Exercise 3.5.3 ($S_n= \langle(1\ 2), (2\ 3), . . . , (n -1\ n) \rangle$)

Proof. Let $n\ge 2$ be an integer, and let $G_n$ be the subgroup of $S_n$ generated by the transpositions $(12),(23),...,(n-1n)$.

We show first that $G_n$ acts transitively on $[[1,n]]$.

Let $i\in \{1,\cdots ,n\}$ and consider the permutation

(using the convention $g=e$ if $i=n$).

Then $g\in G_n$ and $g(n)=i$.

For all $i\in \{1,\cdots ,n\}$, there exists $g\in G_n$ such that $g(n)=i$, so $G_n$ acts transitively on $[[1,n]]$.

We show by induction on $n$ that $G_n=S_n$.

• $S_2=\{(),(12)\}$ is equal to $G_2=⟨(12)⟩$.

• Let us assume that $S_{n-1}=G_{n-1}$.

  We identify the subgroup of $S_n$ of permutations fixing $n$ with $S_{n-1}$.

  Let $\sigma$ be any permutation in $S_n$. Put $i=\sigma (n)$.

  By the first part, there exists $g\in G_n$ such that $g(n)=i$.

  Then $(g^{-1}\circ \sigma )(n)=n$ so $g^{\prime }=g^{-1}\circ \sigma \in S_{n-1}$.

  Therefore $\sigma =g\circ g^{\prime }$, where $g\in G_n,g^{\prime }\in G_{n-1}\subseteq G_n$, hence $\sigma \in G_n$.

  This shows that $S_n\subseteq G_n$, and by definition $G_n\subseteq S_n$, thus $G_n=S_n$.

• The induction is done, so for all $n\ge 2$,

  $$S_n=⟨(12),(23),\cdots ,(n-1n)⟩.$$