Let G be a simple group of order 168. How many elements of order 7 

does it have?

Correct option is C

The order of the group  |G| = 168 , and  G  is simple (as given).

​$$168 = 7 \cdot 2^3 \cdot 3$$​​


Using Sylow's Third Theorem:
- The number of Sylow 7-subgroups in  G  must divide  168  and satisfy

$$n_7 \equiv 1 \pmod{7}$$​.

The divisors of   168/7 = 24  are  1, 3, 8, 24 . Thus,  $$n_7$$​  can be:

​$$n_7 = 1 + k \cdot 7, \quad k = 0, 1, 2, \dots$$​​


Case Analysis:​​

(i)  If  k = 0 ,  $$n_7 = 1 + 0 = 1 (possible).$$​​
(ii) if $$k = 1 , n_7 = 1 + 7 = 8 (possible).$$​​
(iii) If  k = 2 , $$n_7 = 1 + 14 = 15 \text{(not possible since 15 does not divide 24 ).}$$​​

Thus,  G  can have  $$n_7 = 8$$​  Sylow 7-subgroups.

Since 7  is prime, all Sylow 7-subgroups are cyclic. Each cyclic subgroup of order 7 contains 6 elements of order 7. 

​$$\textbf{Number of Elements of Order 7:}$$​​
​$$\text{Number of elements of order 7} = n_7 \cdot (\phi(7)) = 8 \cdot (7 - 1)\\ = 8 \cdot 6 = 48.\\[10pt]\textbf{Final Answer:}\\\textbf{Option (C): 48.}$$​​