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We can decompose any element of $S_p$ into the form $(a_1 \ \ b_1)(a_2 \ \ b_2) \dots (a_i \ \ b_i)$. If for some $1 \le j \le i$, $|a_j - b_j|=1$, it's easy! Because: Let $\sigma = (12 \ldots p)$. Then $\sigma^k (12) \sigma^{-k} = (k+1 \ \ k+2)$. But I can't write $(a_j \ \ b_j)$ as arbitrary products of $(12 \ldots p)$ and $(12)$ if $|a_j - b_j|>1$.

Any answer that I read in MSE was not easy to understand. I would appreciate any simple clear detailed explanation.

  • First you should make sure to use the same labelling. Is $n = p$? – Tobias Kildetoft Mar 16 '16 at 12:10
  • Ok, so you know it suffices to show that all the transpositions of neighboring elements can be written as a product of elements from the given set. Hint for this: Consider what happens if you conjugate the transposition by the $p$-cycle. – Tobias Kildetoft Mar 16 '16 at 12:14
  • @TobiasKildetoft - the problem is that I am very beginner in elementary gp.th and I don't know what "conjugate the transposition" or "p-cycle" means. Is there any simple way to explain the case that I asked? –  Mar 16 '16 at 12:18
  • Conjugating $g$ by $h$ means taking $hgh^{-1}$. I assume you know what transposition and $p$-cycle mean? – Tobias Kildetoft Mar 16 '16 at 12:19
  • @TobiasKildetoft - Yes I remembered both! But how can I apply conjugating transpositions? –  Mar 16 '16 at 12:23
  • Try to compute $hgh^{-1}$ where $g$ is the transposition and $h$ is the $p$-cycle and see what you get. – Tobias Kildetoft Mar 16 '16 at 12:24
  • Start with what you have in the set you want to show is a generating set. – Tobias Kildetoft Mar 16 '16 at 12:27
  • @TobiasKildetoft - That's all what I did! But it works to get (k+1, k+2) not arbitrary (k+1, j+2). g is not arbitrary (ab). It's (12). –  Mar 16 '16 at 12:28
  • Ahh, I misread your comment as if you had already reduced to this. Ok, so now you want to show that any transposition can be written as a product of ones of this form. Try computing $(2\ 3)(1\ 2)(2\ 3)$ and see if you can generalize the pattern. – Tobias Kildetoft Mar 16 '16 at 12:31
  • Yeah it separates! (13). Let me if I can see the pattern in general. Thanks –  Mar 16 '16 at 12:35
  • @TobiasKildetoft - (k+2 k+3)(k+1 k+2)(k+2 k+3)=(k+1 k+3) is valid for any k, I checked. In order to reach, (k+1 k+4) I tried (k+2 k+3)(k+1 k+3)(k+2 k+3) but it failed; also (k+1 k+4)(k+1 k+2)(k+1 k+4) fails though I am not allowed to do that since I haven't derived (k+1 k+4) yet on the other hand. –  Mar 16 '16 at 12:45
  • So now you get all those transpositions differing by $2$. Then try $(3\ 4)(1\ 3)(3\ 4)$. – Tobias Kildetoft Mar 16 '16 at 12:47
  • @TobiasKildetoft - Yes! Thanks a lot. Let me try to understand why (i, i+1)(1,i)(i, i+1)=(1,i+1). I'll come back. –  Mar 16 '16 at 12:50
  • @TobiasKildetoft - Sorry I don't understand (if it's not a guess or experience) how did you come to the point that h=(i, i+1) will change (1,i) to (1,i+1) in conjugation speaking. I mean sure it works, but before knowing that how to arrive to (i,i+1)? –  Mar 16 '16 at 12:58
  • @TobiasKildetoft - It's because $\sigma (ab) \sigma^{-1} = (\sigma (a) \ \ \sigma (b))$. Thank you very much indeed. :) –  Mar 16 '16 at 13:01
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    In general, $h(a\ b\ c,\dots)h^{-1}$ is the cycle $(h(a)\ h(b)\ h(c),\dots)$ (i.e. you apply $h$ to each entry of the cycle to get the new cycle). – Tobias Kildetoft Mar 16 '16 at 13:02
  • Heh, right (you were faster than me). – Tobias Kildetoft Mar 16 '16 at 13:02

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