# Simple Group order 168

• Oct 2nd 2008, 03:14 PM
roporte
Simple Group order 168
Hello,

I need help on proving this problem.

How many elements of order 7 are there in a simple group of order 168?

Thanks a lot!
• Oct 2nd 2008, 07:02 PM
ThePerfectHacker
Quote:

Originally Posted by roporte
How many elements of order 7 are there in a simple group of order 168?

By Sylow's 3rd theorem the number of Sylow's 7-subgroups of a group of order 168 is either 1 or 8. It cannot be 1 because it would imply such a Sylow 7-subgroup must be normal contradicting the fact the group is simple. Therefore there are 8 Sylow 7-subgroups. Given $P_1,P_2$ to be two Sylow-7 subgroups with $P_1\not =P_2$ then it implies $P_1 \cap P_2 = \{ e \}$ (because $P_1\cap P_2$ is a subgroup of $P_1$ and my Lagrange's theorem $|P_1\cap P_2|$ divides $|P_1|$ - a prime). This means the non-trivial elements in $P_1,P_2,...,P_8$ are all distinct. And furthermore each one generates the group i.e. have order 7. There are a total of $6\cdot 8 = 48$ of such elements. And conversely any element of order seven gives rise to a Sylow $7$-subgroup and must be included among those elements. Therefore there are 48 elements of order 7.