Quotient Group question about order of a group

Question

How do I prove that when an element of $G/K$ (where $K$ is normal subgroup of $G$) has an order $n$, there exists an element in $G$ having order $n$? I could only do as the order of elements are $n$. The total number of elements in $G$ are $nm$. How to proceed?

Answer 1

Let the element of $G/K$ of order $n$ be say $rK$. $(rK)^{|r|} = K$, so $n$ divides $|r|$, say $|r| = nk$. Then consider $r^k$. Clearly $|r^k| =n$.

(Edit: Showing $|r^k|=n$: Suppose by way of contradiction $|r^k|\neq n$. Then $|r^k|$ is, say, $l$, and we know $l$ divides $n$. Hence $l<n$. Then $(r^k)^l=1$ so $|r|\leq lk<nk$, a contradiction).