# The number of homomorphisms

• August 16th 2008, 05:11 AM
Aedolon
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!
• August 17th 2008, 11:12 AM
ThePerfectHacker
Quote:

Originally Posted by Aedolon
Find the number of homomorphisms between:
#Hom(S_4, C_4) (intuitively something tells me this is two)

If $\theta : S_4 \to \mathbb{Z}_4$ is a homomorphism then $\theta$ is completely determined by its value on a generating set.
We will use the generating set $\left< (12),(1234) \right>$.
Let $\sigma \in S_4$ and $k = \text{ord}(\sigma)$.
Then $0 = \theta ( \text{id} ) = \theta (\sigma^k) = \theta(\sigma)^k$.
Thus, it is necessary that $\text{ord} (\theta(\sigma))$ divides $\text{ord}(\sigma)$.
Since $\text{ord}(12) = 2$ it means $\theta (12) = 0,2$.
Note that $(234)=(12)(1234)$ thus $\theta(234) = \theta(12) + \theta(1234)$.
Since $\text{ord}(234)=3$ it means (by above) $\theta(234)=e$.
Thus, $\theta(12) + \theta(1234) = e$.
If $\theta(12) = e$ then $\theta(1234)=e$.
This is the trivial homomorphism i.e. everything is mapped to $0$.

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

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

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

*)Definition, $a$ conjugated by $b$ means the product $bab^{-1}$.
• August 17th 2008, 11:33 AM
ThePerfectHacker
Quote:

Originally Posted by Aedolon
#Hom(C_4, S_4)

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

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

A faster way to count the # of such elements is to know the conjugacy class equation for $S_4$.
It is $24 = 1+3+6+6+8$.
The $8$ are the conjugacy classes of $3$-cycles, while everything else is what we are looking for thus altogether we have $1+3+6+6 = 16$ elements.
But if you never learned this then just avoid it.
• August 17th 2008, 11:49 AM
ThePerfectHacker
Quote:

Originally Posted by Aedolon
#Hom(C_6, S_3)

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

(I need more time on the last one).
• August 18th 2008, 03:44 AM
Aedolon
Great! You have helped me a lot on this one :-)!
• August 19th 2008, 12:42 AM
Aedolon
Anyone for the last one?
• August 19th 2008, 10:31 AM
ThePerfectHacker
Quote:

Originally Posted by Aedolon
Anyone for the last one?

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