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Im doing some previous exams in abstract algebra. The last question is

"Prove that the permutations $(12)$ and $(12 \dots n)$ generate $S_n$ for $n ≥ 2.$"

Can someone please give me a hint where to start?

Olba12
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  • Hint: Compose them and see what happens. – Cbjork Apr 07 '15 at 20:03
  • Think about what they mean. The first means you flip the two elements at the beginning of your permutation, the second means you rotate the elements of the permutation to the right. Using these operations, which other permutations can you make? – TravisJ Apr 07 '15 at 20:06
  • This previous question might be of some help to you. – James Apr 07 '15 at 20:12
  • @Olba12: The claim is evident for $n = 2$. So, we assume $n > 2$.

    $\forall r \in {2,...,n-1}: (1,2) \circ (1,2,...,n) \circ (1,r) \circ (1,2,...,n)^{-1} \circ (1,2)^{-1} = (1,2) \circ (2, r + 1) \circ (1,2)^{-1} = (1, r + 1)$

    $\forall r,s \in {2,...,n}: (r \not= s) \supset ((1,r) \circ (1,s) \circ (1,r)^{-1} = (r,s))$

     – Moritz Apr 07 '15 at 20:14

2 Answers2

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With $\tau_{1,2}=(1,2)$ and $\sigma=(1,\dots,n)$ we can show that we can generate $\tau_{i,j}=(i,j)$. Firstly, we can generate $\tau_{i,i+1}$ for all $i=1,\dots,n$ (we take the addition of indices mod $n$), since $\tau_{i+1,i+2}=\sigma\tau_{i,i+1}\sigma^{-1}$. Then, we can generate $\tau_{i,j}=\tau_{i,i+1}\dots\tau_{j-2,j-1}\tau_{j-1,j}\tau_{j-2,j-1}\dots\tau_{i+1,i+2}\tau_{i,i+1}$, here we take $i<j$. The rest is trivial.

Salomo
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An idea:

$$(12...n)(12)(12...n)^{-1}=(23)\;,\;\;(12...n)(23)(12...n)^{-1}=(34)\;,\ldots$$

and now you already have that $\;(12)\,,\,\,(23)\,,\ldots,(n-1\;n)\,,\,(1 n)\in\langle\;(12)\,,\;(12...n)\;\rangle\;$

Try now to take it from here (hint: show any transposition is in the group generated by your two cycles).

Timbuc
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