Question

How do i find the number of group homomorphisms from sn to an?

Answer 1

In case you do not know, kernel of a homomorphism must be a normal subgroup of the inverse image. Simply S6S6has 3 normal subgroups which are {ee}, A3A3, and S3S3.

Let ϕ:S3→Z6ϕ:S3→Z6

Then possible kernels are {ee}, A3A3, and S3S3.

Firstly, try {ee}. By First Isomorphism Theorem; S3/eS3/e which is S3S3 itself, S3S3≃ϕ(S3)≃ϕ(S3). The order of S3S3 is 6 and observe that Z6Z6 has the same order. Thus, it yields ϕ(S3)=Z6ϕ(S3)=Z6. However, S3S3is not abelian although Z6Z6 is. It is contradiction. Therefore, kerϕϕ cannot be {ee}.

Secondly, let's check for S3S3, then S3/S3S3/S3is identity, so you can map every element of S3S3 to the identity of Z6Z6. ϕ(s)=0ϕ(s)=0 , ∀s∈S3∀s∈S3 and 00 is the identity of Z6Z6.

The last option is Kerϕϕ = A3A3, therefore the order of the factor group S3/A3S3/A3 is 22. First isomorphism theorem gives us: S3/A3≃ϕ(S3)S3/A3≃ϕ(S3), then ϕ(S3)ϕ(S3) is {3,03,0}.

ϕ(s)=0ϕ(s)=0 if s∈A3s∈A3

Otherwise, ϕ(s)=3ϕ(s)=3

As a conclusion, the answer is 22.