# Thread: The number of homomorphisms

1. ## The number of homomorphisms

Find the number of homomorphisms between:
#Hom(S_4, C_4) (intuitively something tells me this is two)
#Hom(C_4, S_4)
#Hom(C_6, S_3)
#Hom(D_3, A_4) (D_3 has order 6)

Any help would be appreciated!

2. Originally Posted by Aedolon
Find the number of homomorphisms between:
#Hom(S_4, C_4) (intuitively something tells me this is two)
If $\displaystyle \theta : S_4 \to \mathbb{Z}_4$ is a homomorphism then $\displaystyle \theta$ is completely determined by its value on a generating set.
We will use the generating set $\displaystyle \left< (12),(1234) \right>$.
Let $\displaystyle \sigma \in S_4$ and $\displaystyle k = \text{ord}(\sigma)$.
Then $\displaystyle 0 = \theta ( \text{id} ) = \theta (\sigma^k) = \theta(\sigma)^k$.
Thus, it is necessary that $\displaystyle \text{ord} (\theta(\sigma))$ divides $\displaystyle \text{ord}(\sigma)$.
Since $\displaystyle \text{ord}(12) = 2$ it means $\displaystyle \theta (12) = 0,2$.
Note that $\displaystyle (234)=(12)(1234)$ thus $\displaystyle \theta(234) = \theta(12) + \theta(1234)$.
Since $\displaystyle \text{ord}(234)=3$ it means (by above) $\displaystyle \theta(234)=e$.
Thus, $\displaystyle \theta(12) + \theta(1234) = e$.
If $\displaystyle \theta(12) = e$ then $\displaystyle \theta(1234)=e$.
This is the trivial homomorphism i.e. everything is mapped to $\displaystyle 0$.

The other possibility is $\displaystyle \theta(12) = 2$ and then $\displaystyle \theta(1234) = 2$.
We will prove this defines a homomorphism.
Begin with $\displaystyle \theta(234) = 0$.
Conjugating* $\displaystyle (12)$ by powers of $\displaystyle (234)$ gives $\displaystyle (13)$ and $\displaystyle (14)$.
Therefore, $\displaystyle \theta(12) = \theta(13) = \theta(14) = 2$.
Note that $\displaystyle (x,y) = (1,x)(1,y)(1,x)$.
It follows that $\displaystyle \theta (x,y) = 2+2+2=2$.

This means if $\displaystyle \sigma$ is even then it is a product of even number of transpositions and so $\displaystyle \theta (\sigma) = 0$ since $\displaystyle 2+...+2$ even number of times is $\displaystyle 0$. If $\displaystyle \sigma$ is odd then $\displaystyle \sigma$ is a product of odd number of transpositions and so $\displaystyle \theta(\sigma) = 2$ since $\displaystyle 2+...+2$ odd number of times is $\displaystyle 2$. Therefore, $\displaystyle \theta$ is simply the mapping which sends even permutations into $\displaystyle 0$ and odd permutations into $\displaystyle 2$. It is trivial to proof now that it is indeed a homomorphism.

Thus, $\displaystyle \text{hom}(S_4,\mathbb{Z}_4) \simeq \mathbb{Z}_2$.
And like you guessed it does indeed have only two elements.

*)Definition, $\displaystyle a$ conjugated by $\displaystyle b$ means the product $\displaystyle bab^{-1}$.

3. Originally Posted by Aedolon
#Hom(C_4, S_4)
Let $\displaystyle \theta: \mathbb{Z}_4 \to S_4$ be a homomorphism.
Since $\displaystyle \mathbb{Z}_4 = \left< 1 \right>$ it means $\displaystyle \theta$ is determined by its value on $\displaystyle 1$.
Similarly to above $\displaystyle \text{ord}(\theta(1))$ must divide $\displaystyle \text{ord}(1) = 4$.
However, in case of cyclic group this condition is also sufficient to define a homomorphism.

Any $\displaystyle \sigma \in S_4$ can be written in terms of disjoint cycles.
Therefore, for $\displaystyle \text{ord}(\sigma) = 1,2,4$ the possibilities are:
$\displaystyle \sigma = (xy)$, $\displaystyle \sigma = (xy)(zw)$, $\displaystyle \sigma = (xyzw)$.
Thus, the non-trivial values for $\displaystyle \theta (1)$ can be:
$\displaystyle (12),(13),(14),(23),(24),(34)$
$\displaystyle (12)(34),(13)(24),(14)(23)$
$\displaystyle (1234),(1243),(1324),(1342),(1423),(1432)$
This gives us a total of $\displaystyle 15+1=16$ elements.
----

A faster way to count the # of such elements is to know the conjugacy class equation for $\displaystyle S_4$.
It is $\displaystyle 24 = 1+3+6+6+8$.
The $\displaystyle 8$ are the conjugacy classes of $\displaystyle 3$-cycles, while everything else is what we are looking for thus altogether we have $\displaystyle 1+3+6+6 = 16$ elements.
But if you never learned this then just avoid it.

4. Originally Posted by Aedolon
#Hom(C_6, S_3)
Same idea, $\displaystyle \mathbb{Z}_6 = \left< 1 \right>$.
We also require $\displaystyle \text{ord}(\theta (1))$ to divide $\displaystyle \text{ord}(1)=6$.
But if $\displaystyle \sigma \in S_3$ then $\displaystyle \text{ord}(\sigma)$ divides $\displaystyle |S_3| = 3! = 6$.
Thus, $\displaystyle \theta(1)$ can be any element in $\displaystyle S_3$.
Thus, $\displaystyle |\text{hom}(\mathbb{Z}_6,S_3)| = 6$.

(I need more time on the last one).

5. Great! You have helped me a lot on this one :-)!

6. Anyone for the last one?

7. Originally Posted by Aedolon
Anyone for the last one?
How about you try doing the last one?
First D3=S3. And $\displaystyle S_3 = \{ \text{id},(123),(123)^2,(12),(123)(12),(123)^2(12) \}$. Thus $\displaystyle (123),(12)$ generate $\displaystyle S_3$.
Can you continue from here ?