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

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.

• It should be $g^{n-1}$. – Julian Mejia May 5 at 1:43

$$\gcd(t,n)=n/2$$, 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 $$0\leq t tells you $$t=n/2$$. So, only one solution.
Suppose G is a cyclic group generated by $$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$$.
Not sure why you need the gcd. If a cyclic group with a generator $$x$$ has order $$2m$$, then the element $$x^{m}$$ will have order 2.