Prove that, in any group G, /byb^ -1/ = /y/?

Question

Prove that, in any group G, /byb^ -1/ = /y/?

Answer 1

That's true only for abelian groups.

If by =/= yb, then byb^-1 =/= y.

Edit: Yeah, listen to the guy below me.  His answer is chock full of creamy goodness.

Answer 2

I think the questioner means |byb^(-1)| and |y|, meaning the orders of those elements.  I'm going to use the notation o(y) to denote the order of y.  I'm also going to use e to denote the identity of G.

Certainly it's trivial in the abelian case since byb^(-1) = y.  But there's actually an argument that catches the abelian and nonabelian cases at the same time.

Consider multiplying byb^(-1) by itself.  (byb^(-1))(byb^(-1)) = byb^(-1)byb^(-1) = by[b^(-1)b]yb^(-1) = byyb^(-1) = b(y^2)b(-1).  Multiply this by byb^(-1) again and you get b(y^2)b^(-1)byb^(-1) = b(y^3)b^(-1).

So in general, (byb^(-1))^n = b(y^n)b^(-1).  The b and b^(-1) stay fixed, and the power of y increases.

Observe, then, that (byb^(-1))^(o(y)) = beb(-1) = bb^(-1) = e.  So o(byb^(-1)) is equal to or less than o(y).

[The next step is by contadiction].  Suppose that o(byb^(-1)) < o(y).  Let o(byb^(-1)) = n.  Then b(y^n)b^(-1) = e.  Left-multiplying both sides by b^(-1), we get that (y^n)b^(-1) = b^(-1).  This implies that y^n = e.  But that cannot be since n < o(y).  So it cannot be that o(byb^(-1)) < o(y).

Thus we can conclude that o(byb^(-1)) = o(y).  We never used commutativity in this proof, so the result holds for all groups.

Answer 3

WTF??? This isn't algebra thread, so what are you trying to ask????