Can someone please help me with this proof? I wrote down my idea but I really don't know how to prove this!
Show that if $\sigma$ is not a cycle, then it can be written as a product of at most $n-2$ transpositions.
My work so far:
If $\sigma$ is not a cycle, then it can be written as a product of at least two disjoint transpositions (is this even true?). If $|\sigma |=n$, then $\sigma = (a_{1}a_{2})(a_{3}a_{4})\cdots (a_{n-1}a_{n})$ has a total of $n - 2$ (I honestly don't think so, I just guessed) pairs, since $(a_{n}a_{1})$ cannot be a pair which is of length two.
First note that any cycle $(a_1 a_2 a_3 \ldots a_k)$ can be written in the usual way as a product of $k-1$ transpositions: $(a_1 a_2) (a_1 a_3) \ldots (a_1 a_k)$. Now every permutation can be written as a product of disjoint cycles. Let $\sigma$ be so written; since it is not itself a cycle, it is a product of at least two cycles, e.g $C_1 C_2 ...C_m$. Represent each cycle $C_i$ as a product of transpositions as shown above. Then the total number of transpositions must be $(\lvert C_1\rvert-1) + (\lvert C_2\rvert - 1) + \cdots + (\lvert C_m\rvert - 1) = \lvert\sigma\rvert - m$, where $m$ is at least $2$. QED.