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!

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- August 16th 2008, 06:11 AMAedolonThe 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, 12:12 PMThePerfectHacker
If is a homomorphism then is completely determined by its value on a generating set.

We will use the generating set .

Let and .

Then .

Thus, it is**necessary**that divides .

Since it means .

Note that thus .

Since it means (by above) .

Thus, .

If then .

This is the trivial homomorphism i.e. everything is mapped to .

The other possibility is and then .

We will prove this defines a homomorphism.

Begin with .

Conjugating* by powers of gives and .

Therefore, .

Note that .

It follows that .

This means if is__even__then it is a product of even number of transpositions and so since even number of times is . If is__odd__then is a product of odd number of transpositions and so since odd number of times is . Therefore, is simply the mapping which sends even permutations into and odd permutations into . It is trivial to proof now that it is indeed a homomorphism.

Thus, .

And like you guessed it does indeed have only two elements.

*)Definition, conjugated by means the product . - August 17th 2008, 12:33 PMThePerfectHacker
Let be a homomorphism.

Since it means is determined by its value on .

Similarly to above must divide .

However, in case of cyclic group this condition is also**sufficient**to define a homomorphism.

Any can be written in terms of disjoint cycles.

Therefore, for the possibilities are:

, , .

Thus, the__non-trivial__values for can be:

This gives us a total of elements.

----

A faster way to count the # of such elements is to know the conjugacy class equation for .

It is .

The are the conjugacy classes of -cycles, while everything else is what we are looking for thus altogether we have elements.

But if you never learned this then just avoid it. - August 17th 2008, 12:49 PMThePerfectHacker
- August 18th 2008, 04:44 AMAedolon
Great! You have helped me a lot on this one :-)!

- August 19th 2008, 01:42 AMAedolon
Anyone for the last one?

- August 19th 2008, 11:31 AMThePerfectHacker