Talk:Conjugate of Set by Group Product

I actually think that the link should be for the subset product page because it's a product of singleton sets (although technically the results are the same, ignoring the curly braces). --Cynic (talk) 23:35, 28 May 2010 (UTC)

I think not.

My thinking was: you are make the logical step from:

$\left({b \circ a}\right) \circ S \circ \left({a^{-1} \circ b^{-1}}\right)$

to:

$\left({b \circ a}\right) \circ S \circ \left({b \circ a}\right)^{-1}$

which follows because of the equality / identity:

$\left({a^{-1} \circ b^{-1}}\right) \equiv \left({b \circ a}\right)^{-1}$

which is precisely Inverse of Group Product.

The point about subset products is appropriate elsewhere on the page, agreed. But the point of the proof specifically hinges on the Inverse of Group Product result (which I mucked up the link to when I originally wrote the page).

--Prime.mover 07:26, 29 May 2010 (UTC)