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

Daniiel
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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?
 
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Daniiel said:
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.
 
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
 
Daniiel said:
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.
 
Do you mean that it should be shown what the is?

z= y-1x y ?
 
Daniiel said:
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.
 
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