Uni Stuff > Mathematics
Abstract Algebra
kamil9876:
ahh yes sorry I was high on English beer. I made the problem up myself, I tried to prove in vain that they are not isomorphic ( I first proved that Z and Z^2, Q and Q^2 are not isomorphic). When I come back home on saturday I will post in more detail. I'm not even sure if it is decidable without the continumm hypothesis or axiom of choice. A clue is to think of R and R^2 as vector spaces over Q.
zzdfa:
yeah I thought about it a little more:
The Z->Z^2 case follows from that fact that p(z)=zp(1)
The Q->Q^2 case follows from the fact that any homomorphism is in fact a linear transformation (by fiddling around with the defn of homomorphism), and we know that there is no linear isomorphism Q->Q^2 (different dimensions)
For the R->R^2 case, assuming that bases for R and R^2 exist and have the same cardinality, find a bijection between the bases and we're done.
And it seems like you need AC to find the basis and CH to prove that they have the same cardinality.
Ahmad:
An idea for showing R and R^2 as Q vectorspaces have dimension of the continuum: If R had a countable basis then we can express any element of R uniquely as a finite linear combination of elements of the basis B which means R is the union of all elements expressible by a linear combination of 1 basis element, and those expressible as a linear combination of 2 basis elements, and 3 basis elements and so on. This is a countable union of countable sets (why are they countable?), which is countable but R is uncountable.
So the basis has size more than aleph_0 and at most the continuum (R is a spanning set for itself, of course), now you need CH. Same goes for R^2. Intriguing result.
kamil9876:
yep good
/0:
Let be a cyclic group of order . Can someone please explain how if are coprime then is isomorphic to ? I arrived late to one of the lectures and must have missed the explanation. Thanks
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