Given: A group G of order 231

To find: The number of elements of order 11 in G

Approach:

- Use Sylow's theorems to understand the structure of G
- Determine the number of Sylow 11-subgroups in G
- Use the fact that every element of order 11 lies in a unique Sylow 11-subgroup
- Determine the number of elements of order 11 in each Sylow 11-subgroup
- Add up the number of elements of order 11 from all the Sylow 11-subgroups to get the total number of elements of order 11 in G

Solution:

Using Sylow's theorems:

- The prime factorization of 231 is 3 x 7 x 11
- The number of Sylow 11-subgroups in G is congruent to 1 modulo 11 and divides 3 x 7 = 21
- Thus, the number of Sylow 11-subgroups in G is either 1 or 21

If there is only one Sylow 11-subgroup:

- Let H be the unique Sylow 11-subgroup in G
- H has order 11, and thus has 10 elements of order 11 (since the identity element has order 1)

If there are 21 Sylow 11-subgroups:

- Let H be a Sylow 11-subgroup in G
- H has order 11, and thus has 10 elements of order 11 (since the identity element has order 1)
- Since every element of order 11 lies in a unique Sylow 11-subgroup, there are no repeated elements of order 11 among the 21 Sylow 11-subgroups
- Thus, there are a total of 21 x 10 = 210 elements of order 11 in G

Conclusion:

- The number of elements of order 11 in G is 10, if there is only one Sylow 11-subgroup
- The number of elements of order 11 in G is 210, if there are 21 Sylow 11-subgroups
- Since the number of Sylow 11-subgroups divides the order of G, 21 is not a possible number of Sylow 11-subgroups in a group of order 231
- Thus, the only possibility is that there is only one Sylow 11-subgroup in G, and the number of elements of order 11 is 10.