The 60-element group A₅ (the alternating group on 5 symbols) is depicted, in this case in its isomorphic form as the group of rotational symmetries of the regular icosahedron (which in turn is the symmetry group of the regular dodecahedron, and also the symmetry group of the truncated icosahedron - so A₅ is the symmetry group of the molecule buckministerfullerene, C₆₀).

More specifically, A₅ is rendered as a subgroup of the rotation group SO(3) of R³; SO(3), in turn, is rendered as the ball of radius pi with antipodal boundary points identified. [More details about the solid ball model of SO(3), along with graphics, is available here. Briefly, every rotation is depicted as a point ru, whose distance from the origin r is the amount of counterclockwise about the axis of rotation u (where u is a unit vector).] Note that in the interests of displaying the full symmetry of the figure, 75 points (rather than 60 points) are depicted; 30 points on the (unseen) boundary sphere are to be identified as 15 antipodal pairs, corresponding to the 15 rotations of order two. Boundary points are depicted in red -- as is the origin, corresponding to the identity element of the group, and representing no rotation whatsoever. These 75 points therefore depict precisely the 60 rotational symmetries of the icosahedron.

From our approach to depicting rotations (namely that every rotation is depicted as a point ru pointing along its axis of rotation), note that applying a physical rotation, R, to a figure representing a collection of rotations like the one above, say F, is equivalent to conjugating each group element in F by the element R! (This is simply because the axis of rotation of R h R^-1 is simply (R v), if v is the axis of rotation of h, and the amount of rotation that h represents is unchanged.)

SO:every time the mouse is dragged across the image on this page, one is conjugating each element in A₅ by an element of SO(3). Moreover, every time the rotation accomplished by the mouse is a symmetry of the icosahedron, one is conjugating A₅ by an element of A₅. [For example, pick a blue point, P; it depicts an element of SO(3) corresponding to a symmetry of the icosahedron of order 3, representing a rotation by 2pi/3. Going back to the figure, note the 3-fold symmetry about such a blue point, corresponding to conjugation by the group element corresponding to P.]

Thus, the automorphisms of A₅ that consist of conjugation by elements in SO(3) can be visually deduced, simply by examining the rotational symmetries of the figure.

One can use this figure to "see why" A₅ is simple Basically, one looks for subsets of colored points invariant under the symmetries (rotations) of the icosahedron, which also form a subgroup. Clearly the blue points, for example, are all mapped into each other under symmetry rotations of the icosahedron; so the blue points are an invariant subset, and in fact correspond to a single conjugacy class in A₅. Thus if a normal subgroup H contained a single blue point, it would contain 12 of them, as well as the origin (since that represents the group identity), so H now would have to contain at least 13 points. But then simply by Lagrange's theorem, H would have to contain some other points as well (since the order of H divides 60, it would have to be at least 15). If H also contained a single green point, the entire collection of green points forms another conjugacy class in A₅, since they rotate into each other under the action of A₅. This collection of points cannot yet form a group, for clearly the "product" of a green point (representing a 2pi/5 rotation) with itself is a purple point (representing a 4pi/5 rotation). So if a normal subgroup H contains a single blue point, it must contain them all, and if it also contains a green or purple point it must contain them all as well; but then it would contain over 30 points, so many points that by Lagrange's theorem H would be A₅. Similar conclusions can be reached considering different candidates for elements of H. In the end, any H containing at least one point besides the origin would have to contain all the points; so A₅ is normal.

[For depictions of the full rotation group of R³, i.e. for a fuller depiction of SO(3), try here.]

The graphic may take 10-30 seconds to load, but once it does, you can rotate it and zoom in or zoom out (for the former, on a PC, just hold down the left mouse button while dragging it across the graphic; for the latter, hold down the shift key before you drag, and while you drag, the mouse vertically across the graphic; and finally, if you hold the shift key while dragging the mouse horizontally, the figure rotates about an axis perpendicular to the screen). When examining symmetry of the figure by rotating it, remember that any point on the boundary of the "ball", which would be depicted in red, is really to be identified with the point antipodal to it.

The "Live" rotation of the Mathematica-generated graphic uses the LiveGraphics3D software; for more info on how to rotate Mathematica-generated graphics on the web, visit LiveGraphics3D.

Comments? Send them to Rick Kreminski.