Show that an abelian group of order $2^n$ has odd number of elements of order 2? I am thinking of showing something about the even property of each element and then using the pigeonhole principle at the end. I have a decent understanding of undergraduate level group theory.
 A: Oh, it's very simple. In fact, your hypothesis can be significantly weakened.

Every group of even order has an odd number of elements of order $2$.

Your proof is as follows: from the group of even order, remove the identity, which has order $1$,  to leave an odd number of elements. Next, for every element $h$ which is not of order $2$, note that $h^{-1} \neq h$, hence we remove these pairs of elements also, which is an even number of elements. Odd minus even is odd, hence that leaves us with an odd number of elements $h \neq e$ such that $h^{-1} = h$, or $h$ has order $2$. Thus the number of elements of order $2$ is odd in any even order group. 
A: Hint: For $ G $ an abelian group, the set $ H = \{ x \in G : x^2 = e \} $ is a subgroup of $ G $. What do you know about the order of this subgroup, using Lagrange's theorem?
A: You could appeal to the classification of finitely generated abelian groups and reduce to the case of $\Bbb Z/2^n\Bbb Z$, which is easily seen to have a single element of order $2$.
