Is a group of order 5 always abelian?

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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
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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