Homomorphism between $A_5$ and $A_6$ 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?
 A: According to a certain Stanford problem set, $S_5$ has 6 Sylow-5 subgroups. Indeed, $|S_5|=120 = 5\cdot 24$ so $n_p|24, n_p \equiv 1 (\mod 5)$ that leaves 6,11,16,21.  
These subgroups will be generated by different 5-cycles, and you might try find explicit generators.

I think this is connected to the Outer Automorphisms of $S_6$ and this "nobbly-wobbly" toy:
http://s.petco.com/assets/product_images/7/784369510232C.jpg
5-cycles are even permutations e.g. $(12345) = (12)(13)(14)(15)$ so the same analysis should hold for the alternating group.
A: What do you think of this homomorphism:
f:A5->A6
f(x)=(123)(456)x(654)(321)
This is a homomorphism because 
f(xy)=(123)(456)xy(654)(321)
=(123)(456)x(654)(321)(123)(456)y(654)(321)
=f(x)f(y)
Because (321)(123)=e=(654)(456). 
And it is injective because if f(x)=f(y) then,
(123)(456)x(654)(321)=(123)(456)y(654)(321)
And multiplying both sides with (123)(456) from the right and (654)(321) from the right gives us x=y.
This homomorfism does not leave any points fixed, unless x=e.
