# Four generators of $S(9)$ - A smart way of showing that this generates the entire group?

For the first time, I'm going to post a group theory question.

I have four 4-cycles, given by: $(1452),(2563),(4785),(5896)$. I know that the group generated by these guys are $S(9)$ by brute computation. (Basically, I asked mathematica for the order of the permutation group generated by these guys, and it told me that it is 9!)

I'm pretty sure there is a much more elegant way to show this statement, but unfortunately I'm a complete dunce when it comes to anything algebra.

One thing me and my friend tried to do fruitlessly was to show directly that we can get a 2-cycle and a 9-cycle. Unfortunately, we weren't clever enough to find the right combination of these groups to come up with one.

The motivation for the problem is a bit complicated to describe it in words, but I'll try.

You have 9 squares arranged in a 3 by 3 grid, which I will refer to 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}

etc.

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It's almost like a 2-dimensional Rubik's cube. Maybe you can solve it layer by layer, like it's 3D-brother. –  Myself Jan 6 '12 at 23:04
@Myself: This was implemented as a combinatorial puzzle game named Rotation in earlier cellphone models of my former employer. This exact question made for a nice exercise in a freshman abstract algebra course during those years, when you could count on several students being familiar with the game. –  Jyrki Lahtonen Jan 7 '12 at 7:59

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 $S_9$.

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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.

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I assume that $x^y$ denotes conjugation of $x$ by $y$, i.e. $x^y = yxy^{-1}$? –  Alex Becker Jan 6 '12 at 23:45
@AlexBecker: Other way around: $x^y=y^{-1}xy$. –  user641 Jan 6 '12 at 23:51
One wants to have $(x^y)^z = x^{yz}$. You do see ${}^yx = yxy^{-1}$, though. –  Dylan Moreland Jan 7 '12 at 0:01
But that is a problem with TeX, not with the exponent! –  Mariano Suárez-Alvarez Jan 7 '12 at 2:21
All of these considerations are subject to the the extra ambiguity that some people multiply permutations from left to right and others from right to left, so there are four possible interpretations of $x^y$. –  Derek Holt Jan 7 '12 at 11:51