Classify all groups of order \(28\).
Solution
Let \(|G|=28=4\cdot7\). Sylow gives \[ n_7\equiv1\pmod7,\qquad n_7\mid4, \] so \(n_7=1\). Thus the Sylow \(7\)-subgroup \(P\cong C_7\) is normal. If \(Q\) is a Sylow \(2\)-subgroup, then \(|Q|=4\), \(P\cap Q=1\), and \(PQ=G\). Hence \[ G\cong C_7\rtimes Q, \] where \(Q\cong C_4\) or \(C_2^2\), and the action lands in \[ \operatorname{Aut}(C_7)\cong C_6. \]
1. If \(Q\cong C_4\), there are two possibilities.
The image of \(C_4\to C_6\) has order dividing \(2\). The trivial action gives \[ C_7\times C_4\cong C_{28}. \] The unique nontrivial action sends a generator of \(C_4\) to inversion on \(C_7\), giving one nonabelian semidirect product \[ C_7\rtimes C_4. \]
2. If \(Q\cong C_2^2\), there are two possibilities.
Any image lies in the unique subgroup \(C_2\le C_6\). The trivial action gives \[ C_7\times C_2^2\cong C_{14}\times C_2. \] Every nonzero homomorphism \(C_2^2\to C_2\) is equivalent under \(\operatorname{Aut}(C_2^2)\), so there is one nontrivial semidirect product. It is \[ (C_7\rtimes C_2)\times C_2\cong D_7\times C_2. \] With the convention that \(D_m\) has order \(2m\), this group is also isomorphic to \(D_{14}\).
Thus there are exactly four isomorphism types: \[ C_{28},\qquad C_{14}\times C_2,\qquad C_7\rtimes C_4,\qquad D_7\times C_2\cong D_{14}. \] The two abelian groups are distinguished by cyclicity, and the two nonabelian groups by the isomorphism type of their Sylow \(2\)-subgroups (\(C_4\) versus \(C_2^2\)).