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On Thu, 13 Jan 2005 09:03:56 +1100, Rick Marshall <rjm@z...> wrote: > Peter Hunsberger wrote: > > >On Thu, 13 Jan 2005 08:06:08 +1100, Rick Marshall <rjm@z...> wrote: > ><snip/> <snip/> > > > >Bad analogy Rick. In Group theory groups are, by definition, a set of > >elements (possibly infinite) that is closed over some operator. > > > i think that's what i said - although perhaps i was looking at an > interesting aspect of this closure - you can't tell if an instance of an > element of the set exists in it's own right or as the result of applying > the operator. ie you can't uniquely decompose an element of a group. > > ><snip/> Umm, ok, but I think you're stretching :-) The whole point of a group is that it doesn't matter; there's no such thing as decomposition per se, just elements and an operator, that's why group isomorphism is such a powerful tool. (Closure just guarantees that if a . b is in the group then b . a is also in the group.) BTW, before people start to nitpick I realize that to completely define a group you also need associativity, an identity element and an inverse for every element.... -- Peter Hunsberger
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