Question

Finite group question.

Answer 1

Without Lagrange how can I show that a finite group of odd order has no elements of order 2.

Answer 2

{1,3}
it has order two right O.O

Answer 3

Do you mean no subgroups of order 2?

Assume there is a subgroup of order 2. Because there is more than one element, there is an element a not equal to the identity element e. It has an inverse a^-1. Because there are only two elements and a^-1 is not e, a^-1 = a. Now for any element b of the group except those two, a*b is neither a, b nor e. *long pause for thinking* Aha! Let a*b = c. Multiply by a on the left to get a*a*b = a*c; e*b = a*c; a*c = b. In this manner we can order the rest of the elements in pairs {b, c} with the property that a times any number is the other number. We are relying on the fact that b is not c to ensure that the order of each pair is 2. As the group consists of the two elements e, a and some pairs, its order must be even. You require that the order be odd, so our initial assumption is wrong.

Answer 4

yep ty