Jump to content

Homomorphisms


Recommended Posts

We haven't done conjugate classes yet, so I don't think I can use that information to prove it.

 

This is what I got so far...

In the first part, I found two homomorphisms. All the elements of S4 mapped to the identity. And another one being, all the even permutations to 0 and odd permutations to 1. THe kernels being S4 and A4.

 

The second part I found one permutation, all elements in S4 going to the identity.

 

How do I prove that these are the only ones? My prof hinted that know the composition of Z2, that this are two subgroups would be a good start, but that doesn't make a ton of sense to me.

 

Any suggestions? Please help!!!

Link to comment
Share on other sites

So let's add some more detail:

 

S4 to Z_2 (written additively to match your convention).

 

there's the trivial homomorphisms everything gets sent to 0

 

suppose now that (12) gets sent to 1. Since (123) must get sent to 0 ((123) has order 3, so its image has order divisible by 3) but (123) = (12)(23) so it must be that (23) gets sent to 1 as well. By similar logic (14) must be sent to 1, and similalry we see all 2-cycles must be sent to 1. This describes the homomorphism: s in S_4 is sent to 1 if it is an odd permutation 0 otherwise.

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.