A COUNTER EXAMPLE IN GROUP THEORY

Ramesh Garimella

Northwest Missouri State University

In a first course on basic group theory, one of the standard problems is to show that if G = {x_{1}, x_{2}, ⋯, x_{n}} is an abelian group and n is odd then the product x_{1}x_{2} ⋯ x_{n} = e, where e is the identity element G. In this short note, we give a counter example to show that the above result is not true if we drop the ‘abelianness’ of the group. In looking for an example, we do not need to consider a group of order 3, 5, 7, 11, 13, 17, 19, because these are primes and any group of prime order is cyclic and hence abelian. Also n = 9 does not work, because it is the square of a prime and hence the group is abelian. Also by fairly standard arguments, one can see that group of order 15 is abelian. Therefore the first possible candidate for a counter example is a group of order 21. Apart from the cyclic group of order 21, there is a unique non–abelian group of order 21 (refer to p. 112, problem 11(b) of [1]). We show that this unique non–abelian group G of order 21 works as a counter example. As a matter of fact, we find an arrangement x_{1}, x_{2}, ⋯, x_{20} of non–identity elements of G such that the product x_{1}x_{2} ⋯ x_{20} is non–identity. Let a, bϵG such that order of a and b be 3 and 7 respectively and e be the identity element of G. Let

G = {e, a, a^{2}, b^{i}, ab^{i}, a^{2}b^{i} : 1 ≤ i ≤ 6}.

Since a^{−1} = a^{2}, and {e, b^{i}, 1 ≤ i ≤ 6} is a normal subgroup of G (by Sylow Theorem), and since ab ≠ ba, there exists an i, 2 ≤ i ≤ 6, such that aba^{2} = b^{i}. Now we let,

\begin{matrix}x_{1} = ab^{2}, & x_{5} = ab^{4}, & x_{9} = ab^{6}, & x_{13} = a, & x_{17} = b^{3}, \\ x_{2} = a^{2}b^{2}, & x_{6} = a^{2}b^{4}, & x_{10} = a^{2}b^{6}, & x_{14} = a^{2}b, & x_{18} = b^{4}, \\ x_{3} = ab^{3}, & x_{7} = ab^{5}, & x_{11} = a^{2}, & x_{15} = b, & x_{19} = b^{5}, \\ x_{4} = a^{2}b^{3}, & x_{8} = a^{2}b^{5}, & x_{12} = ab, & x_{16} = b^{2}, & x_{20} = b^{6}. \end{matrix}

Since aba^{2} = b^{i}, the product

x_{1}x_{2} ⋯ x_{10} = b^{2i+2}b^{3i+3} ⋯ b^{6i+6}
= b^{i(2+3+···+6)+(2+3+···+6)}
= b^{20i+20} = b^{20(i+1)}.

Also since a^{3} = e and b^{7} = e, the product x_{11}x_{12} ⋯ x_{20} = b^{2}. Hence the product x_{1}x_{2} ⋯ x_{20} = b^{20(i+1)+2}. Since 7 does not divide 20(i + 1) + 2 for 2 ≤ i ≤ 6, the product x_{1}x_{2} ⋯ x_{20} ≠ e.

\underline{Remark}. By Proposition 6.1, p. 97 [2], it can be seen that the value of i in the above argument can only be either 2 or 4 but not both. But this is not relevant in the above.

\underline{References}

1. I. N. Herstein, Topics in Algebra, John Wiley and Sons, Inc., 1975.

2. T. W. Hungerford, Algebra, Springer–Verlag, New York, 1974.