Transposition-preserving automorphism of finitary symmetric group is induced by conjugation by a permutation

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Statement

Let S be a set, K = \operatorname{Sym}(S) be the Symmetric group (?) on S, and G = \operatorname{FSym}(S) be the subgroup of G comprising the finitary permutations, i.e., H is the Finitary symmetric group (?) on S. Suppose \sigma is an automorphism of H that sends transpositions to transpositions. Then, \sigma is induced by conjugation by some element of G.

Related facts

Stronger facts

Facts used

  1. Transpositions generate the finitary symmetric group
  2. Transpositions commute iff they are disjoint

Proof

Given: A finite set S. G is the finitary symmetric group on S. \sigma is an automorphism of G that sends transpositions to transpositions.

To prove: \sigma is induced by conjugation by a permutation in K, the whole symmetric group on S.

Proof: By fact (1), it suffices to find g \in G such that conjugation by g agrees with \sigma on the set of transpositions. Further, we can assume that S has at least three elements (The statement is obviously true for S having zero, one, or two elements). We now describe how the permutation g can be constructed explicitly.

Also, we use fact (2) at many steps, without explicitly acknowledging it.

  1. Pick distinct elements a,b,c \in S. Then, both \sigma((a,b)) and \sigma((a,c)) are transpositions. We claim that these transpositions have exactly one element in common: If \sigma((a,b)) and \sigma((a,c)) have no element in common, then they commute, and hence, (a,b) commutes with (a,c), which is not true. Thus, \sigma((a,b)) and \sigma((a,c)) have exactly one element in commmon.
  2. Pick distinct elements a,b,c \in S. Then, there are elements a',b',c' \in S such that \sigma((a,b)) = (a',b'), \sigma((b,c)) = (b',c') and \sigma((c,a)) = (c',a'): This follows directly from the previous step.
  3. For any element a \in S, there is a unique element g(a) that is involved in the transposition \sigma((a,b)) for every b \ne a: For any b \ne c distinct from a, there exist a',b',c' as described in the previous step. We now show that \sigma((a,d)) is also a transposition involving a'.
    • Since (a,d) commutes with (b,c), \sigma((a,d)) commutes with (b',c'). In particular, it cannot be a transposition involving b' or c'.
    • Since (a,b) does not commute with (a,d), \sigma((a,b)) does not commute with \sigma((a,d)), so they must have an element in common. Thus, either a' or b' is involved in \sigma((a,d)).
    • These together force a' to be involved in \sigma((a,d)).
    • We can thus define g(a) as the unique element involved in \sigma((a,x)) for every transposition involving a.
  4. The previous step yields a map g:S \to S such that for any transposition (a,b), \sigma((a,b)) is a transposition involving both g(a) and g(b). Step (2) makes it clear that g(a) \ne g(b), so \sigma((a,b)) = (g(a),g(b)). Thus, \sigma is induced by conjugation by the permutation g.