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   Author  Topic: Group existence  (Read 2719 times)
k505rst
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Group existence  
« on: Jan 30th, 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 30th, 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
k505rst
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Re: Group existence  
« Reply #2 on: Jan 30th, 2012, 9:28pm »
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No more than that. Thats my problem. All "finite" groups are isomorphic to permutation group.  http://en.wikipedia.org/wiki/Matrix_group
 
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Michael Dagg
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Re: Group existence  
« Reply #3 on: Jan 30th, 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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Michael Dagg
k505rst
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Re: Group existence  
« Reply #4 on: Jan 31st, 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 31st, 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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Michael Dagg
k505rst
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Posts: 4
Re: Group existence  
« Reply #6 on: Jan 31st, 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 13th, 2012, 8:30pm »
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Did you get it? If so I'd be pleased to see it.
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Michael Dagg
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