List all isomorphism classes of groups of size 120.
Find all isomorphism classes of \(\mathbf{Q}[x]\) modules \(V\) such that \(\dim_{\mathbf{Q}} V=6\) and \(V\) is annihilated by \((x-3)^2(x+5)\).
Find all possible rational canonical forms for a 6 by 6 matrix over \(\mathbf{Q}\) that has minimal polynomial \((x-3)^2(x+5)\).
Solution
The \(47\) groups of order \(120\) in part (a) are the \(39\) semidirect products specified in the table in step 5, the five groups in step 6, and the three groups in step 7.
1. An extension with finite abelian kernel \(N\) and quotient \(Q\) splitswhen \(\gcd(|N|,|Q|)=1\).
Write \(N\) additively and choose representatives \(s(q)\) with \(s(1)=1\). Conjugation defines an action of \(Q\) on \(N\), independent of the representatives because \(N\) is abelian. Put \(f(p,q)=s(p)s(q)s(pq)^{-1}\in N\). Associativity gives \[ f(p,q)+f(pq,r)=p\cdot f(q,r)+f(p,qr). \] Let \(S(p)=\sum_{r\in Q}f(p,r)\) and \(m=|Q|\). Summing the identity over \(r\) gives \[ m f(p,q)=S(p)+p\cdot S(q)-S(pq). \] Multiplication by \(m\) is an automorphism of \(N\). Thus, with \(h(p)=m^{-1}S(p)\), replacing \(s(p)\) by \((-h(p))s(p)\) makes its factor set zero. The new representatives form a subgroup isomorphic to \(Q\), so the extension is \(N\rtimes Q\).
2. A central extension of \(A_4\) by \(C_2\) is either \(C_2\times A_4\) or\(\mathrm{SL}_2(\mathbf F_3)\). A central extension of \(A_5\) by \(C_2\) iseither \(C_2\times A_5\) or \(\mathrm{SL}_2(\mathbf F_5)\).
2.1. We use the presentations\[ T_m=\langle a,b\mid a^2=b^3=(ab)^m=1\rangle, \qquad T_3\cong A_4,\quad T_5\cong A_5. \]Coset enumeration gives the orders of the two presented groups.
F := FreeGroup("a", "b");;
a := F.1;; b := F.2;;
orders := List([3,5], m -> Size(F / [a^2,b^3,(a*b)^m]));;
if orders <> [12,60] then Error("unexpected triangle-group order"); fi;
Print(orders, "\n");
# [ 12, 60 ]For \(m=3\), the permutations \((12)(34)\) and \((123)\) generate \(A_4\): the conjugates of the first generate its Klein four subgroup, and the second supplies the quotient of order \(3\). For \(m=5\), use \((12)(34)\) and \((135)\). Their product has order \(5\), so their subgroup has order divisible by \(2\), \(3\), and \(5\). Its order is therefore \(30\) or \(60\); order \(30\) would give a normal subgroup of index \(2\) in the simple group \(A_5\). Consequently the subgroup is \(A_5\). These surjections and the computed orders establish the two presentations. We use the simplicity of \(A_n\) for \(n\geq5\) here and below [@DF04].
2.2. Let \(z\) generate the central kernel and lift \(a,b\) to \(\alpha,\beta\).Since \(3\) is odd, replacing \(\beta\) by \(z\beta\) when necessary makes\(\beta^3=1\). Since \(m\) is odd, replacing \(\alpha\) by \(z\alpha\) whennecessary makes \((\alpha\beta)^m=1\). This replacement does not change\(\alpha^2\), which is either \(1\) or \(z\).
If \(\alpha^2=1\), the subgroup generated by \(\alpha,\beta\) is a quotient of \(T_m\) that surjects onto \(T_m\), hence maps isomorphically to the quotient. It is a complement to the central kernel, giving the direct product.
2.3. If \(\alpha^2=z\), the extension is a quotient of\[ \widetilde T_m= \langle a,b,z\mid z^2=1,\ [z,a]=[z,b]=1, a^2=z,\ b^3=1,\ (ab)^m=1\rangle. \]Its central subgroup \(\langle z\rangle\) has order at most \(2\), and itsquotient is \(T_m\). Thus \(|\widetilde T_m|\leq 2|T_m|\).For \(p=m\in\{3,5\}\), the matrices\[ A=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad B=\begin{pmatrix}0&1\\-1&-1\end{pmatrix} \quad\text{in }\mathrm{SL}_2(\mathbf F_p) \]satisfy \(A^2=-I\), \(B^3=I\), and\(AB=\left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)\) of order\(p\). They generate \(\mathrm{SL}_2(\mathbf F_p)\): powers of \(AB\) give allupper elementary matrices, conjugation by \(A\) gives the lower ones, andthese perform Gaussian elimination. Diagonal matrices are also generated,as witnessed, for \(t\ne0\), by\[ \begin{pmatrix}1&t\\0&1\end{pmatrix} \begin{pmatrix}1&0\\-t^{-1}&1\end{pmatrix} \begin{pmatrix}1&t\\0&1\end{pmatrix}A =\begin{pmatrix}t&0\\0&t^{-1}\end{pmatrix}. \]There are \(p^2-1\) choices for a nonzero first column and \(p\) choices fora second column having determinant \(1\), so this matrix group has order\(p(p^2-1)=2|T_m|\). It is therefore isomorphic to \(\widetilde T_m\) and tothe extension in this case.
3. Every group \(H\) of order \(24\) is solvable. If its Sylow \(3\)-subgroupis not normal, then\[ H\cong S_4,\qquad C_2\times A_4,\qquad \mathrm{SL}_2(\mathbf F_3). \]
If the Sylow \(3\)-subgroup is normal, step 1 gives \(H\cong C_3\rtimes P\) for a group \(P\) of order \(8\), so \(H\) is solvable. Otherwise there are four Sylow \(3\)-subgroups. Conjugation on them defines \(H\to S_4\) with kernel \(K\) contained in a normalizer of order \(6\). If \(3\) divided \(|K|\), normality would put all four Sylow \(3\)-subgroups inside \(K\), requiring at least \(1+4(3-1)=9\) elements. Thus \(|K|=1\) or \(2\). In the first case \(H\cong S_4\). In the second, the image is \(A_4\), and \(K\cong C_2\) is central. Step 2 gives the other two possibilities. The series \(1\lhd V_4\lhd A_4\lhd S_4\) has abelian factors, and a central extension of a solvable group is solvable, proving the assertion in all cases.
4. A group \(G\) of order \(120\) is solvable if and only if its Sylow\(5\)-subgroup is normal. In that case \(G\cong C_5\rtimes H\) with \(|H|=24\).
We first prove by induction that every solvable group of order dividing \(120\) and divisible by \(5\) has a normal Sylow \(5\)-subgroup. A minimal nontrivial normal subgroup \(N\) of a finite solvable group is elementary abelian: its commutator subgroup must be trivial by minimality and solvability, and its primary subgroups and subgroup of \(p\)-th powers are characteristic. Here \(N\) is therefore \(C_5\), \(C_3\), or \(C_2^r\) with \(1\leq r\leq3\).
If \(N=C_5\), the conclusion follows. Otherwise the induction hypothesis in \(G/N\) gives a normal subgroup \(L/N\) of order \(5\). A Sylow \(5\)-subgroup \(P\) of \(L\) acts trivially on \(N\), because \[ |\operatorname{Aut}(C_3)|=2,\qquad |\operatorname{GL}_r(\mathbf F_2)|\in\{1,6,168\} \] are not divisible by \(5\). Hence \(L=N\times P\), and \(P\) is its unique Sylow \(5\)-subgroup. It is characteristic in \(L\lhd G\), so normal in \(G\). The displayed linear-group orders follow by choosing successive independent columns; for example \((8-1)(8-2)(8-4)=168\).
For \(|G|=120\), a normal Sylow \(5\)-subgroup is cyclic and step 1 splits the extension. Its quotient of order \(24\) is solvable by step 3, so \(G\) is solvable as well.
5. Exactly \(39\) isomorphism classes have a normal cyclic subgroup oforder \(15\). They are the following explicitly specified semidirect products.
5.1. Put \(N=C_3\times C_5=\langle c\rangle\times\langle d\rangle\).Identify its automorphism group with the additive group\(\mathcal A=C_2\oplus C_4\) by letting \((i,j)\) act as\[ c\longmapsto c^{(-1)^i},\qquad d\longmapsto d^{2^j}. \]Write \(u=(1,0)\), \(v=(0,1)\), and \(t=2v=(0,2)\). Thus\(\mathcal A[2]=\{0,u,t,u+t\}\). For each row below form\(N\rtimes_\phi P\), with \(\phi\) given on the displayed ordered generators.Each choice in a set or each explicitly listed tuple is one group.
| \(P\) and ordered generators | Values of \(\phi\) | Number |
|---|---|---|
| \(C_8=\langle r\rangle\) | \(\phi(r)\in\{0,u,t,u+t,v,u+v\}\) | 6 |
| \(C_4\times C_2=\langle r\rangle\times\langle s\rangle\) | \((\phi(r),\phi(s))=(R,0)\), \(R\in\{0,u,t,u+t\}\) | 4 |
| Same \(P\) | \((R,u)\), \(R\in\{0,t\}\) | 2 |
| Same \(P\) | \((R,t)\), \(R\in\{0,u\}\) | 2 |
| Same \(P\) | \((R,u+t)\), \(R\in\{0,u\}\) | 2 |
| Same \(P\) | \((v,0),(u+v,0),(v,u)\) | 3 |
| \(C_2^3=\langle r,s,w\rangle\) | \((0,0,0)\); \((R,0,0)\) for \(R\in\{u,t,u+t\}\); \((u,t,0)\) | 5 |
| \(D_8=\langle r,s\mid r^4=s^2=1,\ srs^{-1}=r^{-1}\rangle\) | \((0,S)\), \(S\in\{0,u,t,u+t\}\) | 4 |
| Same \(P\) | \((u,S)\), \(S\in\{0,t\}\) | 2 |
| Same \(P\) | \((t,S)\), \(S\in\{0,u\}\) | 2 |
| Same \(P\) | \((u+t,S)\), \(S\in\{0,u\}\) | 2 |
| \(Q_8=\langle r,s\mid r^4=1,\ s^2=r^2,\ srs^{-1}=r^{-1}\rangle\) | \((0,0)\); \((R,0)\) for \(R\in\{u,t,u+t\}\); \((u,t)\) | 5 |
The entries sum to \(6+13+5+10+5=39\).
5.2. These are all possibilities for \(P\). An abelian group of order \(8\)is \(C_8\), \(C_4\times C_2\), or \(C_2^3\), according as its exponent is \(8\),\(4\), or \(2\). In the exponent-\(4\) case, choose \(r\) of order \(4\) and anelement outside \(\langle r\rangle\); multiplying the latter by \(r^{-1}\)when its square is \(r^2\) produces an outside involution and hence thedirect product. A nonabelian group of order \(8\) has an element \(r\) oforder \(4\), since a group in which every element squares to \(1\) is abelian.For \(s\notin\langle r\rangle\), conjugation must invert \(r\), and\(s^2\in\{1,r^2\}\). These give exactly the displayed presentations of\(D_8\) and \(Q_8\).
5.3. A normal subgroup \(N\cong C_{15}\) is precisely the set of allodd-order elements of \(G\): the quotient is a \(2\)-group, and every elementof \(N\) has odd order. Thus \(N\) is characteristic, and step 1 splits theextension. Two actions on the same \(P\) give isomorphic groups preciselywhen they differ by precomposition with an automorphism of \(P\).Indeed an isomorphism induces automorphisms of \(N\) and the quotient \(P\);the former conjugates the action, but \(\operatorname{Aut}(N)\) is abelian,so this conjugation does nothing. Conversely, a change of generators in\(P\) realizing the precomposition gives an isomorphism of semidirectproducts. Different types of \(P\) cannot give isomorphic quotients.
5.4. Each row of the table lists one representative of each\(\operatorname{Aut}(P)\)-orbit of actions. For \(C_8\), changing a generator multiplies its imageby an odd integer, hence identifies \(R\) with \(-R\); this gives its sixlisted images. For \(C_4\times C_2\), every automorphism has the form\[ r\longmapsto r^{\varepsilon}s^e,\qquad s\longmapsto r^{2f}s, \quad \varepsilon\in\{1,3\},\ e,f\in\{0,1\}. \]Consequently \((R,S)\) changes to \((\varepsilon R+eS,2fR+S)\).If \(2R=0\), \(S\) stays fixed and \(R\) is taken modulo \(\langle S\rangle\);these are the first ten tuples for this \(P\). If \(R\) has order \(4\), then\(2R=t\). When \(S\in\{0,t\}\), one can set \(S=0\) and reduce \(R\) to either\(v\) or \(u+v\). When \(S\in\{u,u+t\}\), the same operations reduce thepair to \((v,u)\). These are three further, distinct orbits.
For \(C_2^3\), all actions are linear maps to \(\mathcal A[2]\). Changing a basis of the domain identifies two maps exactly when they have the same image; there are the zero subspace, three lines, and the whole plane. For \(D_8\), the relations force \(R,S\in\mathcal A[2]\). Its automorphisms send \(r\) to \(r^{\pm1}\) and \(s\) to \(r^k s\), so \(R\) stays fixed and \(S\) is taken modulo \(\langle R\rangle\), giving the ten listed pairs. Finally, every homomorphism from \(Q_8\) factors through its abelianization \(C_2^2\). Automorphisms permuting the three cyclic subgroups of order \(4\) induce all of \(\operatorname{GL}_2(\mathbf F_2)\). Again the image subspace classifies the action, giving five pairs. This proves both exhaustiveness and irredundancy of the table.
6. The remaining solvable groups of order \(120\) are exactly\[ C_5\times\mathrm{SL}_2(\mathbf F_3),\quad C_5\times S_4,\quad C_5\rtimes_{\mathrm{sgn}}S_4,\quad C_{10}\times A_4,\quad D_{10}\times A_4. \]In the third group, even permutations act trivially on \(C_5\) and oddpermutations act by inversion. Here \(D_{10}\) has order \(10\).
4. If \(H\) has a normal Sylow\(3\)-subgroup, its inverse image in \(G\) has order \(15\) and is cyclic:the action of \(C_3\) on \(C_5\) is trivial because \(3\nmid4\).This is step
5. Otherwise step 3 gives the three possibilities for\(H\). Their abelianizations are respectively \(C_3\), \(C_2\), and\(C_3\times C_2\) for \(\mathrm{SL}_2(\mathbf F_3)\), \(S_4\), and\(A_4\times C_2\).
For the first assertion, abelianize the presentation in step 2 with \(m=3\): \(a^4=b^3=(ab)^3=1\) implies \(a^3=a^4=1\), hence \(a=1\); the quotient onto \(A_4/V_4=C_3\) shows that the abelianization is exactly \(C_3\). Also \([S_4,S_4]=A_4\) and \([A_4,A_4]=V_4\): the respective quotients are abelian, commutators of transpositions give \(3\)-cycles, and a commutator of two \(3\)-cycles in \(A_4\) is a nonidentity double transposition whose conjugates generate \(V_4\).
A homomorphism from these abelianizations to \(\operatorname{Aut}(C_5)=C_4\) is trivial in the first case and is either trivial or the unique nontrivial map from the \(C_2\) factor in each of the other two cases. This gives exactly the five displayed groups. The quotient by the unique Sylow \(5\)-subgroup distinguishes the three types of \(H\); within each type, a trivial action cannot become nontrivial under an automorphism. None has a normal subgroup of order \(15\), since its image in \(H\) would be a normal subgroup of order \(3\).
7. The nonsolvable groups of order \(120\) are exactly\[ S_5,\qquad C_2\times A_5,\qquad \mathrm{SL}_2(\mathbf F_5). \]Thus part (a) has \(39+5+3=47\) isomorphism classes. Its abelian membersare \(C_{120}\), \(C_{60}\times C_2\), and \(C_{30}\times C_2^2\).
7.1. The only nonabelian simple group whose order divides \(120\) is \(A_5\).Here are the order exclusions. Prime-power orders have nontrivial center.Orders \(6,10,15,20,40\) have a normal Sylow \(3\)- or \(5\)-subgroup.For orders \(12\) and \(24\), a nonnormal Sylow \(3\)-subgroup gives an actionon four such subgroups. A simple group would embed in \(A_4\), which isimpossible for order \(24\) and would identify the order-\(12\) group withthe nonsimple group \(A_4\). For order \(30\), absence of normal Sylow\(3\)- and \(5\)-subgroups would require \(10\cdot2+6\cdot4=44\)nonidentity elements of orders \(3\) and \(5\).
For a simple group of order \(60\), Sylow theory gives \(n_5=6\) and \(n_3\in\{4,10\}\). The value \(n_3=4\) would embed it in \(A_4\), so \(n_3=10\). These account for \(24+20=44\) nonidentity elements. The possibilities for \(n_2\) are \(3,5,15\); \(3\) is excluded by the degree-\(3\) conjugation action, and \(5\) embeds the group in \(A_5\), proving the required isomorphism. If \(n_2=15\) and the Sylow \(2\)-subgroups were cyclic, their \(30\) distinct generators of order \(4\) would exceed the \(15\) remaining nonidentity elements. Thus they are Klein four groups. For an involution \(t\), its centralizer has order \(4,12\), or \(20\); order \(60\) is excluded by the trivial center. Order \(20\) gives an impossible faithful action of degree \(3\), while order \(12\) gives an embedding in \(A_5\). If all these centralizers have order \(4\), each involution belongs to a unique Sylow \(2\)-subgroup, so \(15\) Klein four subgroups require \(45\) distinct involutions, again impossible. This proves the order-\(60\) assertion.
Finally, a simple group of order \(120\) would have six Sylow \(5\)-subgroups and hence embed in \(A_6\). Its image would have index \(3\). The action of the simple group \(A_6\) on those three cosets is nontrivial but cannot be injective, since \(360>6\), a contradiction. These cases exhaust all divisors of \(120\). In each use of simplicity, the nontrivial permutation action is faithful and its composite with the sign homomorphism is trivial, so the image lies in the alternating group.
7.2. A composition series of a nonsolvable group of order \(120\) musttherefore have factors \(A_5\) and \(C_2\). The group has either a normalsubgroup \(C_2\) with quotient \(A_5\), or a normal subgroup \(A_5\) withquotient \(C_2\). In the first case the kernel is central, so step 2applies.
In the second case, conjugation gives \(G\to\operatorname{Aut}(A_5)\cong S_5\). To justify the last isomorphism, the five Sylow \(2\)-subgroups of \(A_5\) are the Klein four groups fixing each of the five letters. An automorphism permutes these groups. If it fixes all five, then for each \(g\in A_5\), \(g\) and its image induce the same permutation of them; the faithful conjugation action implies that the automorphism fixes \(g\). Hence \(\operatorname{Aut}(A_5)\) embeds in \(S_5\). Conversely conjugation by \(S_5\) induces all natural permutations of these five groups, so embeds \(S_5\) in the automorphism group. This proves equality.
The kernel of \(G\to\operatorname{Aut}(A_5)\) is the centralizer of its normal \(A_5\) and has trivial intersection with that subgroup. Its order is therefore \(1\) or \(2\). In the first case \(G\cong S_5\); in the second the commuting subgroups give \(G\cong A_5\times C_2\).
7.3. The three groups exist and have a section isomorphic to \(A_5\), soare nonsolvable. They are distinct: \(S_5\) has trivial center, whereasthe other two have a central involution, and\(\mathrm{SL}_2(\mathbf F_5)\) is perfect while \(C_2\times A_5\) hasabelianization \(C_2\). For perfectness, abelianize its presentation fromstep 2: \(4a=0\), \(3b=0\), and \(5a+5b=0\) imply \(a+2b=0\), then\(b=0\) and \(a=0\); also \(z=2a=0\).
5. This gives the stated threeabelian groups.
8. For part (b), put \(R=\mathbf Q[x]\), \(p=x-3\), and \(q=x+5\).The isomorphism classes are\[ V_{a,b,c}=(R/(p^2))^a\oplus(R/(p))^b\oplus(R/(q))^c, \qquad a,b,c\geq0,\quad 2a+b+c=6. \]There are \(7+5+3+1=16\) classes, corresponding to \(a=0,1,2,3\).
Let \(T\) be multiplication by \(x\). Since \(p^2\) and \(q\) are relatively prime and \(p(T)^2q(T)=0\), a Bezout identity decomposes \[ V=\ker(T-3I)^2\oplus\ker(T+5I). \] Indeed, if \(h p^2+kq=1\), then \(w=h(T)p(T)^2w+k(T)q(T)w\), with the first term in \(\ker q(T)\) and the second in \(\ker p(T)^2\). Their intersection is zero by the same identity. On the first summand put \(U=T-3I\), so \(U^2=0\). Choose a basis \(e_1,\ldots,e_a\) of \(\operatorname{im}U\), extend it by \(f_1,\ldots,f_b\) to a basis of \(\ker U\), and choose \(w_i\) with \(Uw_i=e_i\). The vectors \(w_i,e_i,f_j\) form a basis: applying \(U\) proves independence of the \(w_i\) modulo the kernel, and subtracting the appropriate \(w_i\) from any vector puts it in the kernel. Each pair \(w_i,e_i\) gives \(R/(p^2)\) and each \(f_j\) gives \(R/(p)\). On the other summand \(T=-5I\), giving \(c\) copies of \(R/(q)\).
The dimension condition is exactly \(2a+b+c=6\). Conversely every displayed module has that dimension and is annihilated by \(p^2q\). The dimensions of the two primary summands and the rank of \(U\) recover \(a,b,c\), so distinct triples give nonisomorphic modules.
9. For part (c), let \(C(f)\) denote the companion matrix of a monicpolynomial \(f\): it is the matrix of multiplication by \(x\) on \(R/(f)\)in the ordered basis \(1,x,\ldots,x^{\deg f-1}\). The six possiblerational canonical forms are the block diagonal matrices with thefollowing ordered companion blocks.
| \((a,b,c)\) | Invariant factors, in divisibility order |
|---|---|
| \((1,0,4)\) | \(q,\ q,\ q,\ p^2q\) |
| \((1,1,3)\) | \(q,\ pq,\ p^2q\) |
| \((1,2,2)\) | \(p,\ pq,\ p^2q\) |
| \((1,3,1)\) | \(p,\ p,\ p,\ p^2q\) |
| \((2,0,2)\) | \(p^2q,\ p^2q\) |
| \((2,1,1)\) | \(p,\ p^2,\ p^2q\) |
In step 8 the minimal polynomial is exactly \(p^2q\) precisely when \(a\geq1\) and \(c\geq1\). The nonnegative solutions of \(2a+b+c=6\) with these inequalities are exactly the six triples displayed. For each, combine the primary cyclic factors by \(R/(p^i)\oplus R/(q)\cong R/(p^iq)\), arranging the \(p\)-powers in increasing order and pairing the \(q\)-factors with the rightmost ones. This gives each listed divisibility chain. Its degrees sum to \(6\) and its largest factor is \(p^2q\), so its companion blocks have the required size and minimal polynomial. Distinct triples were proved nonisomorphic in step 8, and therefore give distinct similarity classes. Every matrix in question yields one of those modules, proving completeness.