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!
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 .
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.
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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.