We have already met the example of the group D_n, which is the group of symmetries of a regular n-gon in the plane. We have seen that this group is also a subgroup of the automorphism group of the n-gon as a graph. In fact, it is the full automorphism group of the graph. Prove this!

Let P be the Petersen graph. The vertices of P can be identified with the pairs of elements from {1,2,3,4,5}. Two vertices {x, y} and {u, v} are adjacent if the intersection {x, y} {u,v} is empty.

The group S₅ acts on the set {1,2,3,4,5}, but also on the vertex set of P. For, if g S₅, then the map g₂ : {x, y} -> {g(x),g(y)} defines a permutation of the 10 vertices of P. The map

S₅ -> Sym(V),     g -> g₂

is an injective permutation representation of S₅, see a previous example.

Clearly, g₂ also defines an automorphism of P. For if {x, y} and {u, v} are adjacent in the P, then their intersection is empty; because g is a bijection, the intersection of {g(x), g(y)} and {g(u), g(v)} is then also empty. In particular, these vertices are adjacent.

By the above, the automorphism group G of P contains a subgroup, denoted by H, isomorphic to S₅. This subgroup acts transitively on the vertex set of the graph. The triple

[a, b, g] = [{1,2}, {1,3}, {2,4}]

is a basis for G. For, if an element of G fixes the vertices a, b, and c, then it also fixes the unique common neighbour h of a and g, and, similarly, j, the unique common neighbour of a and b. Since each further vertex of the Petersen graph is connected with a unique vertex from the pentagon {a, b, g, h, j}, the element fixes all vertices of P. This argument establishes that G_{a, b, g} is the trivial subgroup of G.

The group G is transitive on the 10 vertices of P. For, H is transitive on the 10 vertices. As the orbit of b under the stabilizer H_a in H of a consists of the 6 vertices of P not connected with a, these 6 vertices also belong to the G_a-orbit of b. If, next to a, we also fix b, then g can only be moved into g or f. The formula for the order of G gives

|G| = 10· 6· 2 = 120.

We conclude that G has exactly as many elements as its subgroup H. Hence G coincides with H and is isomorphic to S₅.

The Fano plane is a geometry with seven vertices and seven lines for which the following axioms hold: each point is on 3 lines; each line contains 3 vertices; each pair of vertices lies on exactly one line; each pair of lines meets in exactly one point.

A model of the Fano plane is the following.
The points are the numbers 1 , ...,7,
the lines are the triples {1, 2, 3}, {1, 4, 5}, {1, 6, 7}, {2, 4, 7}, {2, 5, 6}, {3, 4, 6} and {3, 5, 7}. In a picture:

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An automorphism of the Fano plane is a permutation of the vertices carrying each line into a line. For instance, the permutations (2,3)(4,5), (2,3)(6,7) and (4,5)(6,7) are automorphisms of the Fano plane fixing the point 1.

Let G be the automorphism group of the Fano plane.

1. Show that G acts transitively on the seven vertices and on the seven lines.
2. Show that the stabilizer of a point is transitive on the other six vertices. Thus, the group is 2-transitive on the point set.
3. Which orbits does the group have on unordered triples?
4. Determine a basis for G.
5. Show that the group G has order 168.
6. Prove that G is a subgroup of A₇.

Let G = <g> be a group of order n generated by g. By the previous examples, the order of Aut(G) is (n), the Euler indicator of n.

The group Aut(G) is commutative but need not be cyclic (a counterexample occurs for n = 8).