How many homomorphisms are there from $A_5$ to $S_4$ ?
This is how I tried to solve it.
If there is a homomorphism from $A_5$ to $S_4$ , then order of element of $S_4$ should divide the order of its preimage. Now what are the possible order of elements in $S_4$.1,2,3 and 4. Since $A_5$ contains (12345), which is of order 5.. what could be image of (12345). Definitely Identity element which is of order 1. Similarly all 5 cycles must be mapped to identity. There are 24 elements of 5 cycles. 24 elements out of 60 are mapped to identity .. now only two types of homomorphisms are possible. either 30:1mapping or 60:1 mapping. Consider (12)(34) which belongs to $A_5$. It's image can be element of order 2 or identity.there are 15 elements of order 2 . suppose these 15 elements are mapped to some element 'g' of order 2 of $S_4$, you need another 15 elements to get mapped to 'g' to have 30 :1 mapping. Other type of elements left in $A_5$ is of order 3. None of them can be mapped to g. hence 15 elements of order 2 should be mapped to identity .. so , (24+15=39) elements mapped to identity.As mentioned earlier it should be 30 :1 or 60:1 mapping. So it must be 60:1 mapping.Hence a trivial homomorphism. Answer is 1.
I wanted to know is there any other technique which can be used to find number of homomorphism in the above question ? In general, how to find number of homomorphism between any two arbitrary groups ?