Prove that any finite group of even order contains an element of order 2. (an element g such that g does not equal e and g^2 = e)

Question

anonymous · Answer

oh that is wrong wrong wrong. let me try again

anonymous · Answer

lets try this and see if i can not screw it up so badly.  i think the point is you translate the problem into showing that there is some element of G that is its own inverse,  and you do this by considering the set of elements that are not their own inverses

anonymous · Answer

(if this set is empty you are done, since if every element is its own inverse they all have order 2)
so you prove that the set of elements that are not their own inverses has even order.  that is more or less clear because if you call this set H then 
$$a\in H \text{ and} \neq a^{-1}\text{ then } a^{-1}\in H$$ then

anonymous · Answer

i meant to write 
$$a\in H \text{ and  } a\neq a^{-1}\text{ then } a^{-1}\in H$$

anonymous · Answer

in plain enlish, if a is in H then so is a inverse. so H has an even number of elements. G has an even number of elements as well, and the identity e is not in H, so G must contain an element not in H

anonymous · Answer

oh, ok..I think I am starting to get it.

anonymous · Answer

sorry this was so garbled, what i get for not writing it down ahead of time. if you like i can try to write it like an actual proof

anonymous · Answer

It would be great if you can.

anonymous · Answer

consider the set of all elements in G that do not have order 2, i.e
$$H=\{a\in G| a\neq a^{-1}\}$$
claim that H contains an even number of elements because 
for each 
$$a\in H \hspace {.2cm} \exists a^{-1} \in H $$

anonymous · Answer

Thank you so much for this.

anonymous · Answer

therefore H has even order, (every element in it comes with its inverse)
and since G has even order, and the identity e is in G but not H, then there must be some other element 
g in G such that g is not in H

anonymous · Answer

i.e. there is some element in G that is its own inverse, which is a synonym for having order 2

anonymous · Answer

sorry it took a while.  i am having trouble coming up with the relevant latex commands that makes it look like actual math, but you can write it out.  i think all steps are there

anonymous · Answer

That is just fine. Thank you.

anonymous · Answer

i guess it goes without saying that for each element in H there is a UNIQUE element a inverse in H
then it is just counting

anonymous · Answer

yw