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Nearly all the books I read give $S_n$ $(n \geq 2)$ and the generating set $\{(i,i+1) | 1 \leq i < n \}$ as an example when talking about presentation groups. But is $\{(i,i+1) | 1 \leq i < n \}$ the least set of generators, i.e., is the order of any generating set for $S_n$ equal to or greater than $n-1$? If it is the least, how to prove? Are there any other least set of generators? In general, what do these least sets look like?

Forgive me for so many questions. Thanks sincerely for any answers or hints.

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    See also http://math.stackexchange.com/questions/143999/generators-for-s-n2012-05-23

2 Answers 2

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I think $(1,2), (1,2,\ldots, n)$ is also a set of generators.

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    http://math.stackexchange.com/questions/64848/how-does-12-cdots-n-and-ab-generate-s-n seems to be that $n$ must be a prime.2011-11-27
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    That hypotesis is needed when you want to use an *arbitrary* transposition. For $(1,2)$ it is not needed.2011-11-27
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    The comment by ShinyaSakai is wrong: those generators work for any $n \geq 2$. The link is about using *any* transposition together with $(1,2,...,n)$. But for the *specific* transposition $(1,2)$ and the $n$-cycle $(1,2,...,n)$ you have a generating set for all $n$. See the table at the end of Section 1 at www.math.uconn.edu/~kconrad/blurbs/grouptheory/genset.pdf. A transposition $(a,b)$ and the standard $n$-cycle $(1,2,...,n)$ generate $S_n$ iff $b-a$ and $n$ are relatively prime. So you can use any $(a,a+1)$ as the transposition.2011-11-27
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    actually it works for all n.2011-11-27
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    Thanks for everyone. In fact, if $(1,2,\cdots, n)$ is denoted as $a$, then $a (1,2) a^{-1} = (2,3)$, $\cdots$, $a (n-2, n-1) a^{-1} =(n-1,n)$. Thus, the generating set $\{(i,i+1) | 1 \leq i I referred is obtained.2011-11-28
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It is proved in

J. D. Dixon, The probability of generating the symmetric group, Math. Z. 110 (1969), 199–205.

that the probability that a random pair elements of $S_n$ generate $S_n$ approaches $3/4$ has $n \to \infty$, and the probability that they generate $A_n$ approaches $1/4$.

  • 1
    There's also an article from Isaacs a few years ago (I'll try and find the reference) that proves (when $n\neq4$) that if you give me an element from $S_n$, I can always find a second one so that, taken together, they generate all of $S_n$.2011-11-27
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    Thanks to both of you~ The answer and comment make the generating problem of $S_n$ more clear to me.2011-11-28
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    The article by Isaacs and Zieschang is "Generating Symmetric Groups" in Amer. Math. Monthly 102 (1995), 734-739. A link to it online is http://www.fmf.uni-lj.si/~potocnik/poucevanje/SeminarI2013/T13-JordanovIzrek.pdf2015-01-03