Finding center of a group with one element of order 2

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SUMMARY

This discussion focuses on proving that in a group A with exactly one element of order 2, denoted as c, this element must be the center of the group. The proof involves demonstrating that for all elements a in A, the equation ac = ca holds true. Key steps include manipulating the identity c^2 = 1 and exploring implications of b^2 = 1, where b can either be c or the identity element. The participants emphasize the importance of clearly stating assumptions and constructing logical sequences in the proof.

PREREQUISITES
  • Understanding of group theory concepts, specifically group elements and their orders.
  • Familiarity with the properties of abelian groups and their implications.
  • Knowledge of algebraic manipulation and proof techniques in mathematics.
  • Basic understanding of identity elements in group theory.
NEXT STEPS
  • Study the properties of group centers and their significance in group theory.
  • Learn about abelian groups and how they relate to elements of order 2.
  • Explore proof strategies in algebra, focusing on logical sequences and assumptions.
  • Investigate examples of groups with elements of order 2 to solidify understanding.
USEFUL FOR

This discussion is beneficial for undergraduate students studying abstract algebra, particularly those interested in group theory and proof construction. It also aids educators seeking to clarify concepts related to group centers and element orders.

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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