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Let $G$ be a group and $g,h\in G$ two elements. I need to prove that the equation $xg=h$ has a unique solution.

Applying $g^{-1}$ to both sides of the equation we get : $(xg)g^{-1}=hg^{-1}$

If I consider the Right Hand Side of the equation:

$$(xg)g^{-1}=x(gg^{-1})=x\cdot e=x$$

Now $x=hg^{-1}$ which makes me conclude that $xg=h$ has a unique solution. Can anyone correct me please!!!

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Please notice that you need to place latex code within '$' signs. Place one dollar sign before and one after whatever you want to appear with proper math symbols. –  Ittay Weiss Jan 30 '13 at 2:36
What you did is correct, remembering, of course, that the inverse of an element in a group is unique –  DonAntonio Jan 30 '13 at 2:39
No, this is fine. –  gnometorule Jan 30 '13 at 2:39
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3 Answers

After much thought, I have decided to make significant revisions to my original post.

Firstly, let us try to recreate Timoci’s thought process.

  • As required by the group axioms, $ g $ has an inverse.

  • Without further knowledge of group theory, it is unclear if $ g $ has only one inverse or has many inverses. Whatever the case is, just fix an inverse and denote it by $ g^{-1} $.

  • Multiply both terms in the equation $ xg = h $ on the right by $ g^{-1} $ to obtain $ x = h g^{-1} $.

  • Verify that $ x = h g^{-1} $ is indeed a solution to the equation.

Timoci’s own argument settles the existential part of his problem by furnishing a solution. Concerning the uniqueness of this solution, I would like to make the following point: By fixing an inverse $ g^{-1} $ of $ g $ in the second step, Timoci actually proves that any solution must equal $ h g^{-1} $. Hence, we do not need to assume that $ g $ has a unique inverse in order to prove that the solution is unique. Indeed, we only need to assume that $ g $ possesses at least one inverse. A more formal argument is presented below.

Suppose that $ x_{1} $ and $ x_{2} $ are solutions to the equation $ xg = h $. Then $ x_{1} g = x_{2} g = h $. Pick any inverse of $ g $ (an inverse exists, as stipulated by the group axioms) and denote it by $ g^{-1} $. Apply $ g^{-1} $ to both terms in the equation $ x_{1} g = x_{2} g $ on the right to obtain $ x_{1} = x_{2} $. Therefore, all solutions are identical.

This argument is a demonstration of the phenomenon of group cancellation, so what I have done here is simply an elaboration of ncmathsadist’s solution. With due respect to the other contributors to this thread, we can also solve the problem by translating the assumption that there exist two different solutions into the contradictory statement that $ g $ has two different inverses. However, a satisfactory explanation does not necessarily require prior knowledge of the uniqueness-of-inverse result.

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This is 100% correct. It is an application of the cancellation laws.

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Your answer is correct, as you argued:

\begin{align} xg = h &\iff (xg) g^{-1} = h g^{-1} \\ &\iff x = h g^{-1}. \end{align} Hence, as you say, $ x = h g^{-1} $ solves the equation $ xg = h $.

But be sure to establish that while you found "a" solution to the equation, the solution you found must in fact be the unique solution and why this is so:

$x = hg^{-1}$ is the unique solution to the equation $xg = h$ because $h, g \in G$ and $g^{-1}\in G$ is the unique inverse of $g \in G$, by definition of the fact that $G$ is a group, and each element in a group has a unique inverse within the group. You need only appeal to the definition of the inverse of a group element and its uniqueness.

So there is no other element $j\neq g^{-1}$ such that $x = hj$ is a solution, else $j$ would also necessarily be the inverse of $g$ which is impossible, as the inverse of an element in a group is unique.

In short, it's a minor point, but nonetheless it's worth stating at the conclusion of your argument that the uniqueness of the solution follows from the uniqueness of the inverse of $g$.

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Is this clear now, Timoci? Your work was fine; just back up the uniqueness by asserting that "since the inverse of a group element is unique", the solution $g^{-1}$ is unique. (Any other solution would necessarily be the inverse of g, hence the solution is unique.) –  amWhy Jan 30 '13 at 23:52
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