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Topic: Group existence (Read 2792 times) 

k505rst
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Group existence
« on: Jan 30^{th}, 2012, 7:07pm » 
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Prove there is at least one group that G=Z2 X Z4 X Z3 X Z3 X Z5 isomorphic to.


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Michael Dagg
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Re: Group existence
« Reply #1 on: Jan 30^{th}, 2012, 8:05pm » 
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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(?).


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Michael Dagg
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Re: Group existence
« Reply #3 on: Jan 30^{th}, 2012, 10:08pm » 
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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).


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k505rst
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Re: Group existence
« Reply #4 on: Jan 31^{st}, 2012, 8:36am » 
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Yes


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Michael Dagg
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Re: Group existence
« Reply #5 on: Jan 31^{st}, 2012, 10:05pm » 
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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).


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k505rst
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Posts: 4


Re: Group existence
« Reply #6 on: Jan 31^{st}, 2012, 11:32pm » 
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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.


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Michael Dagg
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Re: Group existence
« Reply #7 on: Feb 13^{th}, 2012, 8:30pm » 
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Did you get it? If so I'd be pleased to see it.


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