Proving Abelianity in Groups with $g^{2} = 1$: A Simple Proof

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Discussion Overview

The discussion revolves around proving that a group where every element satisfies the condition $g^{2} = 1$ is Abelian. Participants are exploring the implications of this condition and discussing various approaches to the proof, including algebraic manipulations and definitions of group properties.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • GreenGoblin seeks assistance in proving that a group with $g^{2} = 1$ for all elements is Abelian, expressing uncertainty about how to demonstrate this.
  • One participant suggests examining the expression $(gh)^2 = 1$ to explore the implications of the group structure.
  • Another participant claims that the condition $g^{2} = 1$ may not be relevant to proving the group is Abelian, raising questions about the meaning of "the group of $g^{2}$."
  • A different participant asserts they have shown algebraically that $gh = hg$ without relying on the condition $(gh)^{2} = 1$.
  • One participant emphasizes that not all groups are Abelian, suggesting that the proof should start by assuming $g^2 = 1$ for all elements and exploring the consequences.

Areas of Agreement / Disagreement

Participants express differing views on the relevance of the condition $g^{2} = 1$ in proving the group is Abelian. There is no consensus on the approach to take or the necessity of this condition in the proof.

Contextual Notes

Some participants are unclear about the implications of the condition $g^{2} = 1$ and the terminology used, leading to potential misunderstandings in the proof process. The discussion reflects various interpretations and approaches to the problem without resolving these ambiguities.

Who May Find This Useful

This discussion may be useful for students or individuals interested in group theory, particularly those exploring properties of groups and the conditions that lead to Abelian structures.

GreenGoblin
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I need to show a group with $g^{2} = 1$ for all g is Abelian. This is all the information given, I do know what Abelian is, and I know that this group is but I don't know how to 'show' it. Can someone help?

Gracias,
GreenGoblin
 
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For a start, see this page from The Math Forum.
 
I did the proof and it doesn't seem to be relevant that $g^{2}=1$. Just that it is the group of $g^{2}$. Is this right?
 
For $$ g,h\in G $$ consider $$ (gh)^2=1 $$ then expand out and see what happens
 
GreenGoblin said:
I did the proof and it doesn't seem to be relevant that $g^{2}=1$. Just that it is the group of $g^{2}$. Is this right?
I am not sure what "the group of $g^{2}$" means. Could you post your proof?
 
I just showed algebraically gh = hg, without using the fact $(gh)^{2}=1$.
 
perhaps you should show your proof.

it is not the case that all groups are abelian, only some groups are.

a typical proof starts out like this:

suppose $g^2 = 1$ for all g in G. let g, h be any two elements of G (which perhaps might be the same element, if |G| = 1).

then gh is an element of G, so

$(gh)^2 = 1$, that is:

ghgh = 1. then,

g(ghgh) = g(1) = g, and...?
 

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