Let \(G = G_1 * G_2\), where \(G_1\) and \(G_2\) are nontrivial groups.
Show that \(G\) is not abelian.
If \(x \in G\), define the length of \(x\) to be the length of the unique reduced word in the elements of \(G_1\) and \(G_2\) that represents \(x\). Show that if \(x\) has even length (at least 2), then \(x\) does not have finite order. Show that if \(x\) has odd length, then \(x\) is conjugate to an element of shorter length.
Show that the only elements of \(G\) that have finite order are the elements of \(G_1\) and \(G_2\) that have finite order, and their conjugates.
Solution
Goal: Prove non-commutativity, word-length reduction under conjugation, and the classification of torsion elements in a free product \(G = G_1 \ast G_2\).
1. Part (a): \(G_1 \ast G_2\) is not abelian.Proof:
1.1. Since \(G_1\) and \(G_2\) are nontrivial, choose \(g_1 \in G_1 \setminus \{e\}\) and \(g_2 \in G_2 \setminus \{e\}\).
1.2. The words \(w_1 = g_1 g_2\) and \(w_2 = g_2 g_1\) are both reduced words of length 2.
1.3. By the Normal Form Theorem for free products, two reduced words represent the same element of \(G_1 \ast G_2\) if and only if they are identical letter-by-letter.
1.4. The first letter of \(w_1\) is \(g_1 \in G_1\), while the first letter of \(w_2\) is \(g_2 \in G_2\). Since \(G_1 \cap G_2 = \{e\}\) and \(g_1, g_2 \neq e\), \(w_1 \neq w_2\).
1.5. Thus \(g_1 g_2 \neq g_2 g_1\), so \(G\) is non-abelian.
2. Part (b): Order of even length words and shortening of odd length words.Proof:
2.1. Even length words have infinite order:- Let \(x = x_1 x_2 \cdots x_{2k}\) with \(2k \ge 2\) be a reduced word of even length.- The letters alternate between \(G_1 \setminus \{e\}\) and \(G_2 \setminus \{e\}\). Because the length \(2k\) is even, \(x_1\) and \(x_{2k}\) belong to different factors (e.g. \(x_1 \in G_1\), \(x_{2k} \in G_2\)).- For any integer \(m \ge 1\), the power \(x^m = (x_1 \cdots x_{2k})(x_1 \cdots x_{2k}) \cdots (x_1 \cdots x_{2k})\) contains no adjacent letters from the same factor at the junctions between copies of \(x\).- Thus \(x^m\) is in reduced form with length \(2km \ge 2m > 0\).- Therefore \(x^m \neq e\) for all \(m \ge 1\), so \(x\) has infinite order.
2.2. Odd length words conjugate to shorter words:- Let \(x = x_1 x_2 \cdots x_{2k+1}\) with \(2k+1 \ge 3\) be a reduced word of odd length.- Since the length is odd, the first letter \(x_1\) and last letter \(x_{2k+1}\) belong to the same factor (say \(G_1\)).- Conjugate \(x\) by \(x_1\):\[y = x_1^{-1} x x_1 = x_2 x_3 \cdots x_{2k} (x_{2k+1} x_1).\]- The product \(h = x_{2k+1} x_1\) is an element of \(G_1\).- If \(h \neq e\), then \(y = x_2 x_3 \cdots x_{2k} h\) is a reduced word of length \(2k < 2k+1\).- If \(h = e\), then \(y = x_2 x_3 \cdots x_{2k}\) is a reduced word of length \(2k-1 < 2k+1\).- In either case, \(x\) is conjugate to an element \(y\) of strictly shorter length.
3. Part (c): Torsion elements are conjugates of torsion elements in factors.Proof:
3.1. Let \(x \in G_1 \ast G_2\) be a non-trivial element of finite order.
3.2. By 2, \(x\) cannot have even length \(\ge 2\).
3.3. If \(x\) has odd length \(\ge 3\), 2 shows that \(x\) is conjugate to an element \(x'\) of strictly shorter length. Since conjugate elements have the same order, \(x'\) also has finite order.
3.4. Applying this length reduction repeatedly, \(x\) is conjugate to an element \(z\) of length 1 having finite order.
3.5. A reduced word of length 1 is an element of \(G_1 \setminus \{e\}\) or \(G_2 \setminus \{e\}\).
3.6. Therefore, every element of finite order in \(G\) is a conjugate of a finite order element of \(G_1\) or \(G_2\). Q.E.D.