prove that in a cyclic group of even order, there is exactly one element of order 2.

Question

prove that in a cyclic group of even order, there is exactly one element of order 2. Please prove the requirements.

Answer 1

Let $G$ be a cyclic group of even order, that is, a group with an order of $2n$ where $n$ is a natural number. We know that every cyclic group G verifies the inverse of Lagrange's Theorem, that is, there exists a unique subgroup H of G with order m, for every divisor $m$ of $n$.

Now, considering that $G=<g>$ where $g$ is a generator of the cyclic group $G$, we have that $g^n$ is an element of order $n$ in $G$, because $(g^n)^2=g^{2n}=1$. This element generates the subgroup $<g^n>=\{1,g^n\}$ of $G$, which has order 2, and is therefore the only subgroup of $G$ with order two (according to the inverse of Lagrange's Theorem, and because $n$ is a divisor of the order $2n$ of our group $G$).

Now let's suppose that there exists another element $x\neq g^n$, $x\in G$, such that the order of $x$ is 2. Then, $<x>=\{1,x\}$ would be a subgroup of order 2 of $G$, which would be different than $<g^n>$, contradicting the uniqueness stated above. This means that $g^n$ is the only element of $G$ with order 2.

Edit: Would anyone mind explaining the downvote? I think my answer is correct and solves the problem.