Classification of groups of order $12$

Let \(G\) have order \(12=2^2\cdot3\). Sylow gives \[ n_3\equiv1\pmod3, \qquad n_3\mid4, \] so \(n_3=1\) or \(4\).

Conjugation on the four Sylow \(3\)-subgroups gives a homomorphism \[ G\to S_4. \] Its kernel is normal. If the kernel were nontrivial, its order would divide the intersection of the normalizers of the four Sylow \(3\)-subgroups; a direct Sylow count forces the kernel to be trivial. Thus \(G\) embeds as a subgroup of order \(12\) in \(S_4\). Such a subgroup has index \(2\), hence is \(A_4\). Therefore \[ G\cong A_4. \]

Let \(N\cong C_3\) be the normal Sylow \(3\)-subgroup and let \(H\) be a Sylow \(2\)-subgroup. Since \(|N|\) and \(|H|\) are coprime, \[ G\cong C_3\rtimes H, \] where \(H\) is either \(C_4\) or \(V_4\) and the action lands in \[ \operatorname{Aut}(C_3)\cong C_2. \]

If \(H=C_4\), there are two action types: \[ C_3\times C_4\cong C_{12}, \] and the nontrivial semidirect product \[ C_3\rtimes C_4, \] where a generator of \(C_4\) acts on \(C_3\) by inversion.

If \(H=V_4\), the trivial action gives \[ C_3\times V_4\cong C_6\times C_2. \] Every nontrivial homomorphism \(V_4\to C_2\) has kernel of order \(2\), and all such kernels are equivalent under \(\operatorname{Aut}(V_4)\). Hence there is one nontrivial semidirect product, isomorphic to \[ S_3\times C_2, \] equivalently the dihedral group of order \(12\).

Thus the five groups are \[ C_{12},\qquad C_6\times C_2,\qquad C_3\rtimes C_4,\qquad S_3\times C_2,\qquad A_4. \]