Decomposition of a cycle into transpositions

Question

Let $S_n$ be the symmetric group of the interval $1\leq i\leq n$. Suppose $\sigma\in S_n$ and $$\langle\sigma\rangle=\{id,\ldots,\sigma^{d-1}\}.$$ Since $\sigma$ is a cycle, there exists $1\leq x\leq n$ such that $${\rm supp}(\sigma)=\{x,\ldots,\sigma^{d-1}(x)\}.$$ For $1\leq i\leq d$, write $a_i=\sigma^{i-1}(x)$. I want to prove that $$\sigma=\prod_{j=1}^{d-1}(a_j\, a_{j+1}).$$ For $y\not\in{\rm supp}(\sigma)$, this equality holds trivially. It is also easy to show it for $y=a_i$ when $1\leq i\leq d-1$. I am having difficulty with case $y=a_d$. Any suggestions?

Answer 1

$\;\;\;\;$ Assume $\sigma$ is a cycle. It appears this is assumed or inferred above even though it is not necessarily the case that a permutation of order $d$ consists of only one cauchy cycle.

$\;\;\;\;$ Note $(a_1a_2)(a_2a_3)=(a_1a_2a_3)$. Using this as our basis proceed by induction and assume $\prod_{j=1}^{n-1}(a_ja_{j+1})=(a_1a_2\dots a_n)$ for all $n<d$ so that we have $$\prod_{j=1}^{d-1}(a_ja_{j+1})=\Bigl(\prod_{j=1}^{d-2}(a_ja_{j+1})\Bigr)(a_{d-1}a_d)=(a_1a_2\dots a_{d-1})(a_{d-1}a_d)=(a_1a_2\dots a_{d})=\sigma$$