Hi,

i'm trying to construct an construct an epimorphism from $\displaystyle S_3 \times \mathbb{Z} \rightarrow \mathbb{Z} $ and from $\displaystyle S_3 \rightarrow C_2 $ but unfortunately i have no idea how to start...

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- Dec 12th 2013, 01:53 PMsurryconstruct an epimorphism
Hi,

i'm trying to construct an construct an epimorphism from $\displaystyle S_3 \times \mathbb{Z} \rightarrow \mathbb{Z} $ and from $\displaystyle S_3 \rightarrow C_2 $ but unfortunately i have no idea how to start... - Dec 12th 2013, 03:30 PMHallsofIvyRe: construct an epimorphism
For a in $\displaystyle S_3$ and n in Z, map (a, n) to n. That's trivial.

For $\displaystyle S_3\to C_2$ what do you mean by "$\displaystyle S_3$" and "$\displaystyle C_2$"? My first thought was that "$\displaystyle S_3$" was the set of permutations on 3 objects and "$\displaystyle C_2$" was the set of pairs of complex numbers but in that case, $\displaystyle S_3$ is**finite**(containing 6 members) while $\displaystyle C_2$ is infinite so there**cannot**be such an epimorphism. - Dec 12th 2013, 03:32 PMSlipEternalRe: construct an epimorphism
You haven't given enough information. The morphism $\displaystyle (\sigma,n) \mapsto n$ is an epimorphism for the first one. The morphism $\displaystyle \sigma \mapsto \text{sign}(\sigma)$ is an epimorphism for the second one. I am not sure what more you are trying to do with this. If you want, you can use $\displaystyle (\sigma,n) \mapsto \text{sign}(\sigma)\cdot n$ which would also be an epimorphism for the first one.

- Dec 12th 2013, 03:34 PMSlipEternalRe: construct an epimorphism
- Dec 12th 2013, 03:40 PMromsekRe: construct an epimorphism

maybe this?

$\displaystyle \text{Let }C_\infty\text{ be the group }(Z,+)\text{, and, for an integer }m\geq1\text{, let }C_m\text{ be the group }(Z/mZ,+).$

from here

That's Integers Z, I didn't notice a Tex code for it. - Dec 16th 2013, 12:17 PMDevenoRe: construct an epimorphism
An explicit epimorphism $\displaystyle \phi: S_3 \to C_2$, where $\displaystyle C_2 = \{e,a\}$:

$\displaystyle \phi(e) = e$

$\displaystyle \phi((1\ 2)) = a$

$\displaystyle \phi((1\ 3)) = a$

$\displaystyle \phi((2\ 3)) = a$

$\displaystyle \phi((1\ 2\ 3)) = e$

$\displaystyle \phi((1\ 3\ 2)) = e$

Proving this is a homomorphism is the tricky part, it may be easier to observe that what we are actually doing is mapping:

$\displaystyle S_3 \to S_3/A_3$, which is isomorphic to any cyclic group of order 2, since it is of order 2 (any two groups of prime order are isomorphic and both are cyclic).