Classify the five groups of order 12.
Solution
Let \(G\) be a group of order \(12=2^2\cdot3\).
1. At least one Sylow subgroup of \(G\) is normal.
Sylow’s theorems give \[ n_3\in\{1,4\},\qquad n_2\in\{1,3\}. \] If \(n_3=4\), the four Sylow \(3\)-subgroups contribute \(4(3-1)=8\) nonidentity elements. The remaining four elements contain every Sylow \(2\)-subgroup, so there is only one Sylow \(2\)-subgroup and \(n_2=1\). Thus either \(n_3=1\) or \(n_2=1\).
2. Hence \(G\) is a semidirect product of a group of order \(3\) and a group of order \(4\).
Let \(P_3\in\operatorname{Syl}_3(G)\) and \(P_2\in\operatorname{Syl}_2(G)\). Their intersection is trivial and \[ |P_3P_2|=\frac{|P_3||P_2|}{|P_3\cap P_2|}=12, \] so \(G=P_3P_2\). By 1 one of the two factors is normal. Also \(P_3\cong C_3\), while \(P_2\cong C_4\) or \(V_4\).
3. If the Sylow \(2\)-subgroup is normal, the possibilities are \(C_{12}\), \(C_3\times V_4\), and \(A_4\).
If \(P_2\cong C_4\), an action \(C_3\to\operatorname{Aut}(C_4)\cong C_2\) must be trivial, so \[ G\cong C_4\times C_3\cong C_{12}. \] If \(P_2\cong V_4\), then \(\operatorname{Aut}(V_4)\cong S_3\). The action of \(C_3\) is either trivial, giving \(C_3\times V_4\), or injective onto a subgroup generated by a \(3\)-cycle on the three nonidentity elements of \(V_4\). All such subgroups are conjugate in \(S_3\), and the resulting semidirect product is \(A_4\).
4. If the Sylow \(3\)-subgroup is normal, the new possibilities are \(C_3\rtimes C_4\) with inversion action and \(C_2\times S_3\).
Now the action is a homomorphism \[ P_2\longrightarrow\operatorname{Aut}(C_3)\cong C_2. \] For \(P_2\cong C_4\), the trivial action again gives \(C_{12}\), while the unique nontrivial action sends a generator of \(C_4\) to inversion on \(C_3\). This gives the nontrivial semidirect product \[ C_3\rtimes C_4, \] the dicyclic group of order \(12\) (often denoted \(\operatorname{Dic}_3\) or \(Q_{12}\)).
For \(P_2\cong V_4\), the trivial action again gives \(C_3\times V_4\). Every nontrivial homomorphism \(V_4\to C_2\) has kernel of order \(2\), and the three such kernels are carried to one another by automorphisms of \(V_4\). Thus there is one nontrivial semidirect-product type. Writing \(V_4=\langle a\rangle\times\langle b\rangle\) with \(a\) acting by inversion and \(b\) acting trivially gives \[ G\cong (C_3\rtimes\langle a\rangle)\times\langle b\rangle \cong S_3\times C_2, \] which is the dihedral group \(D_6\) of order \(12\).
5. These five groups are pairwise nonisomorphic.
The two groups \(C_{12}\) and \(C_3\times V_4\cong C_6\times C_2\) are abelian and are distinguished by the existence of an element of order \(12\). The other three are nonabelian. Among them, \(A_4\) has no element of order \(6\), \(D_6\) has an element of order \(6\) but none of order \(4\), and \(\operatorname{Dic}_3\) has an element of order \(4\). Hence they are pairwise nonisomorphic.
6. Therefore the groups of order \(12\), up to isomorphism, are\[ C_{12},\qquad C_6\times C_2,\qquad A_4,\qquad D_6,\qquad \operatorname{Dic}_3. \]
By <1>2–<1>4 every group of order \(12\) is one of these, and by 5 no two on the list are isomorphic.