I think most students who first learn about (finite) groups, eventually learn about the possibility of classifying certain finite groups, and even showing certain finite groups of a given order can't be simple (I'm pretty sure every beginning algebra text has some exercises like this). Up to order 1000, I think there is one that is considered by far the most difficult: 720.


Does anyone know of a proof that there are no simple groups of order 720, which **avoids** showing, via contradiction, such a group would be $M\_{10}$? [To clarify the **avoids** part, the proof sketched by Derek Holt [here](http://sci.tech-archive.net/Archive/sci.math/2006-12/msg07456.html), while very nice, would not qualify.]


I should also disqualify the inevitable reference to Burnside's article on this very topic, which I am fairly confident is flawed (or, at the very least, incomplete).



No simple groups of order 720?
You can only really ask for a proof that avoids a particular fact or construction, if that fact or construction is difficult enough or distinct enough from the thing that you're proving. By that principle, the comments imply that it would be enough to show that Derek Holt's subgroup $\langle a,b,c,e \rangle$ has index 2 in the candidate group $\langle a,b,c,d,e \rangle$. In fact Derek's proof is good: It is easy to show that $\langle a,b,c,e \rangle$ is isomorphic to $L\_2(9) = \text{PSL}(2,\mathbb{F}\_9)$, which plainly has order 360. ($L\_2(9)$ is also isomorphic to $A\_6$, but this fact is not needed.) Once you know that, you can also quickly construct $M\_{10}$ as well, since you can show that the extra generator $d$ normalizes $L\_2(9)$, and that $d^2 \in L\_2(9)$.


In that Usenet posting, Derek gave these expressions:
$$\begin{matrix} a &=& (2\; 3\; 4)(4\; 6\; 7)(8\; 9\; 10) \\
b &=& (2\;5\;8)(3\;6\;9)(4\;7\;10) \\
c &=& (3\;5\;4\;8)(6\;7\;10\;9) \\
e &=& (1\;2)(5\;8)(6\;7)(9\;10) \end{matrix}.$$
Recall that $\mathbb{F}\_9 = (\mathbb{Z}/3)[i]$. You can define a bijection
$$\alpha:\mathbb{F}\_9 \cup \{\infty\} \to \{1,2,\ldots,10\}$$
by the formula
$$\alpha(x+iy) = 2+x+3y \qquad \alpha(\infty) = 1,$$
using the gauche embedding $\mathbb{Z}/3 = \{0,1,2\} \subseteq \mathbb{Z}$.
Then it is easy to check these expressions (using more normal arithmetic in $\mathbb{F}\_9$):
$$a(z) = z+1 \qquad b(z) = z+i \qquad c(z) = iz \qquad e(z) = 1/z.$$
So, that's $L\_2(9)$.


