Thanks for the reference, I'll take a look when I have the chance. I was thinking that I may have a method to prove what I was asking. The Cayley graph of F(A) is a tree T. Now F(A) acts freely and properly discontinuously on T, and thus any subgroup must also act freely and properly...
Suppose we have a group with presentation G = <A|R> i.e G is the quotient of the free group F(A) on A by the normal closure <<A>> of some subset A of F(A). Is it true that that fundamental group of the Cayley graph of G (with respect to the generating set A) will be isomorphic to the subgroup...
Z is just an infinite cyclic group. So give your copies of Z presentations as follows: Z=<a|-> and Z=<b|-> (the generators are distinct because formally we think of the two copies of Z as being distinct). Then Z*Z=<a,b|-> by the explanation here: http://en.wikipedia.org/wiki/Free_product under...
I am trying to prove that C\otimesC (taken over R) is equal to C^2. The method I have seen is to show the following equivalences:
C\otimesC = C\otimes(R[T]/<T^2+1>) = C[T]/<T^2+1> = C.
(All tensor products taken over R).
The only part I am having trouble with is showing that...
I am fairly certain that when we take the free product of a group with itself, we formally think of the elements as being distinct if they come from a different copy of the group. For example, the free product of Z with itself is the free group F_2. It might be useful when doing something like...
Is it true that for any unital ring (not necessarily commutative), that we have a ring homomorphism for all a = (a_1, ...,a_n)[itex]\in R, from R[X_1,...,X_n] → R given by sending a polynomial f to f(a)? I have only ever seen this for fields. I cannot think of any possible reason it would be...
I was thinking about the following proposition that I think should be true, but I can't pove:
Suppose that F is a group freely generated by a set U and that F is also generated by a set V with |U| = |V|. Then F is also freely generated by V.
This is something that I intuitively think must...
"According to this book I'm reading, if you cut out a closed disc in the projective plane, then the complement of the interior of this disc is topologically a Mobius strip with boundary."
Are you trying to say that removing an open disc in the projective plane renders the Mobius strip?
The...