The problem is to find an injective homomorphism between the alternating groups $A_5$ and $A_6$ such that the image of the homomorphism contains only elements that leave no element of $\{1,2,3,4,5,6\}$ fixed. I.e., the image must be a subset of $A_6$ that consists of permutations with no fixed points. The hint given is that $A_5$ is isomorphic to the rotational symmetry group of the dodecahedron.
I figured out that the permutations in $A_6$ that leave no point fixed are of the forms:
- Double 3-cycle, e.g. (123)(456), in total 40 of them
- transposition + 4-cycle, e.g. (12)(3456), in total 90 of them
Considering that $A_5$ has 60 elements and $A_6$ has 360 elements, how should I proceed in finding a homomorphism whose image is a subset of the 130 elements described above?