|G|=4. Prove the group is either cyclic or g^2=e

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Discussion Overview

The discussion centers on the properties of a group G with an order of 4, specifically exploring whether G is cyclic or if every element squared equals the identity element. The scope includes theoretical reasoning and mathematical exploration of group structure.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that the order of an element x in G must be either 2 or 4, as the order must divide the group's order of 4.
  • Others argue that if the order of x is 2, then it follows that x^2=e, meaning each element is its own inverse.
  • A participant questions the implications if the order of x is 4, expressing uncertainty about the group structure in that case.
  • Another participant confirms that if there is an element of order 4, it generates the group, making G cyclic.
  • One participant reiterates the need to consider the orders of elements, stating that if no elements of order 4 exist, then x^2=e for all x in G.

Areas of Agreement / Disagreement

Participants generally agree on the possible orders of elements in the group but express uncertainty regarding the implications of these orders on the overall structure of G. Multiple competing views remain regarding the nature of the group based on the order of its elements.

Contextual Notes

There are limitations in the discussion regarding the assumptions about the existence of elements of certain orders and the implications of those orders on the group's structure. The discussion does not resolve the mathematical steps necessary to fully prove the claims made.

mathmajor2013
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Let G be a group with |G|=4. Prove that either G is cyclic or for any x in G, x^2=e.
 
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Consider and element x in G of maximum order. What possibilities are there, and what does each of them say about the group structure?
 
So the order of x is 2 or 4, right? Since the order must divide 4. If it's 2, the x^2=e, and so each element is its own inverse. If it's 4...I'm lost.
 
Correct. And if it's 4, what can you say about x, x^2,x^3 and x^4? Are any of them equal? What is x for this group?
 
Damn. They are all unique, and x^4=e. and x is the generator for the group yeah?
 
So the order of x is 2 or 4, right? Since the order must divide 4. If it's 2, the x^2=e, and so each element is its own inverse. If it's 4...I'm lost.
 
You are correct. The order of each element in the group must divide the order of the group. So, the order of each element must be 1,2 or 4. If there are no elements of order 4, then x^2=e for each x in the group (since the order of each element is 1 or 2). If there is an element of order 4, then this element is a generator, as you pointed out, and so the group is cyclic.
 
mathmajor2013 said:
Let G be a group with |G|=4. Prove that either G is cyclic or for any x in G, x^2=e.

a non identity element must have either order 4 or order 2
 

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