Finding center of a group with one element of order 2

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Homework Help Overview

The discussion revolves around proving that if a group contains exactly one element of order 2, then that element is the center of the group. The original poster formulates the problem using group properties and seeks to establish commutativity with all elements in the group.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the element's order, considering cases where elements square to the identity. They discuss manipulating expressions involving the group elements and question the validity of certain algebraic steps.

Discussion Status

Some participants provide hints and suggestions for approaching the proof, while others express uncertainty about specific steps taken. There is an ongoing exploration of the relationships between the elements and the assumptions made, with no clear consensus yet on the proof's structure.

Contextual Notes

Participants note the importance of distinguishing between cases when dealing with elements of order 2 and the implications of the group's structure. The original poster is also new to algebra, which may influence their understanding of the concepts discussed.

Sprooth
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I need to prove that if a group contains exactly one element with order 2, then that element is the center of the group.

Here is how I formulated the problem:
Let A be a group with an element c such that c^2 = 1 (i.e. c = c^-1), and b^2 = 1 implies b=c.

Want either:
ac = ca, for all a in A
ca(c^-1) = a, for all a in A.

After doing this, I tried manipulating the symbols for a while using different identities and properties that I know, but I can't seem to come up with any equations where one side doesn't reduce to the other trivially. I have a feeling that there is one small but crucial thing I am missing. Abelian groups have crossed my mind a few times, but I don't know how to use that.

Does anyone have any suggestions for where to go from here?
 
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b^2=1 actually implies b=c OR b=1.

So the information we have is:
b^2=1 iff b=c or b=1.

And we're trying to show:
ac = ca for all a in A.

So we need to somehow convert our problem into one where we can use ALL the information we have. So can you convert what we are trying to show into b^2=1 (keeping in mind c^2=1) for some b?
 
Try taking the square of aca^{-1}.
 
fantispug: Thanks for pointing out that b^2 = 1 implies either b = 1 or b = c. I hadn't thought of that. Taking that into consideration, I played around with the equations some more, but there was something that confused me a little bit. When considering b with b^2 = 1, when b is 1 or c must be treated as separate cases, right?

jambaugh: I think your hint may have given me the connection I needed, although it still feels like I'm not doing everything correctly.

So is this in the right direction of the proof?
1 = b^2
aa^{-1} = b^2
acca^{-1} = b^2
aca^{-1} * aca^{-1} = b^2
(aca^{-1})^2 = b^2 = c^2
(aca^{-1})^2 = c^2
aca^{-1} = c // I'm not sure about the step leading to this. Can you take roots like that?
ac = ca
So c is the center of the group.

Could someone please tell me if I skipped any steps or made any incorrect steps or if I am just not getting it? I'm just starting algebra this semester, and I think it's interesting, but I'm having a hard time getting into the right mindset so far.

Thanks for the help.
 
I think you have all the steps but you should work on the sequence of inferences. You should include some text especially when invoking the assumptions. I'm thinking something like:

"Observe that __=__."
"Thus by our assumption ___ either ___ or ___ ."
"If (case 1) then ...
"If (case 2) then ...
"Thus it is shown that _____"
 

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