Order of the symmetry group of Feynman Diagrams

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SUMMARY

The discussion focuses on understanding the symmetry factor in Feynman diagrams, specifically the equation $$ \frac{1}{O(G)} = \frac{M}{n!(4!)^n} $$, where O(G) represents the order of the symmetry group. The user attempts to calculate O(G) as 10, considering the asymmetry of internal line pairs, but arrives at a symmetry count S of 12. Clarification on the correct method for counting the order of symmetry in Feynman diagrams is sought.

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  • Understanding of Feynman diagrams and their components
  • Familiarity with group theory concepts, particularly symmetry groups
  • Basic knowledge of combinatorial mathematics
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Hello,

I am learning Feynman diagrams and I still do not understad quite well the symmetry factor idea. The equation is:

$$ \frac{1}{O(G)} = \frac{M}{n!(4!)^n} $$

I was trying the next example:

cIR6Sil.jpg


If I am not wrong it is O(G) = 10 taking care of the asymmetry of each pair of internal lines connecting vertices.
 
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Now, I count S = O(G) = 12.
 
Can anyone at least explain "for dummies" how to count the order of symmetry?
 

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