Show that every group of order \(p^2\) with \(p\) prime is abelian.
State the 3 Sylow theorems.
Show that any group of order \(4225 = 5^2 13^2\) is abelian.
Write down one representative from each isomorphism class of abelian groups of order 4225.
Solution
(a).
1. Let \(G\) have order \(p^2\). Then \(Z(G) \neq 1\) (a \(p\)-group has nontrivial center).
class equation.
2. \(G/Z(G)\) has order \(1\) or \(p\) (since \(|Z(G)|\) is \(p\) or \(p^2\)).
1 and Lagrange.
3. If \(|G/Z(G)| = p\), then \(G/Z(G)\) is cyclic, which forces \(G\) abelian (a group with cyclic center quotient is abelian).
standard fact.
4. If \(|G/Z(G)| = 1\), then \(G = Z(G)\) is abelian.
2.
5. Hence \(G\) is abelian.
3 and 4.
(b).
1. Sylow 1: For each prime \(p\) dividing \(|G|\), there is a Sylow \(p\)-subgroup.Sylow 2: All Sylow \(p\)-subgroups are conjugate, and the number \(n_p\) satisfies \(n_p \equiv 1 \pmod p\) and \(n_p \mid |G|\).Sylow 3: \(n_p \equiv 1 \pmod p\) and \(n_p\) divides \(|G|/p^a\) (where \(p^a\) is the largest power of \(p\) dividing \(|G|\)).
statement of the Sylow theorems.
(c).
1. Let \(G\) have order \(4225 = 5^2 \cdot 13^2\). By Sylow, \(n_5 \equiv 1 \pmod 5\) and \(n_5 \mid 13^2 = 169\), so \(n_5 \in \{1, 169\}\).
Sylow’s theorem.
2. \(n_{13} \equiv 1 \pmod{13}\) and \(n_{13} \mid 25\), so \(n_{13} = 1\).
Sylow’s theorem (the divisors of \(25\) are \(1, 5, 25\), and only \(1 \equiv 1 \pmod{13}\)).
3. Hence \(G\) has a unique normal Sylow \(13\)-subgroup \(P \cong \ZZ/13^2\) or \(\ZZ/13 \times \ZZ/13\).
2.
4. The normal subgroup \(P\) acts by conjugation on the Sylow \(5\)-subgroups, so \(n_5\) divides \(|P| = 169\) and \(n_5 \equiv 1 \pmod 5\); the only such divisor is \(n_5 = 1\).
3 and Sylow (the orbit sizes divide \(169\), and \(n_5 \equiv 1 \pmod 5\) forces \(n_5 = 1\)).
5. Hence \(G\) has a unique normal Sylow \(5\)-subgroup \(Q\) and a unique normal Sylow \(13\)-subgroup \(P\), so \(G = P \times Q\).
3 and 4.
6. \(P\) and \(Q\) are abelian (groups of order \(p^2\) are abelian by (a)), so \(G = P \times Q\) is abelian.
5 and (a).
(d).
1. The abelian groups of order \(4225 = 5^2 \cdot 13^2\) are the products of abelian groups of order \(5^2\) and \(13^2\).
fundamental theorem of finite abelian groups.
2. The abelian groups of order \(5^2\) are \(\ZZ/25\) and \(\ZZ/5 \times \ZZ/5\); the abelian groups of order \(13^2\) are \(\ZZ/169\) and \(\ZZ/13 \times \ZZ/13\).
fundamental theorem.
3. Hence the four isomorphism classes are:\[\ZZ/25 \times \ZZ/169,\ \ZZ/25 \times \ZZ/13 \times \ZZ/13,\ \ZZ/5 \times \ZZ/5 \times \ZZ/169,\ \ZZ/5 \times \ZZ/5 \times \ZZ/13 \times \ZZ/13.\]
1 and 2.
4. Q.E.D.
5 (a), 1 (b), 6 (c), 3 (d).