Proving G Abelian iff G/Z(G) is Cyclic

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In summary, the conversation discusses the concept of Abelian groups and their relation to cyclic factor groups. The direction of implication from a group being Abelian to its factor group being cyclic is considered trivial, while the reverse direction requires further explanation. The conversation also touches on the different interpretations of "P <=> Q" and the importance of understanding its meaning in mathematical contexts.
  • #1
DanielThrice
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This isn't some homework question its more theory based that I'm struggling with from class and we will probably have homework on it.

If G is some arbitrary group, why is G Abelian <---> the factor group of G/Z(G) is cyclic? My professor mentioned something about one direction being trivial but the rest of his proof was unfollowable.
 
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  • #2
Hi Daniel.

The trivial direction is the direct implication ("==>").
Suppose that G is abelian. Then every element commutes with every other element. So by definition then, Z(G) = G, and therefore G/Z(G) = {e}. The group with just one elements is definitely cyclic, so you are done.

Let me just get you started on the reverse, hopefully you can finish it by yourself. So suppose that G/Z(G) is cyclic. We want to prove that G is abelian. Since G/Z(G) is cyclic, this means that there is some a, such that G/Z(G) = {e, a, a2, ..., an} (or possibly infinite). This means that every element from G can be written as ak z for some value of k and some z in Z(G). Now can you show that gh = hg for any two such elements, g = akz and h = ak' z'?
 
  • #3
Thank you compu chip i actually sat down and solved it, i thought the arrows mean't something else...been a late night. I got the other half of it, I can post it if you want it up for other users, unless you'd rather them try on their own
 
  • #4
You can post it, if you would like us to check it.
I am confident that you managed to do it correctly though, and if you are too, you might as well save yourself the trouble of typing it out.

Just out of curiosity, what did you think the arrows meant?
 
  • #5
I thought it mean't implies that, I've had two professors use it diffeerently, which would be unidirectional...yea...one of those nights :)
 
  • #6
Then they were sloppy and/or wrong.

"P <=> Q" always means "P implies Q and Q implies P", so if either is true then so is the other. Other formulations are "P if and only if Q", "P is necessary and sufficient for Q" and "P is equivalent to Q".
 

Related to Proving G Abelian iff G/Z(G) is Cyclic

1. What does it mean for a group to be Abelian?

A group is considered Abelian if its elements commute with each other, meaning that the order of operations does not affect the outcome. Formally, this means that for any two elements a and b in the group, a*b = b*a.

2. How is the commutativity of a group related to its center?

The center of a group, denoted as Z(G), is the set of elements that commute with every element in the group. Therefore, if a group is Abelian, every element in the group will also be in the center, making the center a normal subgroup of the group.

3. What is the significance of G/Z(G) in determining the structure of a group?

The quotient group G/Z(G) represents the cosets of the center in the group. This group is important because it allows us to study the structure of a group by breaking it down into smaller, more manageable subgroups.

4. How does the cyclic nature of G/Z(G) relate to the Abelian property of G?

If G/Z(G) is cyclic, then it means that G/Z(G) is generated by a single element. This element will also generate the entire group G, meaning that G is cyclic as well. Therefore, if G/Z(G) is cyclic, then G must also be Abelian.

5. Can the converse of "Proving G Abelian iff G/Z(G) is Cyclic" also be true?

Yes, the converse of this statement is also true. If a group G is cyclic, then it must also be Abelian. This is because every cyclic group is commutative, and the center of a cyclic group is the entire group itself. Therefore, G/Z(G) will also be cyclic, proving the converse.

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