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).
 * --Matt Westwood 07:26, 29 May 2010 (UTC)