C8":
        return power(t[0], 8) == I
    if name == "C2^3":
        return (all(power(x, 2) == I for x in t)
                and all(mul(x,y) == mul(y,x)
                        for x,y in combinations(t, 2)))
    r, s = t
    if name == "C4xC2":
        return (power(r,4) == I and power(s,2) == I
                and mul(r,s) == mul(s,r))
    if name == "D8":
        return (power(r,4) == I and power(s,2) == I
                and mul(s,r) == mul(power(r,3),s))
    if name == "Q8":
        return (power(r,4) == I and power(s,2) == power(r,2)
                and mul(s,r) == mul(power(r,3),s))
    raise ValueError(name)

def changes(name, t):
    if name == "C8":
        return [(power(t[0], k),) for k in (1,3,5,7)]
    if name == "C2^3":
        def value(v):
            ans = I
            for x, e in zip(t, v):
                ans = mul(ans, power(x,e))
            return ans
        return [tuple(value(v) for v in basis) for basis in bases]
    r, s = t
    if name == "C4xC2":
        return [(mul(power(r,e),power(s,f)), mul(power(r,2*g),s))
                for e,f,g in product((1,3),(0,1),(0,1))]
    if name == "D8":
        return [(power(r,e), mul(power(r,k),s))
                for e,k in product((1,3),range(4))]
    if name == "Q8":
        axes = [[(1,0),(3,0)], [(0,1),(2,1)], [(1,1),(3,1)]]
        def value(v):
            return mul(power(r,v[0]), power(s,v[1]))
        return [(value(u),value(v))
                for i,j in product(range(3), repeat=2) if i != j
                for u in axes[i] for v in axes[j]]
    raise ValueError(name)

def orbit(name, t):
    result = set()
    for changed in changes(name, t):
        for m in GL:
            mi = inverse(m)
            result.add(tuple(mul(mul(m,x),mi) for x in changed))
    return result

for name, reps in table.items():
    candidates = {t for t in product(GL, repeat=len(reps[0]))
                  if valid(name,t)}
    seen = set()
    for rep in reps:
        if not valid(name,rep):
            raise ValueError(("bad representative", name, rep))
        current = orbit(name,rep)
        if current & seen:
            raise ValueError(("repeated orbit", name, rep))
        seen.update(current)
    if seen != candidates:
        raise ValueError(("incomplete", name, len(seen), len(candidates)))
    print(name, len(candidates), len(reps))

Its output gives the number of homomorphisms followed by the number of distinct action orbits:

C8 32 5
C4xC2 88 8
C2^3 344 4
D8 100 8
Q8 100 5

Let \(T\) be a normal Sylow \(2\)-subgroup and \(P\) a Sylow \(3\)-subgroup. Then \(G=T\rtimes P\), with \(P=C_9\) or \(C_3^2\). For the five possible types of \(T\), the respective automorphism-group orders are \[ \begin{array}{c|ccccc} T&C_8&C_4\times C_2&C_2^3&D_8&Q_8\\\hline |\operatorname{Aut}(T)|&4&8&168&8&24. \end{array} \] The cyclic case counts odd generator powers; the generator formulas in step 1 give the two orders \(8\); the order \(168=(8-1)(8-2)(8-4)\) counts ordered bases; and the quaternion count is \(6\cdot4=24\) as in step 2.

In the first two cases \(|K|=18\) or \(9\), and the image is a \(2\)-group. Every Sylow \(3\)-subgroup of \(G\) would then lie in \(K\). But either of these orders gives a unique Sylow \(3\)-subgroup in \(K\), a contradiction. Therefore \(|K|=6\) or \(3\). In the first case the image is the unique index-two subgroup \(A_4\) of \(S_4\); in the second the image is \(S_4\).

We use that \([S_4,S_4]=A_4\) and \([A_4,A_4]=V_4\), where \(V_4\) is the subgroup of double transpositions. The relevant quotients are abelian; commutators of transpositions include \(3\)-cycles and their conjugates generate \(A_4\), while \([(123),(124)]=(12)(34)\), using \([x,y]=xyx^{-1}y^{-1}\). Its conjugates in \(A_4\) generate \(V_4\). This proves both equalities. In particular the only nontrivial map \(S_4\to C_2\) is sign, and there is no nontrivial map \(A_4\to C_2\).

If \(K=C_6\), let \(N\) be its characteristic subgroup of order \(3\). Conjugation on \(N\) factors through \(G/K=A_4\), since \(K\) is abelian. The preceding observation makes this action trivial, so \(N\) is central in \(G\). Let \(U\) be the inverse image of \(V_4\lhd A_4\); then \(U\lhd G\) has order \(24\). A Sylow \(2\)-subgroup \(T\) of \(U\) commutes with \(N\) and gives \(U=N\times T\). In this direct product the elements of \(2\)-power order form exactly \(T\), so \(T\) is characteristic in \(U\) and normal in \(G\). This contradicts the assumption on the Sylow \(2\)-subgroups.

Thus \(K=S_3\). Every automorphism of \(S_3\) is inner: it permutes the three transpositions, giving an injection into \(S_3\), and conjugation already realizes all six permutations because the center is trivial. For every \(g\in G\), choose \(k\in K\) with the same conjugation action on \(K\). Then \(k^{-1}g\in C_G(K)\), so \(G=K C_G(K)\). The intersection is \(Z(K)=1\), and the two factors commute. Consequently \(G\cong K\times C_G(K)\cong S_3\times A_4\).

Set \(H=G/V\). Then \(|H|=18\) and the induced map \(q_H:H\to S_3\) has kernel the image of \(N\), of order \(3\). The map \[ G\longrightarrow H\times S_4,\qquad g\longmapsto(gV,gN) \] is injective because \(V\cap N=1\), and its image lies in the fiber product of \(q_H\) and \(\pi\). That fiber product has order \(18\cdot24/6=72\), so the map is an isomorphism onto it.

If \(P=C_9\), the only automorphisms of order dividing \(2\) are the identity and inversion, and the quotient condition forces inversion. Thus \(H=D_{18}\) with its unique subgroup of order \(3\) as kernel.

If \(P=C_3^2\), the involution on this vector space is diagonalizable: each vector is the sum of \((x+t x)/2\) and \((x-t x)/2\), in the \(+1\) and \(-1\) eigenspaces respectively. The quotient has eigenvalue \(-1\). If the kernel has eigenvalue \(+1\), the two eigenspaces give \(H=C_3\times S_3\) with its central \(C_3\) as kernel. If the kernel has eigenvalue \(-1\), the involution is \(-I\), so \(H=C_3^2\rtimes_{-I}C_2\); all choices of a kernel line are equivalent under \(\operatorname{GL}_2(\mathbb F_3)\).

Any two quotient identifications differ by an automorphism of \(S_3\), which is inner. Conjugation by a lift in \(S_4\) absorbs that change in the fiber product. Therefore no extra classes arise from the choice of the quotient map. The first case gives \(D_{18}\times_{S_3}S_4\). The second gives \(C_3\times S_4\). In the last case split \(C_3^2\) as the kernel line plus another line. Both are inverted by \(t\), so the pullback is \(C_3\rtimes_{\mathrm{sgn}}S_4\). This proves exhaustiveness.

The kernel of the action on the four Sylow \(3\)-subgroups has order \(6\) in \(S_3\times A_4\) and order \(3\) in the other three; it is an isomorphism invariant. To check the asserted kernels, each Sylow \(3\)-subgroup of \(S_4\) or \(A_4\) fixes exactly one of the four letters, with one such subgroup for each letter. The conjugation action on these subgroups is thus their faithful natural action on the four letters. Also \(S_4\) has trivial center, since an element commuting with every transposition preserves every unordered pair of letters and therefore fixes every letter. Thus \(C_3\times S_4\) has center of order \(3\). The other two have trivial center: a central element projects to the trivial center of \(S_4\), hence lies in the kernel \(C_3\), and an odd permutation inverts that kernel. Finally \(D_{18}\times_{S_3}S_4\) has a cyclic Sylow \(3\)-subgroup of order \(9\), the pullback of a \(3\)-cycle subgroup along \(C_9\to C_3\). In \(C_3\rtimes_{\mathrm{sgn}}S_4\) the Sylow \(3\)-subgroup is \(C_3\times C_3\), since a \(3\)-cycle acts trivially on the kernel. This distinguishes the last two groups.

Every group has either a normal Sylow \(3\)-subgroup, a normal Sylow \(2\)-subgroup but not a normal Sylow \(3\)-subgroup, or neither. Steps 1, 3, and 4 classify these disjoint cases, with \(42\), \(4\), and \(4\) distinct classes respectively. Their sum is \(50\).