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Cyclic Groups

by Gear300
Tags: cyclic, groups
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Gear300
#1
Aug18-09, 11:15 AM
P: 1,133
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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robert Ihnot
#2
Aug18-09, 11:44 AM
PF Gold
P: 1,059
A cyclic group is generated by a single element.
Gear300
#3
Aug18-09, 01:15 PM
P: 1,133
Therefore, any one of those elements should be able to generate the others, right?

Elucidus
#4
Aug18-09, 01:44 PM
P: 286
Cyclic Groups

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
Gear300
#5
Aug18-09, 02:47 PM
P: 1,133
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)?
aziz113
#6
Aug18-09, 09:51 PM
P: 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!
Gear300
#7
Aug19-09, 09:39 AM
P: 1,133
Quote Quote by aziz113 View Post
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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