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

Question

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?

Answer 1

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.