A similar question has been asked before but I believe that this question is slightly different. So far,the analyses of simple groups of order 168 have no elements of order 6, need to rely on computing the number of Sylow 3-subgroups, $n_3$.
(Such as
Prove there is no element of order 6 in a simple group of order 168
https://www.youtube.com/watch?v=QrdE7Wt7RZA)
The question reads as follows:
- Let $G$ be a simple group of order 168.
a) Show that $G$ has precisely $8$ Sylow 7-subgroups.
b) Show that $G$ is isomorphic to a subgroup $\hat{G}$ of the alternating group $A_8$ and that no element of order $2$ in $G$ has a fixed point.
c) Show that $G$ has no element of order 6.
d) Find the number of Sylow-3 groups of $G$. Prove your answer.
This question asks us to deduce the value of $n_3$ after showing that there are no elements of order 6. I was able to do (a) and (b) but got stuck from there onwards. From (b), we can deduce that if such an element exists it is of cycle type $(1 2 3 4 5 6) (7 8)$. Other permutations of order 6 either have a fixed point when taken to the 3rd power or it is not an even permutation.
But from here onwards, I could not see how to proceed. Are there alternative ways to deduce the nonexistence of elements of order $6$ than the ones listed above?
Any help would be appreciated.
Best Answer
If there is an element of order $6$ with cycle structure $(6\,2)$ then for each pair of points $\{a\,b\}$ there are at least two elements of order $6$ with $(a\,b)$ as a cycle and at least $56$ elements of order $6$, since $G$ is $2$-transitive. Likewise there are at least $56$ elements of order $3$. As there are $48$ elements of order $7$ there are at most $168-56-56-48=8$ elements of $2$-power order, so the Sylow $2$-subgroup is unique, and normal, contradicting the simplicity of $G$.