[Math] Prove that an abelian group of order 2n contains precisely one element of order 2

Question

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?

Answer 1

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.