Advanced High-School Mathematics

(Tina Meador) #1

234 CHAPTER 4 Abstract Algebra


see that|G|


∣∣
∣24. But there are plenty of permutations of the vertices
1, 2, 3, and 4 which are not graph automorphisms (find one!), and so
|G|<24.


On the other hand, note that the powers e, σ, σ^2 , σ^3 give four au-
tomorphisms of the graph, and the elementsτ, στ, σ^2 τ, σ^3 τ give four
more. Furthermore, since τστ = σ^3 we can show that the set
{e, σ, σ^2 , σ^3 , τ, στ, σ^2 τ, σ^3 τ}is closed under multiplication and hence
is a subgroup ofG. Therefore 8


∣∣
∣|G|and so it follows that|G|= 8 and
the above set is all ofG:


G = {e, σ, σ^2 , σ^3 , τ, στ, σ^2 τ, σ^3 τ}.

Below is the multiplication table for G (you can fill in the missing
elements). Notice that G has quite a few subgroups—you should be
able to find them all (Exercise 3).


◦ e σ σ^2 σ^3 τ στ σ^2 τ σ^3 τ
e e σ σ^2 σ^3 τ στ σ^2 τ σ^3 τ
σ σ σ^2 σ^3 e
σ^2 σ^2 σ^3 e σ
σ^3 σ^3 e σ σ^3
τ τ
στ στ
σ^2 τ
σ^3 τ

Exercises



  1. Use the corollary on page 233 to give another proof of Fermat’s
    Little Theorem; see page 86.

  2. Suppose that Gis a finite group of prime orderp. Prove thatG
    must be cyclic.

  3. Refer to the multiplication table above for the groupGof symme-
    tries of the square and list all of the subgroups.

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