(MOTIVATIONAL CONTEXT) As a part of my solution to part (b) of exercise 3 in section 5 of chapter 11 of Linear Algebra (3e) by Serge Lang, I am desirous to prove that a finite sum of vectors in a vector space is invariant with respect to permutations of vector terms (due to the commutativity and associativity of vector addition in a vector space). And as a part of this project, I would like to prove that every transposition of $J_n$ (for some natural number $n$) can be written as a product of adjacent transpositions — since the commutativity of vector addition, in and of itself, only states that a finite sum of vectors (in a vector space) is unaffected by adjacent transpositions of its vector terms.
But beyond this particular textbook exercise, I would like to direct your attention to a more general (and perhaps somewhat pedantic) issue of "pedantic index error" in mathematical proofs; for example, in equation (1) of my proof provided below, I suspect that, if we have $j=i+1$, then the expression $\theta_n^i\cdots\theta^{j-2}_n\theta_n^{j-1}\cdots\theta_n^{i}$ might be misconstrued as $\theta^i_n\theta^{i-1}_n\theta^i_n\theta^i_n$ although $\theta^i_n$ is the intended meaning; and you can probably find other such examples of what I refer to as "pedantic index error" in the proof provided below... Though this "pedantic index error" may seem as though it is too pedantic and technical to warrant our attention, in the interest of greater mathematical rigour and precision, I believe that it is beneficial to consider and discuss this issue, about which I could not find much on the Internet and on this site, at least once.
(REQUEST) Can you please (i) check if my proof below is valid, with your focus on the acceptability of "pedantic index errors" in this proof? And can you also present (ii) your opinion about the acceptability of these "pedantic index errors" in general, (iii) provide us with some reasons behind your thus presented opinion, and (iv) finally advise us as to how we can deal with these "pedantic index errors" in our mathematical proofs (if we have to at all)? Thank you very much in advance...
(DEFINITION) Let $n\in\mathbb{N}$. And let $m\in\mathbb{N}$ such that $m<n$. Then let $\theta_n^m\in\mathrm{Perm}(J_n)$ such that we have $\theta_n^m(m)=m+1\space,$ $\theta_n^m(m+1)=m\space,$ and $\theta_n^m(k)=k$ if we have $k\ne m$ and $k\ne m+1$; in other words, $\theta^m_n$ denotes a permutation of $\{1\dots n\}$ which swaps $m$ and $m+1\space.$ And we shall refer to $\theta^m_n$ as the $m$-th adjacent transposition of $J_n.$
(NOTATION) For the sake of convenience, if $n\in\mathbb{N}$, then let $J_n$ denote the set $\{1\dots n\}$ within this post.
(LEMMA P.10.2) Let $n\in\mathbb{N}$ such that $n\ge2$; and let $\sigma$ be a transposition of $J_n$. Then $\sigma$ can be expressed as a product of adjacent transpositions of $J_n$.
(PROOF) Since $\sigma$ is a transposition of $J_n\space,$ let $i$ and $j$ be the two distinct elements of $J_n$ that $\sigma$ interchanges; and as $i$ and $j$ are distinct natural numbers, without the loss of generality, assume that $i>j$ ; then we contend that we have $$\begin{equation}\tag{1}\sigma=\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^i_n\space; \end{equation}$$ suppose that $x\in J_n$ such that $x<i<j$ or $i<j<x$. Then because we have $x\ne i$ and $x\ne j$, and because $\sigma$ is a transposition of $J_n$ which swaps $i$ and $j$, it follows that $\sigma(x)=x$. Because the product on the right hand side of equation (1) are purely comprised of $\theta^i_n\dots\theta^{j-1}_n$, and because each $\theta^k_n$ (where $k\in\mathbb{N}$ such that $i\le k \le j-1$) only interchanges $k$ and $k+1$ (and leaves everything else in $J_n$ unchanged), it is rather clear that only elements of $\{k\in\mathbb{N}\space|\space i\le k \le j\}$ might (or might not) be moved by the right hand side of equation (1), and that all other elements of $J_n$ remain fixed under the right hand side of equation (1) ; and therefore, as $x\in J_n$ such that $x<i<j$ or $i<j<x$, we have $$\begin{aligned} (\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^i_n)(x) &= x \\ &= \sigma(x). \end{aligned}$$ As $\sigma$ is a transposition of $J_n$ which interchanges $i$ and $j$, we clearly have $\sigma(i)=j$. And we have $$\begin{aligned} (\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^i_n)(i) &= (\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^{i+1}_n)(i+1) \\ &\space\space\vdots \\ &=(\theta^i_n\cdots\theta_n^{j-2})(j) \\ &= j \\ &=\sigma(i). \end{aligned}$$ As $\sigma$ is a transposition of $J_n$ which interchanges $i$ and $j$, we have $\sigma(j)=i$. And we have $$\begin{aligned} (\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^i_n)(j) &= (\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n)(j) \\ &=(\theta^i_n\cdots\theta_n^{j-2})(j-1) \\ &= (\theta^i_n\cdots\theta_n^{j-3})(j-2) \\ &\space\space\vdots \\ &=\theta^i_n(i+1) \\ &= i \\ &= \sigma(j). \end{aligned}$$ And if $x\in J_n$ such that $i<x<j$, then as $x\ne i$ and $x\ne j$, and as $\sigma$ is a transposition of $J_n$ which interchanges $i$ and $j$, we have $\sigma(x)=x$. And we have $$\begin{aligned} (\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^i_n)(x) &= (\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^{i+1}_n)(x) \\ &\space\space\vdots\\ &=(\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^{x-1}_n)(x) \\ &=(\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^{x}_n)(x-1) \\ &=(\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^{x+1}_n)(x-1) \\ &\space\space\vdots\\ &=(\theta^i_n\cdots\theta_n^{x-1})(x-1) \\ &=(\theta^i_n\cdots\theta_n^{x-2})(x) \\ &=(\theta^i_n\cdots\theta_n^{x-3})(x) \\ &\space\space\vdots\\ &=x \\ &=\sigma(x). \end{aligned}$$ And therefore, if we have $x\in J_n\space,$ then in any case, we have $\sigma(x)=(\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^i_n)(x)\space,$ meaning that we have $\sigma=\theta^i_n\cdots\theta_n^{j-2}\theta^{j-1}_n\cdots\theta^i_n\space.$
Q.E.D.