Show isomorphism between two groups

Click For Summary
SUMMARY

The discussion focuses on proving that a non-abelian group G of order 12, which contains exactly two elements of order 6 and seven elements of order 2, is isomorphic to the dihedral group D12. The initial approach involved analyzing the cyclic subgroup generated by an element of order 6 and its complement, but it was noted that this complement does not form a subgroup due to the absence of an identity element. The correct proof must leverage the non-abelian nature of G and the specific count of elements of order 2 to establish the isomorphism definitively.

PREREQUISITES
  • Understanding of group theory, specifically non-abelian groups
  • Familiarity with dihedral groups, particularly D12
  • Knowledge of subgroup properties and normal subgroups
  • Comprehension of element orders within group structures
NEXT STEPS
  • Study the properties of dihedral groups, focusing on D12 and its structure
  • Learn about normal subgroups and their significance in group theory
  • Explore the implications of element orders in finite groups
  • Review proofs involving isomorphisms between groups
USEFUL FOR

This discussion is beneficial for students of abstract algebra, particularly those studying group theory, as well as mathematicians interested in the properties and classifications of finite groups.

playa007
Messages
25
Reaction score
0

Homework Statement


Suppose G is a non-abelian group of order 12 in which there are exactly two
elements of order 6 and exactly 7 elements of order 2. Show that G is isomorphic to the
dihedral group D12.


Homework Equations





The Attempt at a Solution


My attempt (and what is listed in the official solutions) was to first consider the cyclic group generated by an element of order 6 in group G. Thus, this cyclic group has order 6. Consider the elements in G \ <x> (complement of G and <x>); this subgroup has index 2(but the problem here its not even a subgroup since it has no identity element); so all the elements of G \ <x> has order 2(deduced from the hypothesis) and is a normal subgroup so by definition of normal subgroups, yxy^-1 = x^-1 is satisfied and G can be written as {x^6 = 1 , y^2 = 1 such that yxy^-1 = x^-1} which is precisely the same group structure as D12 => isomorphic.

I'm certain that there is a crucial flaw here and a correct proof or a way to fix the existing proof is very much appreciated.
 
Physics news on Phys.org
Yes that is a crucial flaw. And the solution doesn't seem to use the fact that G is nonabelian or that there are 7 elements of order 2 in G. These are things that you'd probably want to take advantage of!
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K