Prove the following.
Theorem. Let \(G = G_1 * G_2\), where \(G_1\) and \(G_2\) are cyclic of orders \(m\) and \(n\), respectively. Then \(m\) and \(n\) are uniquely determined by \(G\).
Show \(G/[G, G]\) has order \(mn\).
Determine the largest integer \(k\) such that \(G\) has an element of order \(k\). (See E-175IG.)
Prove the theorem.
Solution
Goal: Prove that the orders \(m\) and \(n\) of the cyclic factors in the free product \(G = (\mathbb{Z}/m\mathbb{Z}) * (\mathbb{Z}/n\mathbb{Z})\) are uniquely determined by the isomorphism class of \(G\).
1. Part (a): Order of the abelianization \(G/[G, G]\).Proof:
1.1. By the abelianization property of free products (Theorem 69.2), the abelianization of a free product is isomorphic to the direct sum of the abelianizations of its factors:\[G / [G, G] \cong G_1^{\text{ab}} \oplus G_2^{\text{ab}}.\]
1.2. Since \(G_1 \cong \mathbb{Z}/m\mathbb{Z}\) and \(G_2 \cong \mathbb{Z}/n\mathbb{Z}\) are abelian, \(G_1^{\text{ab}} = G_1\) and \(G_2^{\text{ab}} = G_2\).
1.3. Thus \(G / [G, G] \cong (\mathbb{Z}/m\mathbb{Z}) \oplus (\mathbb{Z}/n\mathbb{Z})\).
1.4. The order of this finite abelian group is \(|G / [G, G]| = |\mathbb{Z}/m\mathbb{Z}| \cdot |\mathbb{Z}/n\mathbb{Z}| = mn\).
2. Part (b): Maximum finite order of an element in \(G\).The largest finite order \(k\) of an element in \(G\) is \(k = \max(m, n)\).Proof:
2.1. By the standard classification of torsion elements in free products (Kurosh Subgroup Theorem / E-175IG), every element of finite order in \(G_1 * G_2\) is conjugate to an element of \(G_1\) or an element of \(G_2\).
2.2. Conjugate elements have identical orders.
2.3. The elements of \(G_1\) have orders dividing \(m\), with the generator having order \(m\).
2.4. The elements of \(G_2\) have orders dividing \(n\), with the generator having order \(n\).
2.5. Any element of \(G\) not conjugate to an element of a factor is cyclically reduced of syllable length \(\ge 2\), hence has infinite order.
2.6. Therefore, the set of finite orders of elements in \(G\) is \(\{d \in \mathbb{Z}_+ \mid d \mid m \text{ or } d \mid n\}\), whose maximum element is \(k = \max(m, n)\).
3. Part (c): Unique determination of \(m\) and \(n\).Proof:
3.1. The abstract group structure of \(G\) determines two algebraic invariants:- The order of its abelianization: \(P = |G / [G, G]| = mn\).- The maximum finite order of an element in \(G\): \(M = \max(m, n)\).
3.2. Assuming without loss of generality that \(m \ge n\), we have:\[m = M \quad \text{and} \quad n = \frac{P}{M}.\]
3.3. Thus the multiset \(\{m, n\}\) is uniquely determined by the group invariants \(P\) and \(M\) of \(G\). Q.E.D.