Failure of the sylow counting argument

Click For Summary
SUMMARY

The discussion centers on the failure of the Sylow counting argument in demonstrating the existence of a normal subgroup of order 5 in a group G of order 30. The Sylow theorem indicates that the number of Sylow 5-subgroups, denoted as n_5, must divide 6 and satisfy the condition n_5 ≡ 1 (mod 5). The possible values for n_5 are 1 or 6. If n_5 equals 1, a normal subgroup exists; however, assuming n_5 equals 6 leads to a contradiction regarding the number of elements of order 5, indicating insufficient room for other elements in the group.

PREREQUISITES
  • Understanding of Sylow theorems in group theory
  • Familiarity with group orders and subgroup properties
  • Knowledge of modular arithmetic and its application in group theory
  • Basic concepts of normal subgroups and their significance
NEXT STEPS
  • Study the implications of Sylow's theorems on group structure
  • Explore examples of groups of order 30 and their subgroup configurations
  • Investigate the relationship between subgroup orders and element orders in finite groups
  • Learn about the classification of finite groups and their normal subgroups
USEFUL FOR

This discussion is beneficial for mathematicians, particularly those specializing in group theory, as well as students and educators seeking to deepen their understanding of Sylow subgroups and their properties in finite groups.

catcherintherye
Messages
47
Reaction score
0
show that there is a normal subgroup of G of order 5 when G is a group of order 30. My friend just called me with this problem, he said the usual method of solution fails. (i.e. using sylow and then showing that the subgroup is unique and deducing that it must therefore be normal), I told him to post it on physics forum so ytou might get this question twice.
 
Physics news on Phys.org
The theorem says that the number of 5-groups, n_5, divides 30/5=6 and is equal to 1 mod 5. So it's either 1 or 6.

If it's 1, you're done, so assume it's 6. What would this imply about the number of elements of order 5 in the group? Try to show that it doesn't leave enough room for the other elements in the group.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K