Question

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

Answer 1

You haven't learned about Sylow theory yet have you?

Answer 2

@KingGeorge i think you can do this without sylow

Answer 3

I think you can too, but sylow trivializes the problem which is always nice.

Answer 4

oh ok

Answer 5

as i remember idea is to pair up all elements of G that are not there own inverses with their inverse, and then count

Answer 6

That would be a good method, since we just need to show that there is at least one element that is its own inverse.

Answer 7

this set would have an even number of elements, and not include the identity
since the order of the group is even, there must be some other element not in this set, and that will be its own inverse

Answer 8

that is the idea anyway, can probably make it look nicer

Answer 9

for my curiosity, how does sylow give it to you instantly?

Answer 10

I think I was actually mis-remembering the first sylow theorem. So ignore my first couple comments.

Answer 11

Sorry Guys I am still here just in middle of postttiinnnggggggg I have 9 minutes left

Answer 12

Thanks guys only posted 5 minutes late. Thought I wld be later