Groups of order 12, and two nonisomorphic subgroups of the same order

There are five groups of order \(12\) up to isomorphism: \[ C_{12},\qquad C_6\times C_2,\qquad A_4,\qquad D_{12},\qquad \operatorname{Dic}_3. \] Here \(D_{12}\) denotes the dihedral group of order \(12\), and \[ \operatorname{Dic}_3 =\langle a,b\mid a^6=1,\ b^2=a^3,\ bab^{-1}=a^{-1}\rangle. \]

To see why these are the possibilities, let \(G\) have order \(12=2^2\cdot3\). Sylow gives \[ n_3\in\{1,4\},\qquad n_2\in\{1,3\}. \] If \(n_3=4\), the four Sylow \(3\)-subgroups contribute eight nonidentity elements, leaving exactly four elements; those four form the unique Sylow \(2\)-subgroup, so it is normal. This case yields \(A_4\).

If \(n_3=1\), let \(Q\cong C_3\) be the normal Sylow \(3\)-subgroup and let \(P\) be a Sylow \(2\)-subgroup. Then \[ G\cong C_3\rtimes P, \] with \(P\cong C_4\) or \(V_4\). Since \[ \operatorname{Aut}(C_3)\cong C_2, \] the possible actions are the trivial action and the unique nontrivial action through a quotient of order \(2\). For \(P=C_4\) these give \(C_{12}\) and \(\operatorname{Dic}_3\); for \(P=V_4\) they give \(C_6\times C_2\) and \(D_{12}\cong C_2\times S_3\). Thus there are exactly five isomorphism classes.

Yes, a group of order \(12\) can have two nonisomorphic subgroups of the same order. In \[ D_{12}=\langle r,s\mid r^6=s^2=1,\ srs=r^{-1}\rangle, \] the subgroup \[ \langle r\rangle\cong C_6 \] has order \(6\), while \[ \langle r^2,s\rangle\cong S_3 \] also has order \(6\). One is abelian and the other is not, so they are not isomorphic.