group theory

Feb 2009
148
10
Chennai
Let \(\displaystyle f,g :\mathbb{Z}/5\mathbb{Z} \to S_{5}\) be 2 non trivial group homomorphisms. Prove that there exists a \(\displaystyle \sigma \in S_{5}\) such that \(\displaystyle f(x)=\sigma g(x) \sigma^{-1}\) for every \(\displaystyle x \in \mathbb{Z}/5\mathbb{Z}\)
 
Oct 2009
4,261
1,836
Let \(\displaystyle f,g :\mathbb{Z}/5\mathbb{Z} \to S_{5}\) be 2 non trivial group homomorphisms. Prove that there exists a \(\displaystyle \sigma \in S_{5}\) such that \(\displaystyle f(x)=\sigma g(x) \sigma^{-1}\) for every \(\displaystyle x \in \mathbb{Z}/5\mathbb{Z}\)

Hints:

1) if f,g are non-trivial homomorphisms then automatically they're monomorphisms (i.e., \(\displaystyle \ker f\,,\,\ker g\neq {0}\)), so what must be their images?

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

Tonio
 
Feb 2009
148
10
Chennai
HI

Hi--

Tonio, is that the only way to go. Actually, i did this but i am not sure whether i am right or wrong.

\(\displaystyle S=\{ f(x) | x \in \mathbb{Z}_{5}\}, \ R=\{\sigma g(x) \sigma^{-1}\ | x \in Z_{5}\}\)

I actually gave a one-one map from S->R. Does that solve this by any means?

--------

Em, as for your question the yeah, their images must always be a 5-cycle in S_5
 
May 2009
1,176
412
Hi--

Tonio, is that the only way to go. Actually, i did this but i am not sure whether i am right or wrong.

\(\displaystyle S=\{ f(x) | x \in \mathbb{Z}_{5}\}, \ R=\{\sigma g(x) \sigma^{-1}\ | x \in Z_{5}\}\)

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 \(\displaystyle S_5\)? What you proved would still hold, but the result you are trying to prove would not.
 
Feb 2009
148
10
Chennai
dindnt get

Could you be a bit more precise.
 
May 2009
1,176
412
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 \(\displaystyle \sigma\) (although if the \(\displaystyle \sigma\) was variable - as in the set \(\displaystyle R\) is actually \(\displaystyle R = \cup R_{\sigma}\) where \(\displaystyle R_{\sigma} = \{\sigma^{-1} f(x) \sigma : x \in \mathbb{Z}_5\}\) 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 \(\displaystyle A\) and you want to show \(\displaystyle B\), but \(\displaystyle A \not\Rightarrow B\) as there exist counter-examples.
 
May 2010
95
38
Let \(\displaystyle f,g :\mathbb{Z}/5\mathbb{Z} \to S_{5}\) be 2 non trivial group homomorphisms. Prove that there exists a \(\displaystyle \sigma \in S_{5}\) such that \(\displaystyle f(x)=\sigma g(x) \sigma^{-1}\) for every \(\displaystyle x \in \mathbb{Z}/5\mathbb{Z}\)
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 \(\displaystyle \sigma \in S_5\) such that \(\displaystyle \sigma b \sigma^{-1}=a\), where b and a are generators of B and A, respectively.

Once we find a \(\displaystyle \sigma\), the bijection is established between B and A. Think of a \(\displaystyle \sigma\) relabels the generator of B to the generator A. Then each element of B corresponds to A in a well-defined manner.