Classify groups of order 35.
Solution
Let \(|G|=35=5\cdot7\).
1. The Sylow \(7\)-subgroup is unique and normal.
Sylow gives \[ n_7\mid5, \qquad n_7\equiv1\pmod7. \] Thus \(n_7=1\).
2. The Sylow \(5\)-subgroup is unique and normal.
Sylow gives \[ n_5\mid7, \qquad n_5\equiv1\pmod5. \] The divisors of \(7\) are \(1\) and \(7\), and \(7\not\equiv1\pmod5\). Hence \(n_5=1\).
3. Therefore \(G\cong C_{35}\).
Let \(P\) and \(Q\) be the normal Sylow subgroups of orders \(5\) and \(7\). Their intersection is trivial and \(|PQ|=35\), so \(G=PQ\). For \(p\in P\) and \(q\in Q\), the commutator lies in both \(P\) and \(Q\) because both subgroups are normal, hence is trivial. Thus \(P\) and \(Q\) commute, so \[ G\cong P\times Q\cong C_5\times C_7\cong C_{35}. \]
Thus there is exactly one group of order \(35\) up to isomorphism, and it is cyclic.