Prove that any element in $S_n$ can be written as a finite product of the following
permutations:
$(a)\ (12),(13), . . . , (1n)$
$(b)\ (12),(23),…,(n−1,n)$
$(c)\ (12),(12\dots n)$.
I have no idea where to start. I know that the products from each category results an element of $S_n$, but no clue for the reverse. I think if some element is n-cycle it must be possible to write as products of transpositions in $(a)$ and depending on the cycle, just order in product varies.
Thank you for guidance.
Best Answer
Every permutation is a product of transpositions. Hence all you have to check is:
We'll use the fact that any two transpositions are conjugate by a transposition:
(1) is easy: $\,(k\,l)=(1\,k)(1\,l)(1\,k)$.
(2) is by induction on $k$: $(1\,2)$ is both of types (a) and (b). Suppose now $(1\,k)$ can be obtained from transpositions of type (b). Then $(1\,k)(k,\,k+1)(1\,k)=(1\,k+1)$ is likewise obtained from type (b).
(3) also by induction on $k$: let $\gamma=(1\,2\,\dots\,n)\,$. We have $\,(2\,3)= \gamma(1\,2)\,\gamma^{-1}$, and more generally: $\,(k\enspace k+1)=\gamma^{k-1}(1\,2)\,\gamma^{-(k-1)}$ for all $\,2\le k\le n-1$.