How many homomorphisms are there from A5 to S4 ?
This is how I tried to solve it.
If there is a homomorphism from A5 to S4 , then order of element of S4 should divide the order of its preimage. Now what are the possible order of elements in S4.1,2,3 and 4. Since A5 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 A5. 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 S4, you need another 15 elements to get mapped to 'g' to have 30 :1 mapping. Other type of elements left in A5 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 ?