Use the Class Equation to prove that any finite \(p\)-group (a group of order \(p^k\) for a prime \(p\) and integer \(k \ge 1\)) has a non-trivial center.
Prove that any group of order \(p^2\) (where \(p\) is prime) is abelian.
Prove that any group of order \(5^2 \cdot 7^2 = 1225\) is abelian.
Write down exactly one representative from each isomorphism class of groups of order \(5^2 \cdot 7^2\).
Solution
Goal: Prove non-triviality of the center for \(p\)-groups via the class equation in (a), deduce abelianness of order \(p^2\) in (b), establish direct product decomposition of order \(5^2 \cdot 7^2\) via Sylow theory in (c), and enumerate isomorphism classes in (d).
1. Part (a): Finite \(p\)-groups have non-trivial centers.
1.1. Let \(G\) be a group of order \(|G| = p^k\) with \(k \ge 1\).
1.2. \(G\) acts on itself by conjugation (\(g \cdot x = g x g^{-1}\)).
1.3. The orbit of \(x \in G\) is the conjugacy class \(\operatorname{Cl}(x)\), and the stabilizer of \(x\) is the centralizer \(C_G(x)\).
1.4. By the Orbit-Stabilizer Theorem, \(|\operatorname{Cl}(x)| = [G : C_G(x)]\).
1.5. The conjugacy class has size 1 if and only if \(g x g^{-1} = x\) for all \(g \in G\), which is equivalent to \(x \in Z(G)\).
1.6. Partitioning \(G\) into disjoint conjugacy classes yields the Class Equation:\[|G| = |Z(G)| + \sum_{i=1}^m [G : C_G(x_i)],\]where \(x_1, \dots, x_m\) are representatives of the conjugacy classes of size strictly greater than 1.
1.7. For each \(i \in \{1, \dots, m\}\), \([G : C_G(x_i)] > 1\) divides \(|G| = p^k\), so \(p\) divides \([G : C_G(x_i)]\).
1.8. Rearranging the Class Equation:\[|Z(G)| = |G| - \sum_{i=1}^m [G : C_G(x_i)].\]
1.9. Since \(p \mid |G|\) and \(p \mid [G : C_G(x_i)]\) for each \(i\), \(p\) divides \(|Z(G)|\).
1.10. Since \(e \in Z(G)\), \(|Z(G)| \ge 1\). Since \(p \mid |Z(G)|\), \(|Z(G)| \ge p > 1\).
1.11. Thus \(Z(G)\) is non-trivial.
2. Part (b): Every group of order \(p^2\) is abelian.
2.1. Let \(|G| = p^2\). By Part (a), \(|Z(G)| \in \{p, p^2\}\).
2.2. If \(|Z(G)| = p^2\), then \(Z(G) = G\), so \(G\) is abelian.
2.3. If \(|Z(G)| = p\), then the quotient group \(G/Z(G)\) has order \(|G/Z(G)| = p^2 / p = p\).
2.4. Every group of prime order is cyclic, so \(G/Z(G)\) is cyclic, generated by some coset \(x Z(G)\).
2.5. Lemma: If \(G/Z(G)\) is cyclic, then \(G\) is abelian:- Every element \(g \in G\) can be written as \(g = x^a z_1\) for some \(a \in \mathbb{Z}\) and \(z_1 \in Z(G)\).- For any two elements \(g = x^a z_1\) and \(h = x^b z_2\):\[g h = (x^a z_1)(x^b z_2) = x^a x^b z_1 z_2 = x^{a+b} z_2 z_1 = x^b x^a z_2 z_1 = (x^b z_2)(x^a z_1) = h g.\]- Thus \(G\) is abelian.
2.6. But if \(G\) is abelian, \(Z(G) = G\), so \(|Z(G)| = p^2 \ne p\), a contradiction.
2.7. Therefore \(|Z(G)| = p^2\), which proves \(G\) is abelian.
3. Part (c): Every group of order \(5^2 \cdot 7^2\) is abelian.
3.1. Let \(|G| = 5^2 \cdot 7^2 = 1225\).
3.2. By Sylow’s Theorems:- Number of Sylow 5-subgroups: \(n_5 \equiv 1 \pmod 5\) and \(n_5 \mid 7^2 = 49\).Divisors of 49 are \(\{1, 7, 49\}\). Among these, only \(1 \equiv 1 \pmod 5\), so \(n_5 = 1\).- Number of Sylow 7-subgroups: \(n_7 \equiv 1 \pmod 7\) and \(n_7 \mid 5^2 = 25\).Divisors of 25 are \(\{1, 5, 25\}\). Among these, only \(1 \equiv 1 \pmod 7\), so \(n_7 = 1\).
3.3. Let \(P\) be the unique Sylow 5-subgroup and \(Q\) be the unique Sylow 7-subgroup.
3.4. Both \(P\) and \(Q\) are normal in \(G\) (\(P \trianglelefteq G, Q \trianglelefteq G\)).
3.5. Since \(\gcd(|P|, |Q|) = \gcd(25, 49) = 1\), \(P \cap Q = \{e\}\).
3.6. Since \(|P Q| = |P||Q| = 25 \cdot 49 = 1225 = |G|\), \(G\) is the internal direct product:\[G \cong P \times Q.\]
3.7. By Part (b), since \(|P| = 5^2\) and \(|Q| = 7^2\), both \(P\) and \(Q\) are abelian.
3.8. The direct product of abelian groups is abelian, so \(G\) is abelian.
4. Part (d): Isomorphism classes of groups of order \(5^2 \cdot 7^2\).
4.1. By Part (c), every group of order \(5^2 \cdot 7^2\) is abelian of the form \(P \times Q\), where \(P\) has order 25 and \(Q\) has order 49.
4.2. By the Fundamental Theorem of Finite Abelian Groups:- Abelian groups of order \(5^2 = 25\): \(\mathbb{Z}/25\mathbb{Z}\) and \(\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}\).- Abelian groups of order \(7^2 = 49\): \(\mathbb{Z}/49\mathbb{Z}\) and \(\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z}\).
4.3. Combining these yields exactly \(2 \times 2 = 4\) non-isomorphic abelian groups:1. \(\mathbb{Z}/25\mathbb{Z} \times \mathbb{Z}/49\mathbb{Z} \cong \mathbb{Z}/1225\mathbb{Z}\)2. \((\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}) \times \mathbb{Z}/49\mathbb{Z}\)3. \(\mathbb{Z}/25\mathbb{Z} \times (\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z})\)4. \((\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}) \times (\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/7\mathbb{Z})\)
5. Conclusion:
$p$-groups have $|Z(G)| \ge p$, order $p^2$ groups are abelian, order $5^2 \cdot 7^2$ groups decompose into $P \times Q$ (hence abelian), with exactly 4 isomorphism classes.