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

Let G be a group and let H,N \leq G with N normal in G. Prove that if |H| and [G:N] are relatively prime then H \leq N.

First we prove a lemma.

Lemma: Let G be a group, H \leq G, and x \in G an element of finite order n. If k is the least positive integer such that x^k \in H, then k|n. Proof: If k does not divide n, we have n = qk+r for some 0 < r < k by the division algorithm. Now 1 = x^n = x^{qk}x^r \in H, and x^{qk} = (x^k)^q \in H. Thus x^r \in H, which contradicts the minimality of k. Thus k|n. \square

Now to the main result.

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, we have |x| divides |H| by Lagrange, so that (by the lemma) k divides |x| and thus divides |H|. Because |H| and [G:N] are relatively prime, then, k = 1. But then |xN| = 1, so xN = N, and we have x \in N. So H \subseteq N. By a previous exercise H \leq N.

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  • Bobby Brown  On September 26, 2010 at 1:39 pm

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

    • nbloomf  On September 26, 2010 at 6:49 pm

      We’re thinking of G/N as a group of order [G:N]. There, xN is an element of order k. So by Lagrange, k divides [G:N].

      We need this in order to take advantage of the fact that |H| and [G:N] are relatively prime.

  • Bobby Brown  On September 29, 2010 at 8:48 pm

    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.

    • nbloomf  On September 29, 2010 at 9:34 pm

      That’s exactly right. No problem!

  • inquisitive  On March 6, 2011 at 8:56 pm

    but G can also be infinite?

    • nbloomf  On March 7, 2011 at 7:40 am

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

  • quadro  On March 9, 2011 at 4:12 am

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

    • nbloomf  On March 9, 2011 at 9:26 am

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

  • quadro  On March 12, 2011 at 7:16 am


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