Theorem 1   The symmetric group on $\{1, 2, \ldots, n\}$ is generated by the permutations $$ (1, 2), (2, 3), \ldots, (n-1, n). $$

Proof. We proceed by induction on $n$ . If $n = 2$ , the theorem is trivially true because the the group only consists of the identity and a single transposition.

Suppose, then, that we know permutations of $n$ numbers are generated by transpositions of successive numbers. Let $\phi$ be a permutation of $\{1, 2, \ldots, n+1\}$ . If $\phi(n+1) = n+1$ , then the restriction of $\phi$ to $\{1, 2, \ldots, n\}$ is a permutation of $n$ numbers, hence, by hypothesis, it can be expressed as a product of transpositions.

Suppose that, in addition, $\phi (n+1) = m$ with $m \neq n+1$ . Consider the following product of transpositions: $$ (n n+1) (n-1 n) \cdots (m+1 m+1) (m m+1) $$ It is easy to see that acting upon $m$ with this product of transpositions produces $+1$ . Therefore, acting upon $n+1$ with the permutation $$ (n n+1) (n-1 n) \cdots (m+1 m+1) (m m+1) \phi $$ produces $n+1$ . Hence, the restriction of this permutation to $\{1, 2, \ldots, n\}$ is a permutation of $n$ numbers, so, by hypothesis, it can be expressed as a product of transpositions. Since a transposition is its own inverse, it follows that $\phi$ may also be expressed as a product of transpositions.