If order of group $G $ is finite, say $|G|=n$, and if $\exists\; x\in G$ such that $|x|=m$, then ...?

Question

Claim: If order of group $G $ is finite, say $|G|=n$, and if $\exists\; x\in G$ such that $|x|=m$, then $\forall d>0\; \text{such that d divides m, we must have } y \in G$ such that $|y|=d$

Proof: $|G|=n \; \text{and} \; \exists \;x\in G, |x|=m$

Let $H=<x>\Rightarrow |H|=m$

If $d>0$ divides $m$ $\Rightarrow \text{d divides order of H} $ and from Fundamental theorem on cyclic groups, there will exist a subgroup of $H$, say $N$, that is of order $d$. Since $N$ will be cyclic, it will be generated by an element of order $d$

Is this correct?

Answer 1

You are correct!

Another way of seeing this is that $x^{m/d}$ has order $d$.