Is there a way to find a solution to the equation ax=b in the group G?

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SUMMARY

The equation ax = b in a group G has a unique solution, which can be expressed as x = a^{-1}b. To demonstrate the existence of this solution, one can multiply both sides of the equation by the inverse of a, a^{-1}, confirming that a^{-1} is also an element of the group G. The uniqueness of the solution follows from the property that inverses in a group are unique, thus proving that if x' is another solution, it must equal x.

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Homework Statement



let a,b [tex]\in[/tex]G show that ax=b has a unique solution in G

The Attempt at a Solution



i know what needs to be done, i just don't know how to do it.

Want to prove:

1. There is a solution
2. solution is unique

to prove uniquness of a soltuion just suppose you have a different solution x' and show that x'=x


to show that there is a solution (im not sure this part is right, cause it seems too simple) simply multiply (left) by [tex]a^{-1}[/tex]

that is [tex]a^{-1}[/tex]*ax=[tex]a^{-1}[/tex]b

we can do this because a is in the group, which implies [tex]a^{-1}[/tex] is in the group

THis is where i am stuck... i really don't know how to proceed. Have i even done the first part right?

any help appreciated
 
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So you showed that there is a solution, namely [itex]x=a^{-1}b[/itex]. To show that x is unique, use the fact that in a group, inverses are unique.
 

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