$$\sigma \in S_n. \ Prove \ that \ \sigma = \alpha \beta, \ if \ \alpha ,\beta \in S_n \ and \ \alpha^2 = \beta^2 = e.$$ Here is my solution:
Let's prove that statement by induction, basis ($n=2$) - evident statement. The induction step: the statement holds for $n = k$. In $S_{k+1}$ there are $k!$ permutations from $S_{k}$ (for example, $(123)$ from $S_{3}$ and $(123)(4)$ from $S_{4}$). The number of "new" transpositions (transpositions with element $k+1$) $= k$. It's obvious that composition of "new" transpositions and permutations from $S_{k}$ (except $e$) will give us another permutations that we haven't considered yet, their number $= k * (k! - 1)$. For all these permutations the statement holds because of the induction step and the way I construct all of them. Let's add them up: $k! + k + k * (k! - 1) = (k + 1)! \ , $$|S_{k+1}| = (k + 1)!$ It means that we have considered all permutations from $S_{k+1}$ and the statement holds for $S_{k+1}$. Are there any mistakes in my solution?
