Lagrange theorem applications.

Question

Let $G$ a abelian group with order $|G|=2n$ where $n$ is a odd integer. I need to show that $G$ has only one element with order $2$ but the unique tool that I can to use is Lagrange's theorem.

Existence: As $G$ has even order, some element of $G$ is itself its inverse, so, it has order two.

I have troubles with the uniqueness part. It follows easily if we use the decomposition of $G$ in product of cyclic groups but, as I did say in the beginning, the only tool that I can to use is Lagrange's theorem.

Can someone give me a hint to complete my proof?

Answer 1

Hint: If $o(a) =o(b) =2$, what is the size of the generated subgroup $\langle a, b\rangle $?

Answer 2

Hint: suppose $x \neq y$ are two elements of order 2. Can you construct a subgroup using them that contradicts Lagrange's theorem?

Answer 3

Hint: Look at the subgroup generated by the element of order 2 $\langle g\rangle$ and take the quotient $^{G}/_{\langle g\rangle}$ (why is $\langle g\rangle$ normal?). Consider a second element of order 2 and see how it behaves in the quotient by using Lagrange !