# All homomorphisms between S3 and Z6

• Oct 7th 2011, 07:17 PM
Bernhard
All homomorphisms between S3 and Z6
Can anyone help with the following problem:

Find all possible homomorphisms between the indicated groups:

$\displaystyle \phi$: $\displaystyle S_3 \rightarrow Z_6$

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Thoughts:

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

Thus (following the case for {e}) we can assume Ker $\displaystyle \phi$ = {e} for a homomorphism $\displaystyle \phi$ (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 $\displaystyle A_3$

Be grateful for some help.

Peter
• Oct 7th 2011, 07:54 PM
Deveno
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.....
• Oct 7th 2011, 08:25 PM
Bernhard
Re: All homomorphisms between S3 and Z6
Thanks!!

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

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

Correct?

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

Peter
• Oct 7th 2011, 08:32 PM
Deveno
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.
• Oct 7th 2011, 09:11 PM
Bernhard
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 $\displaystyle \phi$ : G $\displaystyle \rightarrow$ G' is a homomorphism with kernel Ker $\displaystyle \phi$ = K, then G/K $\displaystyle \cong$ $\displaystyle \phi$ (G)

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

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

Is that a valid argument?
• Oct 7th 2011, 10:55 PM
Deveno
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?)
• Oct 8th 2011, 04:47 PM
Bernhard
Re: All homomorphisms between S3 and Z6
In answer to "why aren't $\displaystyle S_3$ and $\displaystyle Z_6$ isomorphic" it is because $\displaystyle Z_6$ is cyclic and $\displaystyle S_3$ is not. Therefore there is no homomorphism with kernal {e}.

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

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

Still have to work on this one.

Peter