Classify groups of order 14. Why is there a group of order 7? Are all index-2 subgroups normal?
Solution
Let \(|G|=14=2\cdot7\).
1. There is a unique subgroup \(N\) of order \(7\), and it is normal.
Sylow gives \[ n_7\mid2, \qquad n_7\equiv1\pmod7. \] Hence \(n_7=1\). The subgroup exists by Sylow’s first theorem and is cyclic because it has prime order.
2. Every subgroup of index \(2\) in any group is normal.
If \(H\le G\) has index \(2\), there are exactly two left cosets and two right cosets. Both decompositions contain \(H\) itself, so the unique remaining left coset equals the unique remaining right coset. Hence \(gH=Hg\) for every \(g\), so \(H\trianglelefteq G\).
3. Let \(P\) be a Sylow \(2\)-subgroup. Then\[ G\cong C_7\rtimes C_2. \]
The subgroup \(N\cong C_7\) is normal, \(P\cong C_2\), and \(N\cap P=1\). Their product has order \(14\), so it is all of \(G\).
4. There are exactly two isomorphism types of groups of order \(14\).
The action in the semidirect product is a homomorphism \[ C_2\to\operatorname{Aut}(C_7)\cong C_6. \] There are two possibilities for the image: trivial, or the unique subgroup of order \(2\).
The trivial action gives \[ C_7\times C_2\cong C_{14}. \] The nontrivial action is inversion on \(C_7\) and gives the dihedral group of order \(14\).
Thus the groups of order \(14\) are exactly \(C_{14}\) and the dihedral group \(D_{14}\) (often denoted \(D_7\) when the subscript records the polygon rather than the group order).