I am reading Section 2.3 of Dummit and Foote's Abstract Algebra, 3rd edition, and am having a hard time proving the following proposition on page 57:
Let $H=\langle x\rangle$ and assume $|x|=n<\infty$. Then $H=\langle x^a\rangle$ if and only if $(a,n)=1$, where $(a,n)$ denotes the greatest common devisor of $a$ and $n$.
My attempt;
If $H=\langle x^a\rangle$, then by Prop. 2, $|H|=|x^a|$. Since $H=\langle x\rangle$, Prop. 2 implies $|H|=|x|=n$. So we have $|x^a|=n$. By Prop. 5, $|x^a|=\dfrac{n}{(n, a)}$. This and $|x^a|=n$ imply $(n,a)=1$.
Conversely, suppose $(n,a)=1$. Then by Prop. 5, $|x^a|=\dfrac{n}{(n,a)}=n=|x|$. Since $H=\langle x\rangle$, $|x^a|=|x|=|H|$.
From here, how do I proceed to show $H=\langle x^a\rangle$?
Proposition 2. If $H=\langle x\rangle$, then $|H|=|x|$.
Proposition 5. Let $G$ be a group, let $x\in G$ and let $a\in\mathbb{Z}-\{0\}$. If $|x|=n<\infty$, then $|x^a|=\dfrac{n}{(n,a)}$.
Best Answer
Have you heard of Bezout's identity? It states that there exists $r,s$ integers so that $n\cdot r + a\cdot s=1$. Now try to use this to show $x^{as}=x^1=x$, and from that conclude $H=\langle x^a\rangle$.