# construct an epimorphism

• Dec 12th 2013, 02:53 PM
surry
construct an epimorphism
Hi,
i'm trying to construct an construct an epimorphism from $S_3 \times \mathbb{Z} \rightarrow \mathbb{Z}$ and from $S_3 \rightarrow C_2$ but unfortunately i have no idea how to start...
• Dec 12th 2013, 04:30 PM
HallsofIvy
Re: construct an epimorphism
For a in $S_3$ and n in Z, map (a, n) to n. That's trivial.

For $S_3\to C_2$ what do you mean by " $S_3$" and " $C_2$"? My first thought was that " $S_3$" was the set of permutations on 3 objects and " $C_2$" was the set of pairs of complex numbers but in that case, $S_3$ is finite (containing 6 members) while $C_2$ is infinite so there cannot be such an epimorphism.
• Dec 12th 2013, 04:32 PM
SlipEternal
Re: construct an epimorphism
You haven't given enough information. The morphism $(\sigma,n) \mapsto n$ is an epimorphism for the first one. The morphism $\sigma \mapsto \text{sign}(\sigma)$ 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 $(\sigma,n) \mapsto \text{sign}(\sigma)\cdot n$ which would also be an epimorphism for the first one.
• Dec 12th 2013, 04:34 PM
SlipEternal
Re: construct an epimorphism
Quote:

Originally Posted by HallsofIvy
For a in $S_3$ and n in Z, map (a, n) to n. That's trivial.

For $S_3\to C_2$ what do you mean by " $S_3$" and " $C_2$"? My first thought was that " $S_3$" was the set of permutations on 3 objects and " $C_2$" was the set of pairs of complex numbers but in that case, $S_3$ is finite (containing 6 members) while $C_2$ is infinite so there cannot be such an epimorphism.

Typically, $Z_n$ is the cyclic group of order $n$, but I imagine that is the group the OP meant.
• Dec 12th 2013, 04:40 PM
romsek
Re: construct an epimorphism
Quote:

Originally Posted by HallsofIvy
For a in $S_3$ and n in Z, map (a, n) to n. That's trivial.

For $S_3\to C_2$ what do you mean by " $S_3$" and " $C_2$"? My first thought was that " $S_3$" was the set of permutations on 3 objects and " $C_2$" was the set of pairs of complex numbers but in that case, $S_3$ is finite (containing 6 members) while $C_2$ is infinite so there cannot be such an epimorphism.

maybe this?

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

from here

That's Integers Z, I didn't notice a Tex code for it.
• Dec 16th 2013, 01:17 PM
Deveno
Re: construct an epimorphism
An explicit epimorphism $\phi: S_3 \to C_2$, where $C_2 = \{e,a\}$:

$\phi(e) = e$

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

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

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

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

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

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

$S_3 \to S_3/A_3$, 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).