# group theory

#### Chandru1

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 $$\displaystyle x \in \mathbb{Z}/5\mathbb{Z}$$

#### tonio

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

#### Chandru1

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

#### Swlabr

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.

#### Chandru1

dindnt get

Could you be a bit more precise.

#### Swlabr

Could you a bit more precise.
With what you should do, or with the problem with your proof?

#### Chandru1

HI

with the problem in my proof

#### Swlabr

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.

#### TheArtofSymmetry

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.