244 CHAPTER 4 Abstract Algebra
From the above we conclude that the automorphism group of this graph
has 6×2 = 12 elements, meaning that the first graph issix times more
symmetrical than the second!
Exercises
- Compute |G|, where G is the automorphism group of Graph A,
above. IsGabelian? - Compute |G|, where G is the automorphism group of Graph B,
above. IsGabelian? - Find an element of order 6 in the stabilizer of vertex 1 of Graph
C, above. - As a challenge, compute the order of the automorphism group of
the Petersen graph.^15
(^15) The answer is 120=5!. Indeed, the automorphism group is isomorphic with the symmetric group
S 5. Here’s a possible approach. Construct the graph Γ whose vertices are the 2-element subsets of
{ 1 , 2 , 3 , 4 , 5 }and where we stipulate thatAandBare adjacent precisely whenA∩B=∅. One
shows first that this graph is actually isomorphic with the Petersen graph. (Just draw a picture!)
Next, if we letS 5 operate on the elements of{ 1 , 2 , 3 , 4 , 5 }in the natural way, thenS 5 actually acts
as a group of automorphisms of Γ.