Thread: group theory

Let be 2 non trivial group homomorphisms. Prove that there exists a such that for every

Originally Posted by Chandru1

Let be 2 non trivial group homomorphisms. Prove that there exists a such that for every

Hints:

1) if f,g are non-trivial homomorphisms then automatically they're monomorphisms (i.e., ), so what must be their images?

2) Lemma: two cycles in are conjugate iff they both have the very same length.

Tonio

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

Originally Posted by Chandru1

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?

No...I mean, what if you were mapping into an abelian group instead of ? What you proved would still hold, but the result you are trying to prove would not.

Could you be a bit more precise.

Originally Posted by Chandru1

Could you a bit more precise.

With what you should do, or with the problem with your proof?

with the problem in my proof

Originally Posted by Chandru1

with the problem in my proof

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.

Originally Posted by Chandru1

Let be 2 non trivial group homomorphisms. Prove that there exists a such that for every

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.