Prove that $G$ has at most one element of order $2$

Question

Suppose G is a cyclic group generated by $g\ \epsilon\ G$, i.e. $G=\{e,g,g^2,...,g^{n-1}\}$ where $n$ is the order of $g$. It can be shown that if $0\le k\le n-1$, then the order of $g^k$ is $\frac{n}{\gcd(k,n)}$. Prove that $G$ has at most one element of order $2$.

I know that if $n$ is odd then the group will have no elements of order $2$ as in that case $2 \nmid n$. For the case of $n$ even, I assumed that one $t$ exists such that $0\le t\le n-1$ where the order of $g^t$ is 2. So from above we have $2=\frac{n}{\gcd(t,n)}$ which implies that $\gcd(t,n)=\frac{n}{2}$, which means that $t\ge \frac{n}{2}$. I don't really know where to go from here however.

Answer 1

$\gcd(t,n)=n/2$, tells you more than just $t\geq n/2$. It tells you that $t$ is a multiple of $n/2$. The multiples of $n/2$ are $n/2,n,3n/2,\dots$. Now, your restriction $0\leq t<n$ tells you $t=n/2$. So, only one solution.

Answer 2

Not sure why you need the gcd. If a cyclic group with a generator $x$ has order $2m$, then the element $x^{m}$ will have order 2.

Answer 3

Suppose G is a cyclic group generated by $g\ \epsilon\ G$, i.e. $G=\{e,g,g^2,...,g^{n-1}\}$ where $n$ is the order of $g$.

As you indicated, it can be shown that the order of $g^k$ is $\frac{n}{\gcd(k,n)}$.

Assume $g^i$ and $g^j$ both have order $2$. Then $\frac{n}{\gcd(i,n)}=\frac{n}{\gcd(j,n)}=2,$ so $\gcd(i,n)=\gcd(j,n)=\frac n 2.$ This means $i=k\frac n 2$ and $j=l\frac n 2$ for some odd integers $k$ and $l$. Therefore $g^i/g^j=g^{\frac{k-l}2 n}=e.$

Thus, there is at most one element of order $2$.