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

1. Feb 15, 2014

### U.Renko

1. The problem statement, all variables and given/known data
Let $a,b$ be elements of a group $G$. Show that the equation $ax=b$ has unique solution.

2. Relevant equations

none really

3. 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?

Last edited: Feb 15, 2014
2. Feb 15, 2014

### Dick

That's it. It is easy.

3. Feb 15, 2014

### U.Renko

Oh cool.
That is nice!

thanks!

4. Feb 16, 2014

### AlephZero

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.