Show that the free product \(G * H\) of nontrivial groups \(G\) and \(H\) has trivial center, and that the only elements of \(G * H\) of finite order are the conjugates of finite-order elements of \(G\) and \(H\).
Solution
1. Every element of \(G * H\) has a unique reduced word form \(g_1 h_1 g_2 h_2 \cdots\) with \(g_i \in G \setminus \{1\}\), \(h_i \in H \setminus \{1\}\) (alternating, no adjacent factors from the same group).
normal form theorem for free products.
2. Let \(w \in Z(G * H)\) be a nontrivial central element, written in reduced form.
suppose the center is nontrivial.
3. If \(w\) has length \(\ge 2\), then conjugating by a nontrivial element of the group of the first factor changes the word, contradicting centrality.
e.g. if \(w\) starts with \(g_1 \in G\), then for \(h \in H \setminus \{1\}\), \(h w h^{-1}\) has a different reduced form than \(w\) (the first factor changes), so \(h w h^{-1} \ne w\).
4. If \(w\) has length \(1\), say \(w = g_1 \in G\), then for \(h \in H \setminus \{1\}\), \(h g_1 h^{-1} \ne g_1\).
\(h g_1 h^{-1}\) is a reduced word of length \(3\), not equal to the length-\(1\) word \(g_1\).
5. Hence no nontrivial element is central, so \(Z(G * H) = 1\).
<1>2–<1>4.
6. Let \(w \in G * H\) have finite order, with reduced form \(w = a_1 a_2 \cdots a_n\).
take a finite-order element.
7. If \(n \ge 2\), then \(w\) is cyclically reduced (after conjugation) and \(w^k\) has reduced length \(kn\) for all \(k \ge 1\), so \(w^k \ne 1\).
the reduced form of \(w^k\) is the concatenation of \(k\) copies of the cyclically reduced form, of length \(kn > 0\).
8. Hence \(n = 1\), so \(w\) is conjugate to an element of \(G\) or of \(H\).
7 (a finite-order element must have length \(1\), i.e. lie in a single factor, up to conjugation).
9. Therefore the finite-order elements of \(G * H\) are exactly the conjugates of finite-order elements of \(G\) and \(H\).
<1>6–<1>8.
10. Q.E.D.
5 and 9.