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wu :: forums    riddles    hard (Moderators: SMQ, Icarus, william wu, ThudnBlunder, Eigenray, Grimbal, towr)    Group existence « Previous topic | Next topic » Pages: 1  Reply Notify of replies Send Topic Print    Author  Topic: Group existence  (Read 2842 times) k505rst Newbie     Posts: 4 Group existence   « on: Jan 30th, 2012, 7:07pm » Quote Modify Prove there is at least one group that G=Z2 X Z4 X Z3 X Z3 X Z5 isomorphic to. IP Logged Michael Dagg Senior Riddler     Gender: Posts: 500 Re: Group existence   « Reply #1 on: Jan 30th, 2012, 8:05pm » Quote Modify Every group is isomorphic to a group of   permutations -- hence, Cayley's theorem. So that should do it.   In terms of a direct product, your  G  is isomorphic to the finitely generated abelian group  Z_6 x Z_60  .   I have the feeling that you meant to ask a different   question(?). IP Logged Regards, Michael Dagg k505rst Newbie     Posts: 4 Re: Group existence   « Reply #2 on: Jan 30th, 2012, 9:28pm » Quote Modify No more than that. Thats my problem. All "finite" groups are isomorphic to permutation group.  http://en.wikipedia.org/wiki/Matrix_group   IP Logged Michael Dagg Senior Riddler     Gender: Posts: 500 Re: Group existence   « Reply #3 on: Jan 30th, 2012, 10:08pm » Quote Modify Yes of course. Every finite group is isomorphic to a group of permutations. But that is just a weak form of Cayley's theorem, since a finite group is a group -- "every   group" is isomorphic to a group of permutations.   Your group  G  is the direct product of five finite sets,   and so it is a finite group.     Again, that should do it (in fact, just as you say yourself).   IP Logged Regards, Michael Dagg k505rst Newbie     Posts: 4 Re: Group existence   « Reply #4 on: Jan 31st, 2012, 8:36am » Quote Modify Yes IP Logged Michael Dagg Senior Riddler     Gender: Posts: 500 Re: Group existence   « Reply #5 on: Jan 31st, 2012, 10:05pm » Quote Modify Ah. I surmise that you are looking for a proof to Cayley's   theorem, that is, from your dialogue.     There are many because it is an easy result in group theory.  It may not seem so easy if you disguise the question behind a set of direct products which may form a cyclic group (of order 360). IP Logged Regards, Michael Dagg k505rst Newbie     Posts: 4 Re: Group existence   « Reply #6 on: Jan 31st, 2012, 11:32pm » Quote Modify There is term strongly sad. I not strongly sad with what I posted. Yes I do seek alternate proofs to Cayleys formulae. I do know what with what I have do is to show of symmetric group. Of such of confused. It would be of what word is said "in particular" but I do not know surely with each element of group is a permutation to itself and then have the group plowed into all permutations of the group itself.   IP Logged Michael Dagg Senior Riddler     Gender: Posts: 500 Re: Group existence   « Reply #7 on: Feb 13th, 2012, 8:30pm » Quote Modify Did you get it? If so I'd be pleased to see it. IP Logged Regards, Michael Dagg Pages: 1  Reply Notify of replies Send Topic Print Forum Jump: ----------------------------- riddles -----------------------------  - easy   - medium => hard   - what am i   - what happened   - microsoft   - cs   - putnam exam (pure math)   - suggestions, help, and FAQ   - general problem-solving / chatting / whatever ----------------------------- general -----------------------------  - guestbook   - truth   - complex analysis   - wanted   - psychology   - chinese « Previous topic | Next topic »

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