Classify all groups of order 14 up to isomorphism. Prove your classification.
Solution
Let \(G\) have order \(14\).
1. The Sylow \(7\)-subgroup \(P\) is unique and hence normal.
Its number \(n_7\) satisfies \[ n_7\mid2, \qquad n_7\equiv1\pmod7. \] Thus \(n_7=1\). Since \(|P|=7\), we have \(P\cong C_7\).
2. If \(Q\) is a Sylow \(2\)-subgroup, then\[ G\cong C_7\rtimes C_2. \]
We have \(|Q|=2\), \(P\cap Q=1\), and \(|PQ|=14\), so \(G=PQ\). Since \(P\trianglelefteq G\), this is an internal semidirect product.
3. There are exactly two possible actions \(C_2\to\Aut(C_7)\).
Since \[ \Aut(C_7)\cong(\ZZ/7\ZZ)^\times\cong C_6, \] the image of the nontrivial element of \(C_2\) must have order dividing \(2\). The only possibilities are the identity and inversion \(y\mapsto y^{-1}\).
For the trivial action, \[ G\cong C_7\times C_2\cong C_{14}. \] For inversion, \[ G\cong \langle y,x\mid y^7=x^2=1,\;xyx^{-1}=y^{-1}\rangle, \] which is the dihedral group \(D_7\) of order \(14\).
Hence, up to isomorphism, the groups of order \(14\) are exactly \[ \boxed{C_{14}\quad\text{and}\quad D_7.} \]