If $\gcd(b,n)\mid\gcd(a,n)$ then $\langle x^a\rangle\le \langle x^b\rangle$

Question

If $\gcd(b,n)\mid\gcd(a,n)$, $1\le a,b\le n$, then $\langle x^a\rangle\le \langle x^b\rangle$, where $x^a,x^b\in \langle x\rangle$ with $|\langle x\rangle|=n$ , $\langle x\rangle$ means cyclic group of $x$ and $H\le K$ means $H$ is subgroup of $K$.

How to prove this?

My try: $\gcd(b,n)\mid\gcd(a,n)$ implies $\frac{n}{\gcd(a,n)}\mid\frac{n}{\gcd(b,n)}$ so $\langle(x^b)^{\frac{\gcd(a,n)}{\gcd(b,n)}}\rangle\le \langle x^b\rangle$ and I cannot show $\langle x^a\rangle=\langle(x^b)^{\frac{\gcd(a,n)}{\gcd(b,n)}}\rangle$.

Answer 1

Since $\gcd(b,n)$ divides $\gcd(a,n)$. There is $k'$ such that $\gcd(a,n) = k'\gcd(b,n)$

Lemma If $x$ of order $n$. The cyclic subgroups $\langle x^a \rangle = \langle x^{\gcd(a,n)} \rangle, \langle x^b \rangle = \langle x^{\gcd(b,n)} \rangle$.

Suppose $y \in \langle x^{\gcd(a,n)} \rangle$. Then $y = x^{k \gcd(a,n)} = x^{k k' \gcd(b,n)} = x^{\gcd(b,n)^{k k'}}$ for some $k$.

Therefore, $\langle x^a \rangle = \langle x^{\gcd(a,n)} \rangle \subseteq \langle x^{\gcd(b,n)} \rangle = \langle x^b \rangle $ and since $\langle x^a \rangle$ is a group, $\langle x^a \rangle \leq \langle x^b \rangle$.