Proving Non-Abelian Groups Have Unique Elements with Non-Commutative Properties

[Daniiel]

To show that a non-abelian group G, has elements x,y,z such that xy = yz where y≠z,

Is it enough to simply state for non-abelian groups xy≠yx so if you have xy=yz then it is not possible for x=z due to xy≠yx?

Or is more detail required?

 

[Hurkyl]

for non-abelian groups xy≠yx

There does not exist any group satisfying that identity.

In a non-abelian group, only some pairs $(a,b)$ satisfy $ab\neq ba$.

 

[Daniiel]

Thanks for replying Hurkyl

So it would be a similar argument just clarifying that the for some (not all) x,y in G?

For some x, y in a non-abelian group G

xy≠yx

and if

xy = yz for some z in G

then x≠z otherwise

xy≠yx

Is not satisfied for elements x and y of the group

 

[chiro]

Thanks for replying Hurkyl

So it would be a similar argument just clarifying that the for some (not all) x,y in G?

For some x, y in a non-abelian group G

xy≠yx

and if

xy = yz for some z in G

then x≠z otherwise

xy≠yx

Is not satisfied for elements x and y of the group

To follow on from Hurkyls post, think about identity and inverse elements.

 

[Daniiel]

Do you mean that it should be shown what the is?

z= y^-1x y ?

 

[Barre]

Do you mean that it should be shown what the is?

z= y^-1x y ?

If the group is non-abelian, then there sure must be a pair a,b both of which ain't the identity, such that ab != ba. Take such a pair, and play around with the conjugate you proposed.