Thread: All homomorphisms between S3 and Z6

Can anyone help with the following problem:

Find all possible homomorphisms between the indicated groups:

:

==========================================

Thoughts:

The only normal subgroups of are {e} and

Thus (following the case for {e}) we can assume Ker = {e} for a homomorphism (by the First Isomorphism Theorem)

But how do we find the find the homomorphism - or homomorphisms?

============================================

If I can see the situation for {e} then I am hoping a similar analysis will yield the homomorphism(s) for the case of

Be grateful for some help.

Peter

if ker(φ) = {e}, φ must be injective, which would imply S3 and Z6 were isomorphic. why can't this be the case?

suppose ker(φ) = A3. then φ(S3) must be isomorphic to S3/A3, which has order 2. what are all the subgroups of Z6 of order 2?

can you see a possible way to define this homomorphism, based on the parity (even/odd) of an element in S3?

finally, you are overlooking another possiblity for ker(φ), which produces a rather dull homomorphism.....

Thanks!!

I was previously trying to construct a homomorphism between and , but given your post I have concluded that this is not possble since by the First Isomorphism Theorem a homomorphism between and would imply which is not possible since is cyclic and is not.

Therefore there cannot be a homomorphism between and .

Correct?

Will now work on the Ker situation.

Thanks again for your help.

Peter

homomorphism ≠ isomorphism. there cannot be an isomorphism between S3 and Z6 (because why? this is not a hard question....)

but this doesn't mean there cannot be any homomorphisms between S3 and Z6.

Either I am expressing myself badly or I am wrong - both of which bother me!! :-)

I was trying to argue thus:

The First Isomorphism Theorem states that if : G G' is a homomorphism with kernel Ker = K, then G/K (G)

So in the case of and where K = {e} we have that /{e}

But his cannot be the case ... so therefore there is no homomorphism between and

Is that a valid argument?

no, it only shows that there is no homomorphism with kernel {e}. it does not rule out other possible homomorphisms with different kernels.

now, WHY can't it be the case that there is no homomorphism with kernel {e}? that is, why aren't S3 and Z6 isomorphic?

(they both have order 6....they both have subgroups of order 2 and 3.....so what is it, exactly?)

In answer to "why aren't and isomorphic" it is because is cyclic and is not. Therefore there is no homomorphism with kernal {e}.

Sorry, my assertion that therefore there was no homomorphisms (at all) between and was a brain fade!

Obviously there is the case of the subgroup and to investigate!

Still have to work on this one.

Peter