Classify up to isomorphism all groups of order 18. Do the same for orders 20 and 30.
Solution
We classify each order by first finding a normal Hall subgroup and then recording the possible conjugation actions of a complement.
1. Up to isomorphism there are exactly five groups of order \(18\).
Let \(|G|=18=2\cdot3^2\). By Sylow, \[ n_3\mid2,\qquad n_3\equiv1\pmod3, \] so \(n_3=1\). Thus the Sylow \(3\)-subgroup \(N\) is normal and has order \(9\). By Cauchy’s theorem \(G\) has a subgroup \(C\) of order \(2\), and \(N\cap C=1\), so \(G=N\rtimes C\).
There are two possibilities for \(N\).
If \(N\cong\ZZ_9\), then \[ \operatorname{Aut}(N)\cong(\ZZ/9\ZZ)^\times\cong\ZZ_6. \] An action of \(C\cong\ZZ_2\) therefore has image either trivial or the unique subgroup of order \(2\). These give \[ \ZZ_{18} \qquad\text{and}\qquad \ZZ_9\rtimes_{(-1)}\ZZ_2, \] the latter being the dihedral group of order \(18\).
If \(N\cong\ZZ_3^2\), then an action of \(C\) is determined by an involution in \(\operatorname{GL}_2(\FF_3)\). Since \(2\) is invertible in \(\FF_3\), every such operator is diagonalizable with eigenvalues in \(\{1,-1\}\). Up to conjugacy there are exactly three possibilities: \[ I,\qquad \operatorname{diag}(1,-1),\qquad -I. \] They give respectively \[ \ZZ_3^2\times\ZZ_2, \qquad \ZZ_3\times S_3, \qquad \ZZ_3^2\rtimes_{(-I)}\ZZ_2. \] These five groups are pairwise nonisomorphic: the two cases with cyclic Sylow \(3\)-subgroup are separated from the three with Sylow subgroup \(\ZZ_3^2\); within each case the conjugation action of an involution on that characteristic Sylow \(3\)-subgroup has different fixed-space dimension (or is trivial versus nontrivial). Hence exactly five isomorphism types occur.
2. Up to isomorphism there are exactly five groups of order \(20\).
Let \(|G|=20=2^2\cdot5\). Sylow gives \[ n_5\mid4,\qquad n_5\equiv1\pmod5, \] so \(n_5=1\). Thus the Sylow \(5\)-subgroup \[ N\cong\ZZ_5 \] is normal. If \(P\) is a Sylow \(2\)-subgroup, then \(|P|=4\), \(N\cap P=1\), and \(NP=G\). Hence \[ G\cong\ZZ_5\rtimes P, \] where \(P\cong\ZZ_4\) or \(P\cong\ZZ_2^2\).
Now \[ \operatorname{Aut}(\ZZ_5)\cong\ZZ_4. \] For \(P\cong\ZZ_4\), the image of the action homomorphism \(P\to\ZZ_4\) can have order \(1\), \(2\), or \(4\). Up to automorphisms of source and target these yield exactly three groups: \[ \ZZ_5\times\ZZ_4\cong\ZZ_{20}, \] \[ \ZZ_5\rtimes_{(\operatorname{im}=\ZZ_2)}\ZZ_4, \qquad \ZZ_5\rtimes_{(\operatorname{im}=\ZZ_4)}\ZZ_4. \]
For \(P\cong\ZZ_2^2\), the image has order \(1\) or \(2\); all nonzero maps \(\ZZ_2^2\to\ZZ_2\le\ZZ_4\) are equivalent under \(\operatorname{Aut}(\ZZ_2^2)\). Hence there are exactly two further groups: \[ \ZZ_5\times\ZZ_2^2, \qquad \ZZ_5\rtimes_{(\operatorname{im}=\ZZ_2)}\ZZ_2^2. \]
The five are pairwise nonisomorphic. First, their Sylow \(2\)-subgroups distinguish the \(\ZZ_4\) cases from the \(\ZZ_2^2\) cases. Within either family, the kernel (or equivalently the image size) of the conjugation action on the characteristic Sylow \(5\)-subgroup distinguishes the listed possibilities. Thus there are exactly five groups of order \(20\).
3. Every group of order \(30\) has a normal cyclic subgroup of order \(15\).
Let \(|G|=30=2\cdot3\cdot5\). Sylow gives \[ n_5\in\{1,6\},\qquad n_3\in\{1,10\}. \] The alternatives \(n_5=6\) and \(n_3=10\) cannot both occur, because the six Sylow \(5\)-subgroups contribute \(6(5-1)=24\) nonidentity elements and the ten Sylow \(3\)-subgroups contribute \(10(3-1)=20\) distinct nonidentity elements, already more than \(29\).
Thus at least one of the Sylow \(3\)- and \(5\)-subgroups is normal. Let \(P_5\) and \(P_3\) be Sylow subgroups. If both are normal, then \(P_3P_5\) is a subgroup of order \(15\). If only \(P_3\) is normal, then conjugation by \(P_5\) on \(P_3\) gives a homomorphism \[ P_5\to\operatorname{Aut}(P_3)\cong\ZZ_2, \] which is trivial because \(|P_5|=5\); hence \(P_5\) normalizes and centralizes \(P_3\), so again \(P_3P_5\) is a subgroup of order \(15\). Similarly, if only \(P_5\) is normal, conjugation by \(P_3\) on \(P_5\) maps a group of order \(3\) into \(\operatorname{Aut}(\ZZ_5)\cong\ZZ_4\) and is trivial, so \(P_3P_5\) is again a subgroup of order \(15\).
Call this subgroup \(N\). Since \([G:N]=2\), it is normal. Also any group of order \(15\) is cyclic: its Sylow \(5\)-subgroup is normal, and its Sylow \(3\)-subgroup acts trivially on it because there is no element of order \(3\) in \(\operatorname{Aut}(\ZZ_5)\). Hence \[ N\cong\ZZ_{15}. \]
4. Up to isomorphism there are exactly four groups of order \(30\).
By 3, \(G\) has a normal subgroup \(N\cong\ZZ_{15}\). Cauchy’s theorem supplies an involution \(t\in G\), necessarily outside the odd-order subgroup \(N\). Thus \[ G\cong\ZZ_{15}\rtimes\ZZ_2. \]
By the Chinese remainder theorem, \[ \operatorname{Aut}(\ZZ_{15}) \cong\operatorname{Aut}(\ZZ_3)\times\operatorname{Aut}(\ZZ_5) \cong\ZZ_2\times\ZZ_4. \] An action of \(\ZZ_2\) is determined by an element of order dividing \(2\). There are exactly four such elements: the identity, inversion on the \(3\)-part only, inversion on the \(5\)-part only, and inversion on both parts. Therefore the four groups are \[ \ZZ_{30}, \qquad S_3\times\ZZ_5, \qquad \ZZ_3\times(\ZZ_5\rtimes_{(-1)}\ZZ_2), \qquad \ZZ_{15}\rtimes_{(-1)}\ZZ_2. \] The last two may be described as the product of \(\ZZ_3\) with the dihedral group of order \(10\), and the dihedral group of order \(30\), respectively.
These groups are pairwise nonisomorphic because the action of an involution on the characteristic subgroups of orders \(3\) and \(5\) is respectively trivial or inversion in the four distinct combinations. Hence there are exactly four isomorphism types of groups of order \(30\).