Is a group of order 5 always abelian?

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Homework Help Overview

The discussion revolves around the properties of groups, specifically focusing on whether a group of order 5 is necessarily abelian. The original poster seeks to demonstrate this without utilizing Lagrange's Theorem, which sets the context in the field of Abstract Algebra.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of group order and the nature of cyclic groups, questioning whether the original poster can use known properties of groups of prime order. There is also an exploration of constructing the group from its elements and examining the consequences of assuming non-commutativity.

Discussion Status

The conversation is ongoing, with participants providing guidance on how to approach the problem while clarifying the constraints imposed by the original poster's request. There is no explicit consensus yet, as various interpretations and methods are being considered.

Contextual Notes

The original poster has indicated that only the definitions of a group and an abelian group are to be used, which may limit the approaches available for solving the problem.

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Homework Statement



Show that a group of order 5 must be abelian.Please don't use Langrage's Theorem.


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The Attempt at a Solution



I have been working on this problem for a while and I can't seem to
get anywhere on it. Please help.
 
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The problem is that, without you showing any work at all, we don't know what you can and can't use. It is pretty well known that the only group of any prime order is the cyclic group, which is always Abelian, but are you allowed to use that?
 
This problem is from the first set of problems in Abstract Algebra(I N Herstein).Only the definitions of a group and abelian group are to be used in solving this problem and nothing else ,for these are the only things I came across in this book till now.
 
it looks like you will need to build your group up from scratch {e,x1,x2,x3,x4}

assume there is at least two elements x1 and x2 which do not commute and try to show we cannot form a group for all possible outcomes.
I. x1*x2=e
II. x1*x2=x1
III. x1*x2=x2
IV. x1*x2=x3
V. x1*x2=x4
 

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