I am stuck in an exercise of Rose's group theory book, (page 43 exercise 106 (ii) ). The exercise is as follows. Suppose that $K\unlhd G$ with $|K|=m$. Let $x$ be an element of $G$ and $n$ be a positive integer co-prime with $m$. Prove that if $o(xK)=n$ then there is an element $y$ of $G$ such that $o(y)=n$ and $xK=yK$, ($xK$ and $yK$ are viewed as elements of the group $G/K$).
All I could do was to show that there exists an element $y$ where $o(y)=n$ and $\langle yK\rangle=\langle xK\rangle $. For that notice that $o(xK)=n\mid o(x)=b$ because $(xK)^b$$=(x^b)K=1\cdot K=K$. Thus for $λ=\frac{o(x)}{n}$ and $y=x^λ$ we have $o(y)=n$ and because $(n,m)=1$, $o(yK)=n$ too, (thats exercise 106 (i)). Obviously $yK$ is an element of $\langle xK\rangle $ , thus $\langle yK\rangle \subseteq \langle xK\rangle $ and from orders we get the equality. I don't know if thats the right direction to solve the exercise.
It is a general property of a finite group $G$ that if $\langle a\rangle=\langle b\rangle$ for group elements $a,b\in G$, then $a=b^k$ for some $k\in\mathbb{Z}$.
Apply that to what you have already shown for $G/K$ and you get $xK=(yK)^k=y^kK$. Therefore, $y^k$ is the element you are looking for.
Update:
Let $z:=y^k$. Then $z^n=(y^k)^n=(y^n)^k=1$ since $y^n=1$, so $o(z)$ divides $n$. Assume that $m:=o(z)$ is strictly smaller than $n$. Then also $(zK)^m=z^kK=K$, so $o(zK)<n$, a contradiction.
Therefore $o(y)=n$.