Verifying Noncyclic Abelian Subgroup of S4

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SUMMARY

The subgroup {(1), (1 2), (3 4), (1 2)(3 4)} is confirmed as an Abelian, noncyclic subgroup of S4. The discussion clarifies that a cyclic group is generated by a single element, which is not the case here. The proof provided demonstrates that for any element g in the group G, g² = 1, indicating that no single element can generate the entire group of order 4. Therefore, the group is definitively noncyclic.

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  • Understanding of group theory concepts, specifically Abelian and cyclic groups.
  • Familiarity with permutation notation and operations in S4.
  • Knowledge of group order and its implications on cyclicity.
  • Basic proof techniques in abstract algebra.
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  • Learn about the structure and classification of finite groups, particularly in S4.
  • Explore the concept of group generators and their role in cyclic groups.
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Mathematicians, students of abstract algebra, and anyone studying group theory, particularly those interested in the properties of permutation groups and their substructures.

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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?
 
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A cyclic group is generated by a single element.
 
Therefore, anyone of those elements should be able to generate the others, right?
 
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
 
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)?
 
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!
 
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.
 

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