A NOTE ON A COUNTER EXAMPLE IN FINITE GROUPS

Kandasamy Muthuvel

University of Wisconsin–Oshkosh

It is known that if G is a finite abelian group and the number of elements of order 2 is not equal to 1, then the product of all elements of G is the identity (refer to p. 78, problem 43 of [2]). Garimella [1] gave a counter example to show that the conclusion of the above statement is not true for a non–abelian group of order 21. In this short note, we show that if G is a finite non–abelian group, then there is an arrangement x_{1}, x_{2}, …, x_{n} of all elements of G such that the product x_{1}x_{2} ⋯ x_{n} is a non–identity. The proof of this follows by finding elements a and b of G, say a = x_{1} and b = x_{2}, such that ab ≠ ba and abx_{3} ⋯ x_{n} or bax_{3} ⋯ x_{n} is a non–identity. In fact if n is odd, we find such an arrangement by defining x_{1} = a, x_{2} = b, x_{3} = a^{−1}, x_{4} = b^{−1}, x_{2i} = x_{2i−1}^{−1} for 3 ≤ i ≤ (n − 1)/2, and x_{n} = e.

On the other hand if G is a finite group such that

S = {x ∈ G : x^{2} = e}

is a subgroup of G and |S| ≠ 2, (in particular, if G is a group of odd order), there is an arrangement x_{1}, x_{2}, …, x_{n} of all elements of G such that the product x_{1}x_{2}x_{3} … x_{n} is the identity (to prove this first, show that S is abelian, use p.78, problem 43 of [2] and the idea used in this note). The above result is not true for any finite group. For example, the product of all elements of S_{3} (the set of all permutations of {1, 2, 3}) in any order is a non–identity.

\underline{References}

1. R. Garimella, “A Counter Example in Group Theory,” Missouri Journal of Mathematical Sciences, 3 (1991), 77–78.

2. I. N. Herstein, Abstract Algebra, 2nd ed., Macmillan, New York, 1990.