Consider $a,b\in \Bbb Z_n$ such that $o(a)=r,o(b)=s$. If $s\mid r$ then $b\in \langle a\rangle $.

Question

Is the following statement true ?

If yes prove it and if no then give a counterexample?

Consider the finite cyclic group $\Bbb Z_n$.

Consider $a,b\in \Bbb Z_n$ such that $o(a)=r,o(b)=s$. If $s\mid r$ then $b\in \langle a\rangle $.

Initially I thought the statement is correct. But the problem is I am unable to prove it.

How to prove it?

Answer 1

$\langle a\rangle$ is the unique subgroup of order $r$ and so $\langle a\rangle = \{ x \in G : x^r = 1\} $.

Now, $o(b)=s \mid r$ implies $b^r=1$. Therefore, $b\in \langle a\rangle$.