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I keep reading that $A_n$ is "clearly $(n-2)$-transitive" in various places but am actually having a hard time proving it.

So far, I have tried looking at an arbitrary set $${x_1, ... , x_{n-2}}$$ of $(n-2)$ elements. if $\sigma$ is an element of $A_n$, such that $\sigma(x_i) = (y_i)$ for all i, then it's clear that $n-4$ of the $y_i's$ are in the original set. I'm not sure if this helps us.

I also noticed that we can conjugate any element of $A_n$ by any transposition and get another element of $A_n$.

I feel like I am missing something obvious. Any help is appreciated.

Shaun
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1 Answers1

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Note that forcing $\sigma(x_i) = y_i$ for $i = 1, \ldots, n-2$ is possible for exactly two permutations. There are two possible input values not already specified (ie the two values not equal to any of the $x_i$), which must take on either of two possible values (the ones not equal to any of the $y_i$).

For concreteness, if $a,b$ are the two values not equal to any of the $x_i$ and $c,d$ are the two values not equal to the $y_i$, then we have either $\sigma(a)=c, \sigma(b) = d$ or $\sigma(a) = d, \sigma(b) = c$.

Note that exactly one of these two permutations is even, since they differ by a transposition.

Rolf Hoyer
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  • I do not understand why only one of those two transpositions is even. They both seem the same up to labeling to me.

    Wait we don't have to tell which of the two is even, it's just true that only one of them is, since they differ by a transposition.

    Then you're saying whichever it is, (a c) (b d) for example, we can choose c and d arbitrarily. Therefore we can send a to anything we like? But how does that prove n-2 transitivity? doesn't that only show 1-transitivity?

     –  Nov 25 '17 at 02:06
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    it shows $n-2$ transitivity because the two permutations given (one of which is even) satisfy the given constraint $\sigma(x_i) = y_i$ for $i = 1, \ldots, n-2$. – Rolf Hoyer Nov 25 '17 at 02:11