# The order of an element in a quotient group

Suppose $$G$$ is a finite group, that $$H$$ is a subgroup of $$G$$, and that $$N$$ is a normal subgroup of $$G$$. Suppose that $$|H| = n$$ and $$|G| = m|N|$$, where $$m$$ and $$n$$ are coprime. Consider the quotient group $$G/N$$ and let $$h \in H$$. Determine the order of the element $$hN$$ in the group $$G/N$$.

Attempt:

Then $$|G/N|=m$$ , $$|H|=n$$ , $$\operatorname{gcd}(m,n)=1$$

Order of the element $$hN$$ in $$G/N$$ is $$k$$ where $$k$$ is the smallest positive integer such that $$h^k$$ belongs to $$N$$. $$(hN)^k = (h^k)(N)=1$$ iff $$h^k$$ belongs to $$N$$.

I am not sure how to use $$|H|=n$$ and $$gcd(m,n)=1$$ to find the order of $$hN$$.

• The order of $hN$ divides the order of $h$. – Tortoise Nov 19 '18 at 10:16

I got the idea now. $$(|H|,|G/N|)=1$$ is key to do this problem. Notice that $$|hN|||h|$$ (check) and also $$|h|||H|$$ so that $$|hN|||H|$$ but $$|hN||(G/N)$$ since $$|H|$$ and $$|G/N|$$ are relatively prime so that $$|hN|=1$$. Thus $$hN=N$$ and then $$H\subset N.$$ This finishes the proof.