On Tue, Aug 28, 2012 at 12:33 PM, Tom Bachmann <[email protected]> wrote: > On 27.08.2012 22:52, David Joyner wrote: >>
... >> The way you want to do it, they aren't an action: >> >> if g1 = (1,2) and g2 = (2,4,5) then (g1*g2)(2) \not= g1(g2(2)): >> > > I'm not sure I get this. Since you are much more knowledgable than me maybe > I should just shut up, but as far as I know, a (left) action of a group G on > a set S is a map phi: GxS -> S, where I shall write g(s) for phi(g, s) if g > in G and s in S. Then the axioms are: > > - e(s) = s for all s > - g(h(s)) = (gh)(s) for all s in S, g, h in G > > > If we let Sym(S) denote the symmetric group on S, by which I mean the set of > all bijections S->S with the "ordinary" (left, or *) composition, then I > believe it is easy to check that an action is the same as a homomorphism > from G to Sym(S). Anyway, we *obviously* want (g1*g2)(2) = g1(g2(2)). > > Wikipedia agrees with this definition of (left) action, btw. > > We can similarly define a right action. For example, if Sym'(S) denotes > Sym(S) with the other composition law, then a right action is a homomorphism > G->Sym'(S). More explicitly, it is a map phi:SxG -> S, such that > > - (s)e = s for all s in S > - ((s)g)h = (s)(gh) for all s, g, h. > > Where am I wrong? I think you are saying if you use one definition of the product, it is a left action and the other definition of product is a right action. Good point. You are correct. So my argument that the R-L product is not an action is wrong. Thanks for pointing that out. I still take the position that the L-R product is the usual way it is done in computer algebra systems. > > > I think this all just boils down to the fact that for actual computations > with permutations, it is more convenient to work with the "." operation > (i.e. composition on the left) than with the "*" operation, essentially > because we read from left to right. Much of the confusion probably stems > from the fact that these conventions are completely equivalent (Taking > opposite groups interchanges left and right actions, and the opposite group > is naturally isomorphis to the group we started with). > > Thanks, > Tom > > >> sage: G = SymmetricGroup(5) >> sage: g1 = G([(1,2)]) >> sage: g2 = G([(2,4,5)]) >> sage: g2(2) >> 4 >> sage: g3 = g1*g2 >> sage: a = g2(2); a >> 4 >> sage: b = g1(a); b >> 4 >> sage: g3(2) >> 1 >> >> It is very important for many people that the group >> of permutations yields a group action on the set. >> >> >>> Observe now that any permutation can be written uniquely (up to ordering) >>> as >> >> >> ... >> >>> >>> -- >>> You received this message because you are subscribed to the Google Groups >>> "sympy" group. >>> To post to this group, send email to [email protected]. >>> To unsubscribe from this group, send email to >>> [email protected]. >>> For more options, visit this group at >>> http://groups.google.com/group/sympy?hl=en. >>> >> > > -- > You received this message because you are subscribed to the Google Groups > "sympy" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/sympy?hl=en. > -- You received this message because you are subscribed to the Google Groups "sympy" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/sympy?hl=en.
