1. ## Homomorphism

How can I show that there is no homomorphism from $\displaystyle S_3$ to $\displaystyle Z_3$?
Suppose there is one homomorphism $\displaystyle f$. Let $\displaystyle g \in S_3$, then $\displaystyle f(g)=0$ or $\displaystyle f(g)=1$ or $\displaystyle f(g)=2$. I know $\displaystyle ord(f(g))$ must divide $\displaystyle ord(g)$. I have $\displaystyle ord(0)=\infty$, $\displaystyle ord(1)=3$, and $\displaystyle ord(2)=2$. I want to show that there is an element in $\displaystyle S_3$ such that $\displaystyle ord(f(g))$ does not divide $\displaystyle ord(g)$. I'm stuck on this because based on my approach, $\displaystyle ord(f(g))$ can be 1, and 1 divides anything.

Can anyone give me help here?

2. Originally Posted by jackie
How can I show that there is no homomorphism from $\displaystyle S_3$ to $\displaystyle Z_3$?
Suppose there is one homomorphism $\displaystyle f$. Let $\displaystyle g \in S_3$, then $\displaystyle f(g)=0$ or $\displaystyle f(g)=1$ or $\displaystyle f(g)=2$. I know $\displaystyle ord(f(g))$ must divide $\displaystyle ord(g)$. I have $\displaystyle ord(0)=\infty$, $\displaystyle ord(1)=3$, and $\displaystyle ord(2)=2$. I want to show that there is an element in $\displaystyle S_3$ such that $\displaystyle ord(f(g))$ does not divide $\displaystyle ord(g)$. I'm stuck on this because based on my approach, $\displaystyle ord(f(g))$ can be 1, and 1 divides anything.

Can anyone give me help here?

Who says there is no homom.? There is always the trivial homom.: send everything to the unity element. But for this there is no more since:

If there was a non-trivial hom. F: S_3 --> Z_3 it would have to be onto ==> Then S_3/Ker F ~ Z_3, but this'd mean S_3 has a normal sbgp. (namely, ker F) of order 2, which isn't true.

Tonio