Prove that an abelian group of order 2n contains precisely one element of order 2 Assume that n is odd.  I am allowed to also assume, without proof, that every finite group of even order 2n contains element of order 2.  
My proof is as follows: 
The lagrange theorem states that if H is a subgroup of G then the order of H is a divisor of order G.  Let G be our abelian group and since |G|=2n.  Hence the order of |H| must be divisible by 2n.  Hence we know based on what I am allowed to assume that this group contains an element of order 2.  
I feel this makes sense to me but someone told me something's missing and its wrong.  Can anyone help me out here?
 A: I'm not certain what this bit means:

Let G be our abelian group and since |G|=2n. Hence the order of |H| must be divisible by 2n. Hence we know based on what I am allowed to assume that this group contains an element of order 2.

G has even order, so it has an element of order $2$ (you're allowed to assume this, but if you weren't then it follows from Cauchy's theorem).
You want to show that it has exactly one element of order $2$.
(Edit: Mizar has provided some hints. It's probably best that you have a go at the question using the hints first, so I've spoilered the rest of the answer.)

 Assume that there are at least two elements of order $2$. Let $g,h$ have order $2$ with $g \neq h$. Together they generate the four element subgroup $H=\{g,h,gh,e\}$. (If your group weren't abelian (say if it were dihedral), this group might have more than $4$ elements.)
 But Lagrange's theorem says that the order of $H$ must divide the order of $G$. $4$ doesn't divide $2n$ since $n$ is odd, so we have a contradiction.

A: Who is $H$ in your attempted proof? How do you find the element of order 2? 

Anyway, here is a hint for a possible solution: couple any element and its inverse.
When does an element get paired with itself? (Be careful here..) 
Now you should have deduced by parity considerations that at least one element of order 2 exists; for the uniqueness, if $x\neq y$ are two elements of order 2, what does the subgroup $<x,y>$ look like? What does Lagrange's theorem tell you? Conclude.
A: Assume more than one,
Let $a,b$ has order $2$ than notice that $\{e,a,b,ab\}$ is a subgroup of $G$ which means $|G|$ is divisible by $4$ which is a contradiction.
note: you only need to check uniqness part.
