Basic properties of Hall subgroups

Let G be a finite group. A Hall subgroup of G is a subgroup H \leq G such that |H| and [G:H] are relatively prime. Suppose H \leq G is a Hall subgroup and N \leq G a normal subgroup. Prove that H \cap N is a Hall subgroup of N and that HN/N is a Hall subgroup of G/N.

We have [HN:H \cap N] = [HN: H \cap N], so that [HN:N][N:H \cap N] = [HN:H][H:H \cap N]. By the Second Isomorphism Theorem, we have [HN:N] = [H:H \cap N]. Thus [N:H \cap N] = [HN:H]; because HN \leq G, we have that [HN:H] divides [G:H]. Moreover, |H \cap N| divides |H|. Thus |H \cap N| and [N:H \cap N] are relatively prime.

Now [G:H] = [G:HN][HN:H], so that [G:HN] divides [G:H]. By the Third Isomorphism Theorem, [G:HN] = [G/N : HN/N], so that [G:HN] = [G/N : HN/N] divides [G:H]. Now |HN| = |H| \cdot |N|/|H \cap N|, so that |HN/N| = |H|/|H \cap N|. Thus |HN/N| divides |H|, so that |HN/N| and [G/N : HN/N] are relatively prime.

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  • Gobi Ree  On November 28, 2011 at 11:19 pm

    Note that we can obtain directly [N:H \cap N]=[HN:H] without paying attention to normality, second isomorphism, or finiteness since this holds for arbitrary subgroups, i.e. [AB:A]=[B:A \cap B] for any subgroup . (In page 98 above)

  • Gobi Ree  On November 28, 2011 at 11:49 pm

    This is very awesome and elegant proof. (I proved this problem by writing down prime decompositions of orders of all subgroups, which was very messy.)

    But I have a question. To use third isomorphism theorem, is it true that HN \unlhd G?

    • nbloomf  On November 29, 2011 at 12:00 am

      Good question. I don’t have my book handy, but as I recall the Third Isomorphism Theorem does require the intermediate subgroup to be normal. For this proof, though, I don’t think we need it- in fact we don’t need the full force of the Isomorphism Theorem. All we need is that [G:HN] = [G/N:HN/N], which (since all groups are finite) is equivalent to |G|/|HN| = (|G|/|N|)/(|HN|/|N|).

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