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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?

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    $\begingroup$ 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. $\endgroup$ – hmakholm left over Monica Oct 9 '12 at 15:45
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    $\begingroup$ Further hint: A rotation of the dodecahedron permutes the faces -- but opposite faces always stay opposite. There are 6 pairs of opposite faces. $\endgroup$ – hmakholm left over Monica Oct 9 '12 at 15:52
  • $\begingroup$ 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 $n$. $\endgroup$ – Derek Holt Oct 9 '12 at 16:13
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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.
enter image description here

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.

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  • $\begingroup$ It is not necessary to note that the 5-groups are subgroups of $A_5$ -- just that they are all the subgroups of $S_5$ of order 5 means that any subgroup of $S_5$ will act by conjugation to permute them. What remains to be shown here is that the action of $A_5$ is faithful, and that it creates only even permutations of the subgroups. $\endgroup$ – hmakholm left over Monica Oct 9 '12 at 16:41
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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.

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  • $\begingroup$ You could just point out that it's the restriction of an inner automorphism of $A_6$, hence it is injective. $\endgroup$ – rschwieb Oct 18 '12 at 15:13

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