While learning about alternating tensors, I did some searching and found that the parity of the number of inversions for a permutation $\sigma$ is the quals to the parity of the number of transpositions that $\sigma$ can be decomposed to(regardless of how we choose to decompose it).
This is already sufficient for my use, but originally I was wondering if we can always decompose the permutation to exactly the number of inversions. I guess now the oddness question is already resolved, if the number of transpositions is indeed bounded by the number of inversions, we can just make dummy moves i.e. transpose back and forth to get the conclusion.
I'm sorry that unfortunately, I do not have much experience with abstract algebra. I have thought about some sorting algorithms such as selection sort, but I cannot find a concrete counterexample or proof of why this is false or true.
Yes, given a permutation $\sigma$ of $[n]$ for which the number of inversions is $i$ we can express $\sigma$ as a product of $i$ transpositions.
This can be proven by induction immediately.
The base case, when $\sigma$ is the identity is trivially true.
If $\sigma$ is not identity, there must be a number $k$ such that $\sigma(k)>\sigma(k+1)$. Let $$\tau=(\sigma(k),\sigma(k+1))\circ\sigma,$$ i.e., $\tau$ is the same as $\sigma$ except the values of $k$ and $k+1$ are switched. The inversion number for $\tau$ is one less than that for $\sigma$ as $\tau$ misses the inversion on $(k,k+1)$. Using the induction hypothesis for $\tau$, we see the proposition is true for $\sigma$ since $$ (\sigma(k),\sigma(k+1))\circ\tau=\sigma.$$