Advanced High-School Mathematics

(Tina Meador) #1

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



  1. Compute |G|, where G is the automorphism group of Graph A,
    above. IsGabelian?

  2. Compute |G|, where G is the automorphism group of Graph B,
    above. IsGabelian?

  3. Find an element of order 6 in the stabilizer of vertex 1 of Graph
    C, above.

  4. 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.

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