Advanced High-School Mathematics

(Tina Meador) #1

SECTION 4.2 Basics of Group Theory 243


IfHis the stabilizer of the vertex 1, then surely Hmust permute the
three vertices 2, 4, 6 and must permute the vertices 3 and 5. Further-
more, it is easy to see thatanypermutation of 2, 4, 6 and of 3, 5 will
determine an automorphism of the graph which fixes vertex 1. Since
there are 6 = 3! permutations of 2, 4, and 6, and since there are 2
permutations of 3 and 5, we conclude that there are exactly 6×2 = 12
automorphisms which fix the vertex 1. Therefore the full automorphism
has order 6×12 = 72.


We turn now to the second graph considered in our introduction;
again we draw two versions:


IfH is the stabilizer of the vertex 1, thenH must permute the three
vertices 3, 4, 5 and must permute the vertices 2 and 6. However,
in this case, there are some restrictions. Note thatH must actually
fix the vertex 4 (because there’s an edge joining 3 and 5). Thus an
automorphismτ ∈H can only either fix the vertices 3 and 5 or can
transpose them: τ(3) = 5, τ(5) = 3. However, once we know what τ
does to{ 3 , 4 , 5 }we can determine its effect on 2 and 6. Ifτ fixes 3 and
5, then it’s easy to see thatτ also fixes 2 and 6 (verify this!). Likewise,
ifτ transposes 3 and 5, thenτ also transposes 2 and 6, meaning that
there are only two elements inH,eand the element


τ=







1 2 3 4 5 6

↓ ↓ ↓ ↓ ↓ ↓

1 6 5 4 3 2





.
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