Four generators of $S(9)$ - A smart way of showing that this generates the entire group? I have four 4-cycles, given by: $(1452),(2563),(4785),(5896)$.  I know that the group generated by these guys are $S(9)$ by asking mathematica for the order of the permutation group generated by these four 4-cycles, which came out to be 9!
I am looking for an elegant way to show this statement, but I can't come up with anything. We tried to show directly that we can get a 2-cycle and a 9-cycle without success.
The motivation for the problem is as follows:
9 squares are arranged in a 3 by 3 grid. I will refer to this grid as the "big square".  
You have some kind of a picture drawn in the big square.
The individual squares are scrambled in some weird manner.
Is it possible to get the original picture back, using only the operation given by rotating four squares with the the center of the rotation on the vertex of the central square?
So basically:
\begin{pmatrix}
1 & 2 & 3 \\\
4 & 5 & 6 \\\
7 & 8 & 9
\end{pmatrix}
can become
\begin{pmatrix}
2 & 5 & 3 \\\
1 & 4 & 6 \\\
7 & 8 & 9
\end{pmatrix}
which corresponds to the cycle (1452).
 A: Here's a more "theoretic" approach:
Let $a=(1452)$, $b=(2563)$, $c=(4785)$, and $d=(5896)$.  Now note that $b$ and $c$ act transitively on $\lbrace 2,3,4,5,6,7,8\rbrace$ (you can show this directly or note that $bc$ is a $7$-cycle).  Thus the subgroup $\langle b,c\rangle$ acts transitively - and thus primitively - on $7$ points.  It contains a $3$-cycle $[b,c]=(4,6,5)$, and thus by a theorem of Jordan is isomorphic to $S_7$; that is, it is the full symmetry group on $\lbrace 2,3,4,5,6,7,8\rbrace$.  
From here it is not hard to see $\langle b,c,d\rangle$ is isomorphic to $S_8$ on $\lbrace 2,3,4,5,6,7,8,9\rbrace$, and a similar result for $\langle a,b,c\rangle$.  Together then, they must all generate $S_9$.
A: Apparently, If I denote $w=(1452)$ ,$x=(2563)$ and $y=(4785)$ then $(12) = x^y x^{-1} w^{-1}$. Other permutations of two adjacent elements on the border can of course be found from this one, by rotating or reflecting the square.
This observation allows one to "solve the square". First put the number 5 in the middle, then use these involutions to solve the border.
Please let me know if you find this too cryptic.
A: You can view this rotation puzzle as a graph with 4 faces:
1-2-3
| | |
4-5-6
| | |
7-8-9

Since at least one of the faces has an even number of pieces (and the graph is not one of the special cases), all permutations are possible. If all faces have an odd number of pieces, then the number of possible permutations would be exactly half and correspond to the alternating group.
More details of this theory can be found here (along with what are the 2 special cases).
A: Let $a=(1452)$, $b=(2563)$, $c=(4785)$ and $d=(5896)$. Notice that $G=\langle a,b,c,d\rangle$ is transitive, so all point stabilizers are conjugate. $H=\langle b,c,d\rangle$ is transitive on $\{2,3,4,5,6,7,8,9\}$, all two-point stabilizers are conjugate. Now $J=\langle b,c\rangle$ is transitive on $\{2,3,4,5,6,7,8\}$, so all three-point stabilizers are conjugate. Next, $K=\langle a,b\rangle$ is transitive on $\{1,2,3,4,5,6\}$, so all four-point stabilizers are conjugate. Next, $a^d=(1482)$, so $L=\langle a,a^d\rangle$ is transitive on $\{1,2,4,5,8\}$ and centralizes four points. Of course, $M=\langle a\rangle$ is transitive on $\{1,2,4,5\}$. As was mentioned above, $[b,c]$ is a $3$-cycle, and so $|G|$ is a multiple of $9.8.7.6.5.4.3$, and is either $A_9$ or $S_9$. Finally, $G$ has an odd permutation, so $G=S_9$.
