Show that every element of Sn can be written as an arbitrary product of the elements (01) and (01...n-1).
I understand that this can be solved using induction, and I've set up my base cases. However, I'm not sure how I can prove that if the proposition is true for k, it must also be true for k+1.
Any help would be appreciated.
Try using a direct algorithmic proof instead.
Sketch:
Given any $\alpha \in S_n$, recall that we can write $\alpha$ as a product of transpositions. Hence, if we know how to express any $2$-cycle as a product of $(0,1)$ and $(0,1,\ldots, n-1)$, then we are done.
Given any transposition $(a, b) \in S_n$ with $a < b$, notice that we can write it as a product of consecutive transpositions. For example: $$ (2, 5) = (4, 5)(3, 4)(2, 3)(3, 4)(4, 5) $$ Hence, if we know how to express any consecutive $2$-cycle as a product of $(0,1)$ and $(0,1,\ldots, n-1)$, then we are done.
Given any consecutive transposition $(x, x + 1) \in S_n$ with addition modulo $n$, use conjugations to write it as a product of $(0,1)$ and $(0,1,\ldots, n-1)$. For example, if $x = 3$ and $n = 9$, then notice that: $$ (3, 4) = (0, 1, \ldots, 8)^3(0, 1)(0, 1, \ldots, 8)^{-3} $$
