How to know the number of iso/homo morphisms?

[LCSphysicist]

Homework Statement

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Relevant Equations

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Hello everyone. I have been study a little of group theory, i am a little stuck in how to answer question like this:

"How many homomorphism #f: S_{3}\to Q_{8}# are there? $S_{3}$ and $Q_8$ are the permutations group and the quaternion group, respectively."

A homomorphism is a map from A to B such that $\phi(a') \phi(a'') = \phi(a' a '')$, but how to apply this definition to answer the question?

I could construct the multiplicative table for $Q_{8}$, maybe call i j k 1 -i -j -k -1 as 1 2 3 ... 8, but yet have no idea what to do. Maybe seek a "little table" inside this modified table of Q8 that looks like S3?

 

[Office_Shredder]

$S_3$ is a group of order 6 and $Q_8$ is a group of order 8, right off the bat that should tell you there are no isomorphisms.

As far as homomorphisms, I think that statement alone restricts what kind of images and kernels if can have.

 

[fresh_42]

These groups are finite, so all elements are of finite order. $1=\phi(a^n)=(\phi(a))^n$ shows, that an element of order $n$ is mapped to an element of order $k|n$.

Another way is to look at the normal subgroups. The kernel of a homomorphism is a normal subgroup.

All such things have to happen for a homomorphism.

 

[LCSphysicist]

Thank you. Okay so let's see...
We have in S3 elements of order 1,2,3 (three are order 2)
In Q8, we have one element order 1, one element order 2 and six elements order 4. This means that we can make a map for the elements of order 1, and 1 map for the elementa of order 2? So the number of homom/ is 2?

So following in this way, and for the sake of clarity, the nunber of homomorphism of another map, which do the opposite, that is, go from Q8 to S3, would be 1 (o1) + 3 (o2) + 3 (o4) = 7?

 

[Office_Shredder]

No, there are more homomorphisms that are at least potentially available given your analysis. Given an element of order 2 in $S_3$, how many elements in $Q_8$ can you map it to?

 

[LCSphysicist]

No, there are more homomorphisms that are at least potentially available given your analysis. Given an element of order 2 in $S_3$, how many elements in $Q_8$ can you map it to?

I am not sure for what question was your no, sorry. I will suppose it was to the second question, so i will go on:

S3 has six elements.
S3 ((123),(213),(132),(321),(231),(312))
(123) has order 1
(213),(132),(321) has order 2
(231),(312) has order 3

Q8 has eight elements
1 has order 1
+-i,+-j,+-k has order 4
-1 has order 2

"an element of order n is mapped to an element of order k/n"
Since homomorphism is a group, we can in principle guess this:
I will make an analogy with functions
$\phi(-1) = (213)$,$\phi(-1) = (321)$,$\phi(-1) = (132)$
For each map cited, we can say yet another map, $\phi(1)$, it can be mapped to any elements of S3, so that we have: 3*6 = 18 type of homomorphism in principle.

Now i think we need to check if it is in fact an homomorphism?

There is another way, that i don't know very well how to do too (All books i searched do not have theory of counting homomorphism :S and i am an idiot), that i am supposed to count the kernels. I think we can let this way for later

 

[fresh_42]

Your notation is confusing. You write down the images of $(123)$, but usually we use another notation:

$(123)$ means $(1\to 2\to 3 \to 1)$. With this more common notation, we have $S_3=\{(1),(12),(13),(23),(123),(132)\}.$

You have forgotten the element $1\in Q_8$ which can also be the image of an element of order $2$. You also confused the order: are we talking about homomorphisms $S_3\longrightarrow Q_8$ or $Q_8\longrightarrow S_3.$ This is not a symmetric property, hence direction matters.

Let's stick with the original question $\phi : S_3\longrightarrow Q_8$.

Thus we can map all three elements of order $2$, $(12),(23),(13)$ to either $1$ or $-1$.
This in mind, where do the elements of order three, $(123),(132)$ go to?

 

[LCSphysicist]

Your notation is confusing. You write down the images of $(123)$, but usually we use another notation:

$(123)$ means $(1\to 2\to 3 \to 1)$. With this more common notation, we have $S_3=\{(1),(12),(13),(23),(123),(132)\}.$

You have forgotten the element $1\in Q_8$ which can also be the image of an element of order $2$. You also confused the order: are we talking about homomorphisms $S_3\longrightarrow Q_8$ or $Q_8\longrightarrow S_3.$ This is not a symmetric property, hence direction matters.

Let's stick with the original question $\phi : S_3\longrightarrow Q_8$.

Thus we can map all three elements of order $2$, $(12),(23),(13)$ to either $1$ or $-1$.
This in mind, where do the elements of order three, $(123),(132)$ go to?

I realized now that i was misinterpreting your citation in the other comment...
n is mapped to k|n, i was reading n is mapped to k/n...

In s3 we have elements of order 1,2,3.
In q8 we have elements of order 1,2,4.

Elements of order 2 in S3 can be mapped, as you observed, to elements of order 1 and 2 in Q8.
Elements of order 3 in S3 can be mapped to elements of order k|3, namely 1.

$(123),(132)$ go to 1.

Ok let me try.

Suppose (12) is mapped to 1, (23) to -1.

$\phi((12)(23)) = \phi(12)\phi(23) = 1*(-1) = -1$
But $(12)(23) = (123)$
So that $\phi((123)) = -1$
But it is not allowed...
Maybe we can generalized that (i didn't prove that)
I think we can see for another cases that $\phi(e_{2}(o2)) = \phi(e_{1}(o2))$
where $e_{i}(o2)$ are elements of order 2 in S3
That is, elements of order 2 in S3 is mapped to equal elements in Q8.

So that there are two maps.

 

[fresh_42]

I realized now that i was misinterpreting your citation in the other comment...
n is mapped to k|n, i was reading n is mapped to k/n...

In s3 we have elements of order 1,2,3.
In q8 we have elements of order 1,2,4.

Elements of order 2 in S3 can be mapped, as you observed, to elements of order 1 and 2 in Q8.
Elements of order 3 in S3 can be mapped to elements of order k|3, namely 1.

$(123),(132)$ go to 1.

Yes.

Ok let me try.

Suppose (12) is mapped to 1, (23) to -1.

$\phi((12)(23)) = \phi(12)\phi(23) = 1*(-1) = -1$
But $(12)(23) = (123)$
So that $\phi((123)) = -1$
But it is not allowed...

We already know that there are three elements in the kernel: $(1),(123),(132),$ and as just shown by you, that in case one transposition (2-cycle) is mapped to $1$, all others have to map onto $1,$ too. This means, as soon as one transposition is in the kernel, then the entire group is, which is the trivial homomorphism. The only question left is: Can we map all three transpositions on $-1$?

And: Do you know the name of this homomorphism?

Maybe we can generalized that (i didn't prove that)
I think we can see for another cases that $\phi(e_{2}(o2)) = \phi(e_{1}(o2))$
where $e_{i}(o2)$ are elements of order 2 in S3
That is, elements of order 2 in S3 is mapped to equal elements in Q8.

So that there are two maps.