TITLE: Let $(G,\cdot)$ be a group of order $2n$ with $n$ elements of order $2$. Prove $n$ is odd and $G$ has an abelian subgroup of order $n$.
QUESTION [6 upvotes]: Let $(G,\cdot)$ be a group of order $2n$ which has exactly $n$ elements of order $2$. Prove that $n$ is odd and that $G$ has an abelian subgroup of order $n$.

For the first part, the elements whose order is greater than $2$ can be grouped in pairs of the form $\{x,x^{-1}\}$. Now it easily follows that $n=\text{even}-\text{odd}=\text{odd}$.    
For the second part, I though about considering the set $H=\{x_1,x_2,\dots ,x_n\}$ where $x_i$ is an element of order $2$ $\forall i=\overline{1,n}$.      
I observed that for $i\neq j$ we have that $x_i x_j \in G\setminus H$, because otherwise $\{e,x_i,x_j, x_i x_j\}$ would be a subgroup of order $4$ of $G$, which would contradict Lagrange's theorem. I didn't know how to proceed from here.

REPLY [2 votes]: Suppose that $|G| = 2n$ and that the number of elements of order $2$ in $G$ is $n$.
You can see that the number of $x \in G$ such that $x^2 = 1$ is even (for a proof, partition elements of $G$ into sets $\{x,x^{-1}\}$). This implies that the number of elements of order $2$ in $G$ is odd, so $n$ is odd.
Now consider the regular permutation representation of $G$, that is, $G$ acting on itself by left multiplication). In this action, an element $x \in G$ of order $2$ acts with cycle structure $$(x_1 x_2) (x_3 x_4) \cdots (x_{2n-1} x_{2n}).$$
In particular $x$ acts as an odd permutation, so $G$ has a normal subgroup $H$ of index $2$. (In general if $G \leq S_n$ and $G \not\leq A_n$, then $G \cap A_n$ is a normal subgroup of $G$ with index $2$.)
Let $x \in G$ have order $2$ and let $H \leq G$ be a subgroup of index $2$. Now $H$ has no elements of order $2$ since $|H| = n$ is odd, so every element of $xH$ has order $2$.
Then for all $h \in H$, we have $$xhx^{-1} = xhx = (xh)^2h^{-1} = h^{-1}.$$ So conjugation by $x$ gives an automorphism $h \mapsto h^{-1}$ of $H$. The inverse map is an automorphism if and only if the group is abelian, so $H$ must be abelian.