Exercise 3.5.4 ($S_n = \langle (1 \ 2), (1\ 2\ 3\ \ldots \ n) \rangle$)

Proof. (of lemma) Put $\gamma =(a_1{a}_2\cdots a_k)$, and $\delta =(\sigma (a_1)\sigma (a_2)\cdots \sigma (a_k))$.

• If $1\le i<k$, then

  $$(\mathit{\sigma \gamma }{\sigma }^{-1})(\sigma (a_i))=\sigma (\gamma (a_i))=\sigma (a_{i+1})=\delta (\sigma (a_i)).$$

• If $i=k$, then

  $$(\mathit{\sigma \gamma }{\sigma }^{-1})(\sigma (a_k))=\sigma (\gamma (a_k))=\sigma (a_1)=\delta (\sigma (a_k)).$$

• Finally, if $x\notin \{\sigma (a_1),\cdots ,\sigma (a_k)\}$, then ${\sigma }^{-1}(x)\notin \{a_1,\cdots ,a_k\}$, therefore $\gamma ({\sigma }^{-1}(x))={\sigma }^{-1}(x),$ so

  $$(\mathit{\sigma \gamma }{\sigma }^{-1})(x)=x.$$

This shows that $\mathit{\sigma \gamma }{\sigma }^{-1}=\delta$, so

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Proof. (of Ex. 3.5.4)

We have proved in Exercise 3 that

Put $\tau =(12)$ and $\sigma =(12...n)$.

Let us apply the Lemma to $\tau$ and ${\sigma }^{k-1},1\le k<n$:

Therefore $⟨\sigma ,\tau ⟩\supseteq ⟨(12),(23),\cdots ,(n-1n)⟩=S_n$, where $⟨\sigma ,\tau ⟩\subseteq S_n$, so $⟨\sigma ,\tau ⟩=S_n$.

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