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