Basic Group Theory Proof. Looks easy, might not be.

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Homework Help Overview

The problem involves demonstrating that the equation ax = b has a unique solution within the context of group theory, specifically focusing on the properties of elements in a group.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of group properties, particularly the existence of inverses, to derive the solution. Some express uncertainty about the rigor of their reasoning and whether they might be overlooking important details.

Discussion Status

There is an ongoing exploration of the proof's simplicity and the implications of the result. Some participants acknowledge the need for a deeper understanding of why such results are significant in distinguishing different mathematical structures.

Contextual Notes

Participants note that while the proof may seem straightforward, it is essential to consider examples where similar results do not hold, such as in the case of multiplication of real numbers or matrices, which raises questions about the assumptions underlying the proof.

U.Renko
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Homework Statement


Let a,b be elements of a group G. Show that the equation ax=b has unique solution.

Homework Equations



none really

The Attempt at a Solution



ax = b. Multiply both sides by a^{-1}. (left multiplication). a is guaranteed to have an inverse since it is an element of a group.
Then a^{-1}ax = a^{-1}b and therefore the equattion has solution x=a^{-1}b.
Since in a group, every element has an unique inverse element, it follows that the solution is unique.
I don't know, it just looks too obvious, I may be missing something:
(also: I'm not a math major. I like doing proofs just for fun, and I don't really have that much of practice (yet), so forgive any lack of rigor or something like that. )
Is that it?
 
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U.Renko said:

Homework Statement


Let a,b be elements of a group G. Show that the equation ax=b has unique solution.

Homework Equations



none really

The Attempt at a Solution



ax = b. Multiply both sides by a^{-1}. (left multiplication). a is guaranteed to have an inverse since it is an element of a group.
Then a^{-1}ax = a^{-1}b and therefore the equattion has solution x=a^{-1}b.
Since in a group, every element has an unique inverse element, it follows that the solution is unique.



I don't know, it just looks too obvious, I may be missing something:
(also: I'm not a math major. I like doing proofs just for fun, and I don't really have that much of practice (yet), so forgive any lack of rigor or something like that. )
Is that it?

That's it. It is easy.
 
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Oh cool.
That is nice!

thanks!
 
These "fundamental" proofs are often short and "easy". If you are just starting to do proofs, the hard part may be that the result is so "obvious" you don't understand what needs to be proved!

But these results do need proving, because often they are what make different mathematical structures have different properties. Thinking up examples of situations where a result like this is NOT true can help you understand what is special about a "group". For example, the theorem is not true for multiplication of real numbers. If a = 0, there are no solutions or an infinite number of solutions, depending on whether b is 0 or not. So the integers under multiplication are not a group.

If a,x and b are 2x2 matrices, you can find examples where there are multiple solutions when a and b are both non-zero. So whatever sort of mathematical animal 2x2 matrices are, it's not the same sort of animal as real numbers, and neither of them are groups.
 
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