Subgroup of Sym(n) Isomorphic to S_(n-1) Except n=6

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Discussion Overview

The discussion revolves around the properties of subgroups of the symmetric group S_n, specifically focusing on the conditions under which a subgroup is isomorphic to S_(n-1) and fixes a point in the set {1, 2, ..., n}, with an exception noted for n=6.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant seeks a proof that any subgroup of S_n isomorphic to S_(n-1) must fix a point unless n=6.
  • Another participant questions the validity of a proposed approach regarding the relationship between isomorphic groups and subgroup orders.
  • A different participant suggests a condition that if n ≠ 6, then any subgroup Y of S_n with |S_n:Y|=n must fix a point.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of proposed proofs and conditions, indicating that the discussion remains unresolved with multiple competing perspectives.

Contextual Notes

Participants have not reached consensus on the validity of the proposed proofs or the implications of subgroup orders, and there are unresolved assumptions regarding the nature of the mappings and subgroup properties.

dimuk
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I need a proof of any subgroup of S_n which is isomorphic to S_(n-1) fixes a point in {1, 2,..., n} unless n=6.
 
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the standard answer: what have you tried?
 
I defined a map psi: S_n to S_(n-1) and took a subgroup H={pi \in S_n | pi(n)=n}. And proved that H is a subgroup of S_n, but I want to prove that which is isomorphic to S_(n-1) and fixes a point in {1, 2..., } unless n=6.

I thought to prove this

If X is isomorphic to S_n and Y is isomorphic to X with |X:Y|=n then Y is isomorphic to S_(n-1).

But still I don't know how to prove.
 
No that second one is not correct.
 
I should proof this

If n \neq 6 then any subgroup Y of S_n with |S_n:Y|=n actually fixes a point..?
 

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