Verifying Noncyclic Abelian Subgroup of S4

  • Thread starter Gear300
  • Start date
  • Tags
    Subgroup
In summary, the conversation discusses how to show whether a given group is cyclic or noncyclic. The group presented, {(1), (1 2), (3 4), (1 2)(3 4)}, is shown to be Abelian through pairing the permutations. However, in order to show that it is noncyclic, one must find an element that cannot be generated by any other element. Through a proof, it is determined that the group is indeed noncyclic. The conversation also touches on the inclusivity of negative integers in identifying inverses in a group.
  • #1
Gear300
1,213
9
The problem is to verify that {(1), (1 2), (3 4), (1 2)(3 4)} is an Abelian, noncyclic subgroup of S4.

I was able to show that it is Abelian through pairing the permutations, but my mind stopped at the noncyclic part. When showing that a group is cyclic or noncyclic, what exactly do I have to show?
 
Mathematics news on Phys.org
  • #2
A cyclic group is generated by a single element.
 
  • #3
Therefore, anyone of those elements should be able to generate the others, right?
 
  • #4
Yes, if it were cyclic. In order for a group to be cyclic then there must exist a member a so that for all members b, there exists a non-negative integer n so that an=b.

In order to show a group is cyclic, one must find such a member a. To show it is non-cyclic, one must show that there is a member b which cannot be the power of any other member (it is obviously the 1st power of itself).

I'd look at (1 2)(3 4) and see if one can show whether it is a power of any of the others.

--Elucidus
 
  • #5
Doesn't seem as though (1 2)(3 4) is a power of any of the other elements.
Does n have to be non-negative (in order for it to be a group, shouldn't n also be inclusive of negative integers - to identify the inverses)?
 
  • #6
The group G that you've presented is certainly noncyclic. Here is a proof: For any element g in G, g2=1. However, the order of the group is 4, and so no single element can generate the group. Thus the group is not cyclic.

Hope that helps!
 
  • #7
aziz113 said:
The group G that you've presented is certainly noncyclic. Here is a proof: For any element g in G, g2=1. However, the order of the group is 4, and so no single element can generate the group. Thus the group is not cyclic.

Hope that helps!

Thanks for the help.
 

What is a noncyclic abelian subgroup of S4?

A noncyclic abelian subgroup of S4 is a subset of the group S4 that is closed under the group operation and follows the commutative property, but does not contain any elements that generate a cyclic subgroup. In other words, it is a subgroup that does not have a single element that can be repeatedly multiplied to generate all the elements in the subgroup.

How do you verify that a subgroup of S4 is noncyclic and abelian?

To verify that a subgroup of S4 is noncyclic and abelian, you can perform the following steps:

  1. Check that the subgroup is closed under the group operation, meaning that the result of multiplying any two elements in the subgroup is also in the subgroup.
  2. Check that the subgroup follows the commutative property, meaning that the order of multiplication does not change the result.
  3. Check that the subgroup does not contain any elements that generate a cyclic subgroup, meaning that no single element can be repeatedly multiplied to generate all the elements in the subgroup.

Why is it important to verify that a subgroup of S4 is noncyclic and abelian?

Verifying that a subgroup of S4 is noncyclic and abelian is important because it ensures that the subgroup is a proper subset of the group S4 and follows certain properties. This can help in understanding the structure of the group and its subgroups, and can also be useful in solving problems involving group theory.

Can a subgroup of S4 be both cyclic and abelian?

No, a subgroup of S4 cannot be both cyclic and abelian. This is because a cyclic subgroup is one that can be generated by a single element, while an abelian subgroup follows the commutative property. Since the commutative property requires the elements to be able to be multiplied in any order, it is not possible for a single element to generate the entire subgroup.

What are some examples of noncyclic abelian subgroups of S4?

Some examples of noncyclic abelian subgroups of S4 include the trivial subgroup {e}, which only contains the identity element, and the subgroup {e, (12)(34), (13)(24), (14)(23)}, which consists of all the even permutations of S4. Another example is the subgroup {e, (12)(34), (13)(24), (14)(23), (1234), (1324)}, which consists of all the elements that fix the number 5 in the permutation (12345).

Similar threads

  • General Math
Replies
1
Views
700
  • Linear and Abstract Algebra
Replies
1
Views
719
Replies
4
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
616
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • General Math
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
Replies
19
Views
865
  • Linear and Abstract Algebra
Replies
3
Views
2K
Back
Top