# Math Help - All homomorphisms between S3 and Z6

1. ## All homomorphisms between S3 and Z6

Can anyone help with the following problem:

Find all possible homomorphisms between the indicated groups:

$\phi$: $S_3 \rightarrow Z_6$

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Thoughts:

The only normal subgroups of $S_3$ are {e} and $A_3$

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

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

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If I can see the situation for {e} then I am hoping a similar analysis will yield the homomorphism(s) for the case of $A_3$

Be grateful for some help.

Peter

2. ## Re: All homomorphisms between S3 and Z6

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.....

3. ## Re: All homomorphisms between S3 and Z6

Thanks!!

I was previously trying to construct a homomorphism between $S_3$ and $Z_6$, but given your post I have concluded that this is not possble since by the First Isomorphism Theorem a homomorphism between $S_3$ and $Z_6$ would imply $S_3$ $\simeq$ $Z_6$ which is not possible since $Z_6$ is cyclic and $S_3$ is not.

Therefore there cannot be a homomorphism between $S_3$ and $Z_6$.

Correct?

Will now work on the Ker $\phi$ situation.

Peter

4. ## Re: All homomorphisms between S3 and Z6

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.

5. ## Re: All 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 $\phi$ : G $\rightarrow$ G' is a homomorphism with kernel Ker $\phi$ = K, then G/K $\cong$ $\phi$ (G)

So in the case of $S_3$ and $Z_6$ where K = {e} we have that $S_3$ /{e} $\cong$ $S_3$ $\cong$ $Z_6$

But his cannot be the case ... so therefore there is no homomorphism between $S_3$ and $Z_6$

Is that a valid argument?

6. ## Re: All homomorphisms between S3 and Z6

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?)

7. ## Re: All homomorphisms between S3 and Z6

In answer to "why aren't $S_3$ and $Z_6$ isomorphic" it is because $Z_6$ is cyclic and $S_3$ is not. Therefore there is no homomorphism with kernal {e}.

Sorry, my assertion that therefore there was no homomorphisms (at all) between $S_3$ and $Z_6$ was a brain fade!

Obviously there is the case of the subgroup $A_3$ and $S_3$ to investigate!

Still have to work on this one.

Peter