## 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.

• 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 $|G|/|HN| = (|G|/|N|)/(|HN|/|N|)$.