Permutation Conjugation and Order

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SUMMARY

The discussion centers on proving the non-existence of a permutation \( a \) such that \( (a^{-1})(1,2)(a) = (3,4)(1,5) \). Key insights include the order of the permutation \( (1,2) \), which is 2, leading to the conclusion that squaring \( (a^{-1})(1,2)(a) \) results in the identity permutation. The conversation also touches on the invariance of cycle types under conjugation, emphasizing the need to analyze how elements are mapped by the permutation \( a \).

PREREQUISITES
  • Understanding of permutation groups and their properties
  • Knowledge of cycle notation in permutations
  • Familiarity with the concept of conjugation in group theory
  • Basic grasp of permutation orders and their implications
NEXT STEPS
  • Study the properties of permutation conjugation in group theory
  • Learn about cycle types and their invariance under conjugation
  • Explore the implications of permutation orders in group structures
  • Investigate examples of non-existence proofs in permutation groups
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Mathematics students, particularly those studying abstract algebra, group theory enthusiasts, and anyone interested in the properties of permutations and their applications.

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Homework Statement


Prove that there is no such permutation a such that
(a-inverse)*(1,2)*(a) = (3,4)(1,5)


The Attempt at a Solution


Does it have something to do with the order of (1,2)? I know the order is 2, so if we square (a-inverse)*(1,2)*(a), then we get the identity...how else can I think about it?
 
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Do you know anything about cycle types? In particular, can you prove that they are invariant under conjugation?

If not, notice that 1->5 on the RHS. Convince yourself that (1,2) must send a(1) to a(5). Do the same for 3.
 
By a(1) and a(5) do you mean the numbers 1 and 5 in whatever the permutation a is?
 

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