Ask Your Question
0

abstract algebra

asked 2013-11-18 20:19:38 +0200

anonymous user

Anonymous

updated 2013-11-18 21:13:32 +0200

ppurka gravatar image

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.

edit retag flag offensive close merge delete

1 Answer

Sort by ยป oldest newest most voted
1

answered 2013-11-21 01:39:40 +0200

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?

edit flag offensive delete link more

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

Stats

Asked: 2013-11-18 20:19:38 +0200

Seen: 494 times

Last updated: Nov 21 '13