Suppose that a group $G$ has an involution, then it has an odd number of involutions 
Suppose that a group $G$ has an involution, then it has an odd number of involutions

My attempt:
Suppose that $g\in G $ is an involution: $g^2=e.$ Consider an element $h$ ($\ne g, e$) in $ G$ of order 2. Then we know that $ \langle g \rangle =\langle h \rangle$ or $\langle g \rangle \cap \langle h \rangle = 1$. So $S:=\{\text{all involutions of } G\} $ is the union of cyclic subsets of order 2 that intersect at the identity element. Therefore, $ |S|=\frac{|G|}2 + 1.$ (adding an extra 1 for the identity element). Order of $ G$ is even, since it contains an involution, this the first term is an even positive integer. This proves that there are an odd number of involutions. 
Now, I'm not completely sure about the first term in my solution. This term should represent the number of different subsets of order 2 in $ G$, but I don't think that this is the way to compute it. Does anyone know how to?
Thanks. 
 A: No, your proof does not work. In general the number  of involutions in a group is not equal to  $\frac{|G|}{2}+1$ (btw this is not an odd number in general). See LINK for the case of the symmetric group (you have to subtract $1$, since there the identity $e$ is considered an involution).
Hint. By Lagrange's theorem, a group containing an element of order 2, i.e. a strict involution, has even order. Now $G$ can be split into three disjoint sets:
$$G=\{e\}\cup\{\text{elements of $G$ of order$=2$}\}\cup \{\text{elements of $G$ of order$>2$}\}.$$
Notice that if $g$ has order $>2$ then $g\not=g^{-1}$ which implies that the set $\{\text{elements of $G$ of order$>2$}\}$ can be furtherly split into two sets of the same cardinality.
Can you take it from here?
A: Why would this prove that $\lvert S\rvert$ is odd? It would be so if $\lvert G\rvert$ was a multiple of $4$, but that doesn't have to be the case. In fact, it is not true that $\lvert S\rvert=\frac{\lvert G\rvert}2+1$. If $G=S_3$, then $\lvert S\rvert=3$ and $\lvert G\rvert=6$.
