Sylow theorems, and groups of order 1225

The Sylow theorems state:

1. If \(p^a\) is the largest power of \(p\) dividing \(|G|\), then \(G\) has a subgroup of order \(p^a\).
2. Every \(p\)-subgroup is contained in a Sylow \(p\)-subgroup, and all Sylow \(p\)-subgroups are conjugate.
3. The number \(n_p\) of Sylow \(p\)-subgroups satisfies \[ n_p\equiv1\pmod p, \qquad n_p\mid |G|/p^a. \]

Now \[ 1225=5^2\cdot7^2. \] For Sylow \(7\)-subgroups, \[ n_7\equiv1\pmod7, \qquad n_7\mid25. \] The divisors \(1,5,25\) leave only \(n_7=1\). Similarly \[ n_5\equiv1\pmod5, \qquad n_5\mid49, \] and among \(1,7,49\) only \(1\) is congruent to \(1\) modulo \(5\). Thus both Sylow subgroups \(P_5\) and \(P_7\) are normal.

Their intersection is trivial because their orders are coprime. Since both are normal, for \(x\in P_5\) and \(y\in P_7\) the commutator \([x,y]\) lies in both subgroups, hence is \(1\). Therefore \[ G=P_5\times P_7. \] Every group of order \(p^2\) is abelian, so both factors are abelian and therefore \(G\) is abelian.

The abelian groups of order \(1225\) are obtained by choosing one of the two abelian types of order \(25\) and independently one of the two types of order \(49\). Thus exactly one representative of each isomorphism class is \[ C_{25}\times C_{49}, \qquad C_{25}\times C_7\times C_7, \qquad C_5\times C_5\times C_{49}, \qquad C_5\times C_5\times C_7\times C_7. \]