How many $A_5$ are there inside $A_6$?

Question

I am reading a paper that says:

There are $12$ versions of $A_5$ in $A_6$:

$1) $ the permutations that leave one thing unmoved.

$2)$ the permutations of the six pairs of antipodal iocasahedral vertices resulting from icosahedron's rotational symmetries.

First version is completely clear to me. I don't understand the second version. But I know that icosahedron's rotational symmetries is $A_5$.

Please explain part $2)$, and also how can I see that there are just $12$ versions of $A_5$ in $A_6$. Thanks!

Answer 1

The rotation group of the icosahedron acts on various sets. For example, any rotation obviously permutes the $20$ faces of the icosahedron, so the group has an action on a $20$-element set. But if two faces are opposite each other prior to rotation, they are still opposite each other after rotation. So we also have an action of the group on the $10$-element set whose members are pairs of opposite faces.

What goes for faces also goes for edges and vertices. So there is an action of the group on the $30$-element edge set, but also on the $15$-element set whose elements are pairs of opposite edges. Even more can be said: each pair of opposite edges forms a rectangle whose short sides are the two edges. The long sides of the rectangle pass through the body of the icosahedron and have length which is the golden ratio times the length of the short sides. So we have $15$ such "golden rectangles". These come in five sets of three mutually orthogonal rectangles. (The dihedral angles between the planes containing the rectangles in such a set are right angles.) Since the rotation group of the icosahedron is isomorphic to $A_5,$ it's not surprising that there is some five-element set on which it acts.

Moving on to vertices, the rotation group of the icosahedron permutes the $12$ vertices, but can also be regarded as acting on the six pairs of antipodal vertices. Equivalently, it permutes the six five-fold symmetry axes of the icosahedron.

To see why $A_6$ should contain six copies of $A_5$ that act in this way, we count the ways of labeling the six five-fold symmetry axes using the numbers $1$ through $6.$ Because we are going to act on the set of labeled axes with the rotation group of the icosahedron, we consider labelings equivalent if there is a rotation of the icosahedron that turns one into the other. We can always rotate the icosahedron so that axis $1$ is vertical and so that axis $2$ is also oriented in a specified direction. There are then $4!=24$ ways to assign the labels $3,$ $4,$ $5,$ $6$ to the remaining four axes. We can think of axis $1$ as joining the north and south poles of the icosahedron. There are five vertices that are nearest neighbors of the north pole, and each has an axis passing through it. One of these axes is $2,$ and the labels of the other four, proceeding clockwise, say, around the north pole, are an arbitrary permutation of $3,$ $4,$ $5,$ $6.$

One can then construct a copy of $A_5$ that represents the action of the rotation group on each of these $24$ labelings. These $24$ $A_5$s are not all distinct, however. Consider the labeling in which the five axes proceeding clockwise around the north pole are $2,$ $3,$ $4,$ $5,$ $6.$ A counterclockwise rotation of $2\pi/5$ around the north pole corresponds to the permutation $(23456).$ Consequently the copy of $A_5$ acting on that labeling contains $(23456).$ But then it also contains powers of this cycle: $(24635),$ $(25364),$ and $(26543).$ One can check that if we label the axes proceeding clockwise around the north pole by any of these four five-cycles, the $A_5$s that act on these labelings are all the same. Hence instead of $24$ $A_5$s, we have only six.

To see that $A_6$ contains only $12$ copies of $A_5,$ note that $A_5$ contains elements of order $5.$ The only elements of order $5$ in $A_6$ are $5$-cycles. On the other hand, $A_5$ contains no elements of order $4.$ Now to build a copy of $A_5$ in $A_6,$ you have to put in $5$-cycles. If you put in two $5$-cycles that contain the same five elements (and aren't powers of each other), you get the first type of $A_5.$ If you put in two $5$-cycles that contain different elements, say $1,$ $2,$ $3,$ $4,$ $5$ and $1,$ $2,$ $3,$ $4,$ $6,$ it is not hard to check that unless you make an $A_5$ of the second type, you inevitably generate elements of order $4.$

Answer 2

I'm not sure how to translate this to the symmetries of an icosahedron, but IMHO the following is a simple way of finding the other six subgroups of $A_6$, all isomorphic to $A_5$.

Consider the Sylow $5$-subgroups of $A_5$. It is easy to see that there are exactly six of them. The group $A_5$ acts on them by conjugation. That action is transitive, so this gives us a non-trivial homomorphism $\phi:A_5\to S_6$. Because $A_5$ is simple, the homomorphism is injective. For the same reason the image $\phi(A_5)$ must be contained in $A_6$. Because the action is transitive, the image does not stabilize any of the six points. Therefore $\phi(A_5)$ cannot be any of the point stabilizer from part 1).

Furthermore, $\phi(A_5)$ is of index six in the simple group $A_6$. This implies that $\phi(A_5)$ is its own normalizer and has six conjugates in $A_6$. Thus there are six subgroups of this type filling up the quota for part 2).

To show that there are no other takes a bit more work.

Answer 3

Another way of seeing it is that the group ${\rm PSL}(2,5)$ is isomorphic to $A_5$ (there is a unique simple group of order $60$), and the natural permutation representation of ${\rm PSL}(2,5)$ is transitive of degree $6$. There are $6$ each of Types $1$ and $2$, so $12$ altogether.

Answer 4

Let $H$ and $K$ be distinct subgroups of order $60$ in $A_6$ such that $H \cong K \cong A_5$.

Then $[H : H \cap K] \leq [A_6 : K] = 6$. Because $H$ is simple, it follows that $H \cap K$ has order $10$. Therefore $H$ and $K$ have no elements of order $3$ in common. We can use this observation to prove the claim.

A point stabilizer is isomorphic to $A_5$ and contains a $3$-cycle $(abc)$, this gives you a conjugacy class of $6$ subgroups isomorphic to $A_5$.

You get a different conjugacy class of $6$ subgroups from a transitive copy of $A_5$ in $A_6$. A construction is described in the answer by Jyrki. You can check that it will contain a permutation of cycle type $(abc)(def)$.

We know that $3$-cycles form a single conjugacy class in $A_6$, as do the permutations of type $(abc)(def)$. Thus the subgroups in these two conjugacy classes will contain all the elements of order $3$ in $A_6$. By the discussion in the beginning, a subgroup of $A_6$ isomorphic to $A_5$ must be contained in one of these two conjugacy classes.