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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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2 Answers

up vote 12 down vote accepted

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

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math.stackexchange.com/questions/64848/… seems to be that $n$ must be a prime. –  ShinyaSakai Nov 27 '11 at 18:02
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That hypotesis is needed when you want to use an arbitrary transposition. For $(1,2)$ it is not needed. –  Mariano Suárez-Alvarez Nov 27 '11 at 18:09
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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. –  KCd Nov 27 '11 at 18:11
    
actually it works for all n. –  user641 Nov 27 '11 at 18:11
    
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 $\{(i,i+1) | 1 \leq i <n \}$ I referred is obtained. –  ShinyaSakai Nov 28 '11 at 15:05
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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 $1/4$.

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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 $S_n$. –  user641 Nov 27 '11 at 21:16
    
Thanks to both of you~ The answer and comment make the generating problem of $S_n$ more clear to me. –  ShinyaSakai Nov 28 '11 at 15:17
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