LetGbe a group and leta, b ∈ G. Using only the definition of a group show that the two equationsxa=banday=bhave unique solutions inG. Is it always true thatx=y?

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- Oct 6th 2009, 01:56 PMamm345Groups- 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*? - Oct 6th 2009, 02:00 PMartvandalay11
- Oct 6th 2009, 02:02 PMtonio

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 - Oct 6th 2009, 02:58 PMamm345
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?

- Oct 6th 2009, 03:07 PMartvandalay11
- Oct 6th 2009, 03:08 PMtonio

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