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Math Help - automorphisms of symmetric group.

  1. #1
    Senior Member abhishekkgp's Avatar
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    automorphisms of symmetric group.

    Prove that for each \sigma \in Aut(S_n)

    \sigma: (1 \, 2) \mapsto (a \, b_2), \, \sigma: (1 \, 3) \mapsto (a \, b_3), \ldots, \sigma: (1,\, n) \mapsto (a \, b_n).

    I know that an automorphism sends a transposition to a transposition so \sigma: (1 \, 2) \mapsto (a \, b_2) is fine. How to prove the rest. I tried to use contradiction. denote \sigma((1 \, 3))= \tau. Assume \tau (a)=a, \tau (b_2)=b_2. Then we should get a contradiction but i couldn't arrive at one.
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Re: automorphisms of symmetric group.

    i just found something. let \sigma((1 \, 2))=(a \, b), \sigma((1 \, 3))= (c \, d). Consider \sigma((1 \, 2)(1 \, 3))=\alpha. So \sigma((1 \, 3 \, 2))= \alpha \Rightarrow |\alpha|=3 since \sigma is an isomorphism. Also \sigma((1 \, 2)(1 \, 3))=\sigma((1 \, 2)) \sigma((1 \, 3))=(a \, b)(c \, d)=\alpha. if (a \, b) \, & (c \, d) are disjoint cycles then |\alpha|=2 which is not possible. Hence there is one and only one element common in the cycles (a \, b) \, & \, (c \, d) so WLOG (c \, d)=(a \, d). I guess this will prove the proposition. Please comment.
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