## If the index and order of a normal subgroup and subgroup are relatively prime, then the subgroup is contained in the normal subgroup

Let be a group and let with normal in . Prove that if and are relatively prime then .

First we prove a lemma.

Lemma: Let be a group, , and an element of finite order . If is the least positive integer such that , then . Proof: If does not divide , we have for some by the division algorithm. Now , and . Thus , which contradicts the minimality of . Thus .

Now to the main result.

Suppose , and let be the least positive integer such that . ( exists since is finite.) By a previous exercise, as an element of , , so that divides . Moreover, we have divides by Lagrange, so that (by the lemma) divides and thus divides . Because and are relatively prime, then, . But then , so , and we have . So . By a previous exercise .

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## Comments

I think you may want that k divides the order of x, why would k divide the index of N in G?

We’re thinking of as a group of order . There, is an element of order . So by Lagrange, divides .

We need this in order to take advantage of the fact that and are relatively prime.

I think I’ve found my error. So the order of xN is not the number of elements in that coset, rather the number of operations required in G/N to reach the identity coset? Sorry, and thanks so much for the explanations.

That’s exactly right. No problem!

but G can also be infinite?

We started out with the assumption that is finite. However, it looks like the proof still works for infinite …

in the solution parag no.2,|x| divides |H| clearly but why should k divide |H|?

This is using the lemma and Lagrange’s theorem. divides the order of (by the lemma) which divides (By Lagrange).

thanks..