# Question about proof of index of subgroups

This is a question in "Abstract Algebra(Dummit & Foote)", pg.96 Question 18. The problem is

Let G be a finite group, let $$H$$ be a subgroup of $$G$$, and let $$N$$ be normal in $$G$$. Prove that if $$|H|$$ and $$|G:N |$$ are relatively prime, then $$H \leq N$$.

One proof of this question started with these sentences,

Suppose $$x \in H$$ and let $$k$$ be the least positive integer such that $$x^k \in N$$($$k$$ exists since $$H$$ is finite.) By a previous exercise, as an element of $$G/N$$,$$|xN|=k$$, so that $$k$$ divides $$[G:N]$$. Moreover, ...

and so on. However, I don't get why the order of $$xN$$ must divide the index of $$G$$ in $$N$$, and I don't remember whether there was an exercise regarding this. Can anyone explain why this sentence is true?

• @Bungo Oh....it was actually a dumb question. Thanks for helping me. Apr 13, 2021 at 11:23
• No problem. I'll go ahead and post an answer even though it's very simple, just so the question can be closed.
– user169852
Apr 13, 2021 at 11:24

$$xN$$ is an element of the quotient group $$G/N$$, and the size of $$G/N$$ is $$[G:N]$$. So by Lagrange's theorem, the order of $$xN$$ must divide $$[G:N]$$.