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        riddles >> hard >> Group existence
        
(Message started by: k505rst on Jan 30^th, 2012, 7:07pm)

Title: Group existence
Post by k505rst on Jan 30^th, 2012, 7:07pm

Prove there is at least one group that G=Z2 X Z4 X Z3 X Z3 X Z5 isomorphic to.

Title: Re: Group existence
Post by Michael Dagg on Jan 30^th, 2012, 8:05pm

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(?).

Title: Re: Group existence
Post by k505rst on Jan 30^th, 2012, 9:28pm

No more than that. Thats my problem. All "finite" groups are isomorphic to permutation group. http://en.wikipedia.org/wiki/Matrix_group

Title: Re: Group existence
Post by Michael Dagg on Jan 30^th, 2012, 10:08pm

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).

Title: Re: Group existence
Post by k505rst on Jan 31^st, 2012, 8:36am

Yes

Title: Re: Group existence
Post by Michael Dagg on Jan 31^st, 2012, 10:05pm

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).

Title: Re: Group existence
Post by k505rst on Jan 31^st, 2012, 11:32pm

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.

Title: Re: Group existence
Post by Michael Dagg on Feb 13^th, 2012, 8:30pm

Did you get it? If so I'd be pleased to see it.