# No simple groups of order 720?

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, 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).

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Why do you want to avoid discussing M_10? – Noah Snyder Oct 13 '10 at 3:45
Presumably, you can avoid mentioning $M_{10}$ in Holt's argument by showing that the subgroup $\langle a,b,c,e \rangle$ has index two in the whole group and is therefore normal. – S. Carnahan Oct 13 '10 at 4:46
@Noah: I am just curious if such a proof exists. It is quite possible the answer to my question is "no known proof avoiding M_10 is known". Of course, showing the group is M_10 is quite computational (as in, no one has probably ever done it by hand), but that is a minor point. – Steve D Oct 13 '10 at 4:55
@Scott: I am curious what you mean by "show". If it a matter of enumerating all possible elements in <a,b,c,e>, then it is really just the same as showing <a,b,c,d,e> is M_10. But if there was some more elegant argument that <a,b,c,e> has index 2, I would be very interested in hearing it. – Steve D Oct 13 '10 at 4:58
I believe that Turing was asing whether a computer could be used to determine if there is a simple group of order 720. – user38185 Aug 5 '13 at 1:29

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 $L_2(9)$.

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While I completely disagree with your first sentence, the rest is very nice indeed! – Steve D Oct 14 '10 at 22:39
I agree with Steve D. (Counterexample: in Serre's Course in Arithmetic, he makes a point of proving Dirichlet's theorem on primes in arithmetic progressions without using the fundamental theorem on finite abelian groups. Of course the latter is much easier than the former and he could just include a proof of it if he wanted, but he prefers to show that it is not necessary.) – Pete L. Clark Oct 15 '10 at 1:41
Sure, easier than I can accept...but avoiding something outright easy in a proof? The problem is that you can be like Molière's character, speaking prose without knowing it. A proof that uses elementary fact X and a proof that seems to avoid X could at second glance be equivalent proofs. To take your example, the classification of finite abelian groups is hard enough (and distinct enough) that you could plausibly be far away from proving it after accepting Serre's proof. But okay, I'll edit the answer a bit. – Greg Kuperberg Oct 15 '10 at 4:26
I don't know what reciting prose has to do with it, but I think my request was a reasonable one. For example, you can prove there is only one simple group of order 360 in a similar manner to Holt's proof: the details are in Suzuki's group theory books. But you can also prove it using some character theory, as in Isaacs's book; the argument is more involved, but it's an essentially different argument. I don't see what the problem with asking for more than one argument is. – Steve D Oct 15 '10 at 17:41
There is nothing wrong with asking for more than one argument. But you have to accept that it's a fuzzy question, especially if a complete argument exists that is reasonably simple. – Greg Kuperberg Oct 15 '10 at 17:57

How about if we follow Derek Holt's beautiful argument until we establish the following two facts?

1) $G$ has 10 Sylow 3-subgroups.

2) Let $P_3$ be a Sylow 3-subgroup of $G$. In the action of $G$ on $Syl_3(G)$, every nonidentity element of $P_3$ has cycle type (3,3,3,1).

Since $G$ is simple, the action on $Syl_3(G)$ embeds $G$ in $A_{10}$. In $A_{10}$, no element of type (3,3,3,1) commutes with an element of order two. Thus if $Q$ is a Sylow 2-subgroup of $N_G(P_3)$, no nonidentity element of $Q$ centralizes any nonidentity element of $P_3$. Since $|Q|=8$, it follows that all eight nonidentity elements of $P_3$ are conjugate in $N_G(P_3)$. Therefore, $G$ contains exactly one conjugacy class of elements of order three. Let $g_3$ be an element of this class. No element of order 5 in $A_{10}$ commutes with $g_3$, and it follows now that $|C_G(g_3)|=9$, so $G$ has 80 elements of order 3.

Since 5 does not divide $|N_G(P_3)|$, we see that every element of order 5 in $G$ has cycle type (5,5) in the action on $Syl_3(G)$. Using Sylow's Theorem (and the fact that $G$ is not $S_6$), we see that G has either 16 or 36 Sylow 5-subgroups. Let $P_5$ be a Sylow 5-subgroup of $G$. By Burnside transfer, we cannot have $|N_G(P_5)|=45$. Therefore, $|N_G(P_5)|=20$. No element of order two in $A_{10}$ commutes with an element of cycle type (5,5). Therefore, $N_G(P_5)$ must induce all of $Aut(P_5)$, so all nonidentity elements of $P_5$ are conjugate in $N_G(P_5)$ and therefore, $G$ has one conjugacy class of elements of order 5. There are 144 such elements. Let $g_5$ be one such element.

Now let us consider the character table of $G$. By the arguments above, each of $g_3$ and $g_5$ are conjugate to all of their nonidentity powers, and it follows that all irreducible characters of G take integer values on $g_3$ and $g_5$. Moreover, these integers have absolute value at most two, as can be seen using the orthonormality conditions on the character table and the sizes of the given conjugacy classes.

Now $N_G(P_3)$ has one more conjugacy class than $Q$ (as defined above). It follows from this that the irreducible characters of $N_G(P_3)$ are those with kernel containing $P_3$ and one more, call it $\chi$. Using orthonormality conditions on the character table, we get $\chi(1)=8$, $\chi(g_3)=-1$ and $\chi(q)=0$ for all $q$ not of order 1 or 3. Induce $\chi$ up to $G$ to get a character $\psi$ that takes the value 80 on 1, -1 on the class of $g_3$ and 0 elsewhere. For any irreducible character $\eta$, the inner product $\langle\psi,\eta\rangle$ is

$$\frac{80(\eta(1)-\eta(g_3))}{720}$$.

It follows that $\eta(1)-\eta(g_3)$ is divisible by 9.

Now $N_G(P_5)=Z_5.Z_4$ has a character $\alpha$ such that $\alpha(1)=4$, $\alpha(g_5)=-1$ and $\alpha(q)=0$ if $q$ does not have order 1 or 5. Induce $\alpha$ up to $G$ to get $\beta$. Arguing as we did with $\psi$, we see that for every irreducible character $\eta$, $\eta(1)-\eta(g_5)$ is divisible by 5.

Now using basic facts about irreducible characters, we see that for any irreducible character $\eta$ of G, the triple $(\eta(1),\eta(g_3),\eta(g_5))$ is one of

$$(1,1,1),(8,-1,-2),(9,0,-1),(10,1,0),(16,-2,1),(18,0,-2) or (20,2,0)$$.

Any class function $\eta$ satisfying $\eta(1)=18$ and $\eta(g_5)=-2$ has norm larger than 1. If $\eta$ is a class function of norm 1 satisfying $\eta(1)=20$ and $\eta(g_3)=2$ then $\eta(q)=0$ for all q not of order 1 or 3. But then the inner product of $\eta$ and the trivial character is positive. Similarly, if $\eta$ has norm 1 and $(\eta(1),\eta(g_3),\eta(g_5))=(8,-1,-2)$ then $\eta$ is zero on all remaining classes and the inner product of $\eta$ and the trivial character is negative. We see now that all nontrivial irreducible characters of $G$ have degree 9, 10, or 16. We get a contradiction when trying to add up squares of these degrees to get 719.

This proof has a clear disadvantage of greater length when compared with the argument of Holt, as clarified beautifully by Greg Kuperberg. However, it has the advantage of alerting us to the important fact that, given a purported simple group $G$, it can be profitable to consider large subgroups of $G$ whose characters we understand, in particular those that are Frobenius groups.

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I'm most likely going to accept Greg Kuperberg's great answer, but I must say this is a beautiful argument! – Steve D Oct 15 '10 at 18:31