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?

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

1 Answer 1


$xN$ 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 $[G:N]$.


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