If you find any mistakes, please make a comment! Thank you.

Basic properties of Hall subgroups


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.10

Solution: 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.


Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu