5
$\begingroup$

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?

  • 3
    The image of a homomorphism must always contain the identity, which leaves _all_ elements fixed. The best you can hope for is that _every_ element of $\{1,2,3,4,5,6\}$ is moved by _some_ permutation in the image.2012-10-09
  • 3
    Further hint: A rotation of the dodecahedron permutes the faces -- but opposite faces always stay opposite. There are 6 _pairs_ of opposite faces.2012-10-09
  • 0
    Just to stress that you have definitely misunderstood the problem. Any subgroup of $S_n$ in which no non-identity element fixes any points has order dividing $n$.2012-10-09

2 Answers 2