# $A_6 \simeq \mathrm{PSL}_2(\mathbb{F}_9)$ only simple group of order 360

How can one show, without the use of character theory, that $A_6 \simeq \mathrm{PSL}_2(\mathbb{F}_9)$ is, up to isomorphism, the only simple group of order 360?

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You can find a proof in Suzuki's Group Theory. The proof is not easy, however. Perhaps if I have time later tonight I will write up a summary. The general idea is to assume I have a simple group $G$ of order 360, and show there is a homomorphism to $S_{10}$, whose image is generated by explicit generators. Thus two simple groups $G$ and $H$, both of order 360, are isomorphic to this subgroup of $S_{10}$, hence to each other. – user641 Sep 19 '11 at 21:21
Thanks Steve D, I found an outline of a proof as an exercise in Suzuki's Group Theory II, page 136, exercise 3 (it doesn't seem easy). No mention of $S_10$ though. Is your reference different? – Nathan Portland Sep 19 '11 at 21:50
You can find a variety of approaches on this sci.math thread: sci.tech-archive.net/Archive/sci.math/2005-03/5687.html – Alon Amit Sep 19 '11 at 22:26

Here is a proof that any simple group $G$ of order $360$ is isomorphic to a specific subgroup of $A_{10}$ (and hence there can be only one [insert Highlander pun]).

Let $G$ be a simple group of order $360$, and let's ask how many Sylow 3-subgroups there can be. A quick check shows the possibilities are : $1,4,10,40$. $1$ and $4$ can easily be ruled out, and $40$ is ruled out because then the Sylow 3-group is self-normalizing. But a group of order $9$ is abelian, and hence by Burnside's Transfer Theorem, $G$ can't be simple. [You can avoid BTT by showing a subgroup of order 3 has normalizer of size at least 72, and getting a contradiction that way.]

Thus, there are 10 Sylow 3-groups; let's pick one and call it $P$. Then the conjugation action of $G$ on these ten Sylows gives an embedding of $G$ into $A_{10}$. So let's assume for the rest of this post that $G$ actually lives inside $A_{10}$. Note that $N_G(P)$ has order $36$, and is a point stabilizer in $G$ (let's say the point stabilizer of $10$).

Now if $P$ was cyclic, then elements of $N_G(P)$ would basically be elements of $A_9$ normalizing a 9-cycle. 9-cycles in $A_9$ are self-centralizing however (count conjugates), and thus $N_G(P)/P$ would embed in $\operatorname{Aut}(P)$; this is a contradiction because the former group has order $4$ and the latter order $6$.

So $P$ is non-cyclic of order $9$, generated by two elements $a$ and $b$ of order $3$. Each of these is a product of 3 3-cycles in $\{1,2,\ldots,9\}$. We can assume that $$a = (1,2,3)(4,5,6)(7,8,9);$$ $$b = (1,4,7)(2,5,8)(3,6,9).$$

This is because we can renumber the points so $a$ looks has the required form, and take an appropriate element of the form $a^ib^j$ to give us the form for $b$.

Now consider the point-stabilizer of $1$ in $N_G(P)$. since $P$ acts transitively on $\{1,2,\ldots,9\}$, the orbit-stabilizer theorem shows this point stabilizer has order $4$. It is thus a Sylow 2-subgroup $Q$ of $N_G(P)$, and $N_G(P)=PQ$. Again, the centralizer of $P$ in $A_9$ is a 3-group, and hence $Q\cong N_G(P)/P$ embeds in $\operatorname{Aut}(P)$; this implies $Q$ is cyclic of order $4$. Let $Q$ be generated by a permuation $c$ of order $4$. We know $c$ fixes both $10$ and $1$, so for $c$ to be an even permutation it must be the product of two 4-cycles.

Note also that $c$ is almost completely determined by where it sends $2$; this is because every element of $P$ is determined by where it sends $1$ [if $x,y\in P$ both sent $1$ to the same point, then $xy^{-1}$ would fix $1$, hence be in $Q$, so we would have $xy^{-1}=1$.]. So for example, if $c$ sends $2$ to $3$, then it must send $a$ to $a^2$, and there are only two ways to do this (basically it sends $4$ to either $6$ or $9$). One can easily check that permutations sending $2$ to one of $\{2,3,5,6,8,9\}$ don't have order $4$, and thus $c$ sends $2$ to either $4$ or $7$. One will simply give the inverse of the other, and so we can assume that $$c=(2,4,3,7)(5,6,9,8).$$

It's important to note that no non-trivial power of $c$ fixes any point of $\{2,3,4,5,6,7,8,9\}$; that is, no element of $G$ fixes more than $2$ points.

Now let $S$ be a Sylow 2-subgroup of $G$ containing $Q$; note that this means there's an element $d$ such that $S=\langle c,d\rangle$. It is an easy exercise to show the Sylow 2-subgroup of a simple group can't be cyclic, so that $d$ has order either $4$ or $2$. Also since $d\notin N_G(P)$, it cannot fix the point $10$. Now suppose that $d$ sends the point $1$ to the point $p\notin\{1,10\}$; then $c^d$ would not fix $1$, and yet $c^d\in Q$. Similarly, if $d$ sent $10$ to a point $q\notin\{1,10\}$, $c^d$ would not fix $10$. Thus $d$ must permute $1$ and $10$ amongst each other, and since it can't fix $10$, it contains the 2-cycle $(1,10)$ [it's a 2-cycle because $d^2\in Q$].

But if $d$ had order $4$, then - ignoring that $(1,10)$ cycle - it would be an odd permutation on $8$ points fixing $c$. Since it can fix at most $2$ points, it would be a 4-cycle $m$ multiplied by a 2-cycle $n$. Now if it sent $2$ to one of $\{5,6,8,9\}$, it could fix no points at all; thus $m$ must normalize one of $(2,4,3,7)$ and $(5,6,9,8)$. But 4-cycles are only normalized by their own powers (at least restricting to other 4-cycles on the same 4 points), and thus $m$ centralizes its 4-cycle. However, $n$, a 2-cycle, must then centralize its 4-cycle, which is impossible. Thus $d$ can't have order $4$.

So $d$ must be order $2$, and in fact, every element of $S-Q$ has order $2$ [so $S$ is dihedral]. The same analysis above shows - ignoring once again the $(1,10)$ cycle - $d$ is the product of 3 2-cycles. Thus it is the product of $m$ and $n$, except this time $m$ looks like $(\cdot,\cdot)(\cdot,\cdot)$ and $n$ is a 2-cycle. Again, $m$ must invert one of the two 4-cycles making up $c$, and $n$ inverts the other. So the 8 possibilities for $d$ [ignoring $(1,10)$] are a product of one of $(2,3)$,$(4,7)$, $(2,4)(3,7)$, and $(2,7)(3,4)$, together with one of $(5,9)$, $(6,8)$, $(5,6)(9,8)$, and $(5,8)(6,9)$. [There are not 16 possibilities, because for example (2,3)(5,9) is not of the required $mn$ form.]

Now if $d=(1,10)(2,3)(5,6)(8,9)$, then it's routine to check that $cd$, $c^2d$, and $c^3d$ give three other acceptable products from the above 8. If we set $\hat{d}=(1,10)(4,7)(5,8)(6,9)$, then we can check that the other four are given by $\hat{d}$, $c\hat{d}$, $c^2\hat{d}$, and $c^3\hat{d}$. However, a direct computation shows $ab\hat{d}$ has order $21$. Thus, up to factors of $c$ (which we can safely ignore), we have $$d = (1,10)(2,3)(5,6)(8,9).$$

Now we are done: the subgroup $\langle a,b,c,d\rangle\le G$ has order at least $72$; but $G$ is simple, and so we must have $\langle a,b,c,d\rangle=G$.

EDIT - Here is the argument to avoid Burnside's Transfer Theorem:

Assume $G$ has $40$ Sylow 3-groups. Since $40\not\equiv 1\pmod{9}$, there are two Sylow 3-groups $A$ and $B$ such that $D=A\cap B$ in non-trivial (and hence order 3). Now the normalizer $N_G(D)$ has more than one Sylow 3-group, and thus has order at least $36$. If $|N_G(D)|>36$, we would have $|N_G(D)|\ge72$, and that gives a subgroup of index $5$ in $G$, which implies (via the right coset action) that $G$ embeds in $A_5$, contradiction. Thus we can assume $N_G(D)$ has order 36, and since it does not have a normal Sylow 3-group (remember they were self-normalizing), it must have a normal Sylow 2-group (for this implication see the proof here). Thus this subgroup $T$ of order $4$ is normalized by a Sylow 3-group, and since "normalizers grow" in p-groups, its normalizer also has order divisible by $8$. That is, $|N_G(T)|\ge72$, and once again we have a contradiction. Thus there cannot be $40$ Sylow 3-groups.

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Of course, I tried to keep this as low-tech as possible; assuming more group theory the proof does get somewhat shorter. – user641 Sep 20 '11 at 0:26
That's excellent, but you assumed that all nontrivial elements of $P$ have no fixed points. If not, then an element of order 3 lies in two distinct Sylow 3-subgroups, and then its centralizer has order at least 36, and you get essentailly the same contradiction to the one in your final paragraph. Do you know some easier way of showing that? – Derek Holt Sep 20 '11 at 19:26
@DerekHolt: Yes, of course, you're right! However, I believe I only assumed that $P$ could be generated by elements with no fixed points. And I think this can be seen as follows (of course your argument works just as well): $P$ is generated by $a$ and $b$, elements of order 3; let $\operatorname{Fix}(a)$ be the fixed points of $a$, similarly for $\operatorname{Fix}(b)$. Since $a$ and $b$ commute, $b$ fixes $\operatorname{Fix}(a)$ setwise, and vice versa. Also, $\operatorname{Fix}(a)\cap\operatorname{Fix}(b)=\emptyset$. – user641 Sep 21 '11 at 1:56
... if we define $\operatorname{Mov}(a)$ and $\operatorname{Mov}(b)$ to be the non-fixed points of $a$ and $b$, then I think it's not too hard to show $\langle a,b\rangle$ is generated by fixed-point-free elements. The key is that the fixed points for $ab$ come from $\operatorname{Mov}(a)\cap\operatorname{Mov}(b)$. For example, if $a$ is a 3-cycle, then $b$ fixes no points. Now if $ab$ fixes any points, its fixed set must be $\operatorname{Mov}(a)$. But then it's easy to see $a(ab)=a^2b$ has no fixed points. – user641 Sep 21 '11 at 1:59
No never mind, that doesn't work - your argument is best! – user641 Sep 21 '11 at 2:06