Problem 7 15 points Suppose G is a simple group with G 168
Problem 7 (15 points). Suppose G is a simple group with |G| = 168. How many elements of order 7 are there in G? Prove your answer.
Solution
Since 168=2^3*3*7
, there exists a subgroup of order 7. Let n7 be the number of Sylow 7-subgroups. Since n7|24 and n71 mod 7, n7=8. Since every group of prime order is cyclic, there exist 8 elements of order 7
| Since 168=2^3*3*7 |
