Show that if G is a finite group of even order then there is

Show that if G is a finite group of even order, then there is an a in G such that a is not the identity and a^2 = e.

Solution

solution

=======Since G is a group, then for any element A, there must be an A^-1 such that A * A^-1 = A^-1 * A = e. We need to show that there\'s an element that not equal to e, but still its own inverse.

G must contain e plus an odd number of elements. Let A, B, C, D, E, F, etc.... be these non-identity elements. Since G is a group, that means A^2, B^2, C^2, etc. are all contained in the group too. Assume that none of these are e. That means their values have to all be duplicated in the set {A, B, C, ...}.

Let\'s just assume we have nothing but {e, A, B, C} in the group. Then {A^2, B^2, C^2} c {A,B,C}. A^2 can\'t be A, because that would imply A is the identity element. So either A^2 = B or A^2=C. If A^2=B, then B^2=A and C^2=C, the latter of which can\'t happen. So we must have A^2=C then B^2=A and C^2=B. Though this implies that:
A^2 = C
A^2 C = C^2
A^2 C = B
A^2 CB = B^2
A^2 CB = A
A(ACB) = A

This imples ACB = e, but that shouldn\'t happen since AC should be in the set {A,B,C} and AC*B should be in the set too, and we said the said shouldn\'t include e.

See if you can use inductive reasoning or similar to show that it doesn\'t work for any other odd number of elements, not just 3.