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let f=x^3*y^3+2y^2, f1=2xy^2+3x+4y^2, f2=y^2-2y-2 in Q[x,y] using lex with x>y divide f by f1,f2 to obtain a reminder r. Repeat this exercise reversing the role of f1 and f2.

let f=x^3*y^3+2y^2, f1=2xy^2+3x+4y^2, f2=y^2-2y-2 in Q[x,y] using lex with x>y divide f by f1,f2 to obtain a reminder r. Repeat this exercise reversing the role of f1 and f2.

There are (at least) three ways to define a cyclic of order 12 in Sage:

sage: C12 = groups.permutation.Cyclic(12)
sage: G12 = AbelianGroup([12])
sage: H12 = AdditiveAbelianGroup([12])

Now evaluate C12.gens() (for example): how many generators does this group have? Any homomorphism is determined by where the generators go. So what are the possible endomorphisms (= homomorphisms from the group to itself) of this group? Can you figure out which of them are actually automorphisms?

An automorphism is an isomorphism between a group and itself. The identity function (x -> x) is always an isomorphism, which we consider trivial. Use Sage to construct a nontrivial automorphism of the cyclic group of order 12. Check that the mapping is both onto and one-to-one by computing the image and kernel and performing the proper tests on these subgroups. Now construct all of the possible automorphisms of the cyclic group of order 12.