Thread: Groups- Proof using on the defination of a group

Let G be a group and let a, b ∈ G. Using only the definition of a group show that the two equations xa = b and ay = b have unique solutions in G. Is it always true that x = y?

Originally Posted by amm345

Let G be a group and let a, b ∈ G. Using only the definition of a group show that the two equations xa = b and ay = b have unique solutions in G. Is it always true that x = y?

all elements of a group have inverses and the inverse is commutative

Originally Posted by amm345

Let G be a group and let a, b ∈ G. Using only the definition of a group show that the two equations xa = b and ay = b have unique solutions in G. Is it always true that x = y?

Multiply xa = b by the inverse of a from the right and ay = b by the inverse of a from the left and you'll get uniqueness (since x (y) is expressable by means of a,b)

No, in general the solutions aren't equal (if the group is non-abelian).
Take for example, in the symmetric group S_3, x(12) = (123), (12)y = (123).

The first equation's solution is x = (123)(12) = (13), whereas the second one's is y = (12)(123) = (23).

Tonio

I'm sorry, but I just don't see how multiplying by inverse of a on the right of the first and the left of the second proves uniqueness.. Please explain more?

Originally Posted by amm345

I'm sorry, but I just don't see how multiplying by inverse of a on the right of the first and the left of the second proves uniqueness.. Please explain more?

so so so

multiplying is well definited so the unique solution is , the product cannot be two different elements of g

Originally Posted by amm345

I'm sorry, but I just don't see how multiplying by inverse of a on the right of the first and the left of the second proves uniqueness.. Please explain more?

I already did: xa = b ==> xa*a^(-1) = b*a^(-1) ==> x = ba^(-1) and you're expressing the solution x as a function of a,b and this means x is unique since a, b are unique (arbitrary. but once they're given they're fixed and unique).
The same is true with the other one.

Tonio