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

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Discussion Overview

The discussion revolves around the theoretical relationship between a group G being Abelian and the factor group G/Z(G) being cyclic. Participants explore the implications of these properties, with a focus on understanding the proof and its components.

Discussion Character

  • Theoretical exploration
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant questions why G being Abelian is equivalent to G/Z(G) being cyclic, noting difficulty in following the professor's proof.
  • Another participant explains that if G is Abelian, then Z(G) equals G, leading to G/Z(G) being the trivial group, which is cyclic.
  • There is a proposal to demonstrate the reverse implication, suggesting that if G/Z(G) is cyclic, then G must be Abelian, and asks the original poster to show that elements commute under this condition.
  • A participant expresses gratitude for assistance and mentions they have solved the problem, indicating a misunderstanding of notation regarding implications.
  • There is a discussion about the meaning of the notation "P <=> Q," with one participant asserting that it represents a biconditional relationship.

Areas of Agreement / Disagreement

Participants generally agree on the trivial direction of the proof but do not reach a consensus on the reverse implication, as it remains an open question for further exploration.

Contextual Notes

There are unresolved issues regarding the notation used in the discussion, which may lead to different interpretations of implications. The understanding of the relationship between G and Z(G) is also dependent on the definitions and properties of groups that may not have been fully articulated.

Who May Find This Useful

This discussion may be useful for students studying group theory, particularly those interested in the properties of Abelian groups and factor groups.

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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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'?
 
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
 
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?
 
I thought it mean't implies that, I've had two professors use it diffeerently, which would be unidirectional...yea...one of those nights :)
 
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".
 

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