$\quad$In the February edition of The Mathematical Association of America Monthly there is a article called "$\mathit{Rubik's\ on\ the\ Torus}$". Where they are dealing with solving problems involving the electronic puzzle game called the Rubik's Slide.
$\mathbf{\underline{NOTE:}}$ I've given all the details about this topic that I believe will help give you an idea of what I'm working with. I've also tried to make this as clean as possible, but If you believe I should somehow break this up into a smaller question or multiple questions tell me and I will do it.
Important: Permutation Multiplication Goes From Left to Right in This Article!
$\mathbf{\underline{Introduction:}}$ So we have: $$\text {Vertical Shift} = (147)(258)(369)$$$$\text{Horizontal Shift} = (123)(456)(789)$$$$\text {Clockwise rotation around center}=(12369874)$$Where my $3\times3$ Rubik's Slide looks like this: $$\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix}$$ $\mathbf{\underline{Lemma 1:}}$ The subgroup $<h,v>$ generated by $h$ and $v$ is abelian and of order $S_9$.
$\underline{Proof:}$ By computation we can see that $$hv=(159)(267)(348)=vh.$$
We also know that $$h^3=\varepsilon=v^3$$
I believe now I can make the assumption that $$\text{"If a subgroup is generated by two cyclic groups is commutative that the subgroups is also abelian."}$$ I think it's a theorem, I just can't find it.
Thus $$<h,v>=\{h^iv^j:i,j=0,1,2\}=S_9$$ $\mathbf{\underline{Lemma 2:}}$ $$c^3h=(14)(2739568) \ \text{and} \ (c^3h)^7=(14)$$
$\underline{Proof:}$ By Computation.
The one that I really stuck on is this one,
$\mathbf{\underline{Theorem 3:}}$ The group $<h,v,c>$ contains every transposition in $S_9$; hence $<h,v,c>=S_9$
The proof that is given in the article states that to transpose two squares, $x$ and $y$
1) Repeatedly apply $h$ and $v$ to move $x$ to position $5$ and to call this series of moves $\sigma$
2) Repeatedly apply $c$ to move $y$ to position $2$ and call this series of moves $\tau$
3) Now apply $h^{-1}$ to move $x$ to position $4$ and $y$ to position $1$.
4) Now swap $x$ and $y$ using Lemma 2.
5) Lastly apply $h\tau^{-1}\sigma^{-1}$
6) Thus the resulting product of moves, $\sigma\tau h^{-1}(c^3h)^7h\tau^{-1}\sigma^{-1}=\ \text{transposition} \ (x,y)$
$\mathbf{\underline{What\ I\ Need\ Help\ With:}}$
I completely believe it works and have proven multiple cases. The problem I'm having is how to prove that it works for all choices of $x$ and $y$.
$\quad$So far I have been looking into conjugacy classes, but I can't seem to find a way to use them to prove this formula.
$\quad$The best Idea I've come up with is to prove that I can transpose any two neighbors with each other without effecting any other square. The reason I'm doing this is because once I prove that I can do that with every neighbor ($12$ cases) I can prove the big transposition equation by saying that any transposition is just the product of transpositions. Therefore since I can transpose every neighbor I can transpose any two squares on the Rubik's slide. So far I've only figured out a couple of the cases.
$\quad$This is where the symmetry comes in. If I can find the vertical and horizontal symmetries of the Rubik's Slide I believe I can narrow down my cases to like $4$ instead of $12$.
If you have trouble arranging your understanding of the proof, I suggest splitting it into two parts.
Lemma, for any permutation group. For any $x,y,a,b$, if $\rho$ is a permutation such that $\rho(x)=a$ and $\rho(y)=b$, then $\rho^{-1}(a\,b)\rho=(x\,y)$ -- or, depending on whether you compose left-to-right or right-to-left, $\rho(a\,b)\rho^{-1}=(x\,y)$.
You can prove the lemma completely algebraically, by verifying that the two permutations it asserts are equal take the same elements to the same results.
Observation. For any two different positions $x$ and $y$ on the slide, there is a sequence of moves that take $x$ to position $4$ and $y$ to position $1$.
This is just step 1,2,3 in the proof sketch, steps you say you do trust.
Corollary. Any transposition is realizable in the slide group, by the Lemma with $a=4, b=1$ and $\rho$ being the sequence of moves from the observation.