Answer:
6*8=48 groups with elements of order 7
Step-by-step explanation:
For this case the first step is discompose the number 168 in factors like this:
[tex] 168 = 8*3*7= 2^3 *3*7 [/tex]
And for this case we can use the Sylow theorems, given by:
Let G a group of order [tex]p^{\alpha} m[/tex] where p is a prime number, with [tex]m\leq 1[/tex] and p not divide m then:
1) [tex]Syl (G) \neq \emptyset[/tex]
2) All sylow p subgroups are conjugate in G
3) Any p subgroup of G is contained in a Sylow p subgroup
4) n(G) =1 mod p
Using these theorems we can see that 7 = 1 (mod7)
By the theorem we can't have on one Sylow 7 subgroup so then we need to have 8 of them.
Every each 2 subgroups intersect in a subgroup with a order that divides 7. And analyzing the intersection we can see that we can have 6 of these subgroups.
So then based on the information we can have 6*8=48 groups with elements of order 7 in G of size 168
