Let be 2 non trivial group homomorphisms. Prove that there exists a such that for every
Hi--
Tonio, is that the only way to go. Actually, i did this but i am not sure whether i am right or wrong.
I actually gave a one-one map from S->R. Does that solve this by any means?
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Em, as for your question the yeah, their images must always be a 5-cycle in S_5
You haven't done anything, really. You have said that there is a bijection between the set of images and their conjugates under fixed (although if the was variable - as in the set is actually where then you would be fine).
However, what you have proven will always hold although the result you are trying to prove does not hold. You have proven and you want to show , but as there exist counter-examples.
As tonio hinted, f and g are embeddings, i.e., the images of f and g are cyclic subgroups of order 5 in S_5.
Let A be the image of f such that A=<(a_1, a_2, a_3, a_4, a_5)> and let B be the image of g such that B=<(b_1, b_2, b_3, b_4, b_5)>. Since the generator of A and B have the same cycle type, they are in the same orbit under the action of conjugation. That means, there exists such that , where b and a are generators of B and A, respectively.
Once we find a , the bijection is established between B and A. Think of a relabels the generator of B to the generator A. Then each element of B corresponds to A in a well-defined manner.