Re: Binary operator, (x + y)^2 = x^2 + x*y + y*x + y^2

cbrown@xxxxxxxxxxxxxxxxx wrote:

<snip>

The function q you describe doesn't actually change the problem; the
action of q^-1 K q acts like an (inner) automorphism on (S, o) because
q is a bijection.

I meant, "because q is invertible"

Cheers - Chas

.

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