The number of homomorphisms

**Aedolon** · August 16th 2008, 05:11 AM

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!

**ThePerfectHacker** · August 17th 2008, 11:12 AM

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 $bab^{-1}$ .

**ThePerfectHacker** · August 17th 2008, 11:33 AM

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.

**ThePerfectHacker** · August 17th 2008, 11:49 AM

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).

**Aedolon** · August 18th 2008, 03:44 AM

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

**Aedolon** · August 19th 2008, 12:42 AM

Anyone for the last one?

**ThePerfectHacker** · August 19th 2008, 10:31 AM

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 ?

Thread: The number of homomorphisms

LinkBack

Thread Tools

Display

The number of homomorphisms

Similar Math Help Forum Discussions

Number of different homomorphisms

Number of homomorphisms between two groups

Number of Homomorphisms from Zn to Zm

Number of group homomorphisms

Homomorphisms

Search Tags