# $A_6 \simeq \mathrm{PSL}_2(\mathbb{F}_9)$ 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?

• 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, 2011 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? Sep 19, 2011 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 Sep 19, 2011 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.

• 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, 2011 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? Sep 20, 2011 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, 2011 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, 2011 at 1:59
• No never mind, that doesn't work - your argument is best!
– user641
Sep 21, 2011 at 2:06

Thanks to user641 for this concise proof. If I'm not wrong, it is a variant of Cole's proof (1893), which can be found here :

http://www.jstor.org/stable/2369516?seq=1#page_scan_tab_contents

Perhaps the average student can find some difficulties in the proof given by user641, so I take leave to write a second answer, only to make things more explicit.

"Then the conjugation action of $G$ on these ten Sylows gives an embedding of $G$ into $A_{10}$."

Perhaps it should be stressed that $G$ embeds in $A_{10}$ not only as a group, but as an operating group. I don't know if this fact is mentioned in many textbooks, so I will give a proof. If this proof is too complicate, please say it.

Lemma 1. Let $G$ be a group, let $N$ be a normal subgroup of $G$ and $S$ a simple subgroup of $G$. Then $S$ is contained in $N$ or isomorphic to a subgroup of $G/N$.

Proof. $N \cap S$ is normal in $S$. Since $S$ is simple, $N \cap S$ is thus equal to $S$ or to $1$. In the first case, $S$ is contained in $N$ and the statement is true. In the second case, $S$ is isomorphic to $SN/N$ (second isomorphism theorem), which is a subgroup of $G/N$, so the statement is also true in the second case.

Lemma 2. Let $E$ a finite set. If $G$ is a simple subgroup of the symmetric group $S_{E}$, if $\vert G \vert > 2$, then $G$ is contained in the alternating group $A_{E}$.

Proof. $A_{E}$ is normal in $S_{E}$. Thus, in view of lemma 1, $G$ is contained in $A_{E}$ or isomorphic to a subgroup of $S_{E}/A_{E}$. But the latter is impossible, since $S_{E}/A_{E}$ has order $\leq 2$ and $\vert G \vert > 2$ by hypothesis.

Definition. If $\cdot : G \times X \rightarrow X$ denotes an action of a group $G$ on a set $X$, if $\star$ denotes an action of a group $H$ on a set $Y$, let us define an isomorphism from the first action onto the second action as an ordered pair $(f, \sigma)$, where $f$ is a bijection from $X$ onto $Y$ and $\sigma$ a group isomorphism from $G$ onto $H$ such that, for each $x$ in $X$ and for each $g$ in $G$,

$f(g \cdot x) = (\sigma (g) ) \star f(x)$.

Lemma 3. Let $\cdot$ denote an action of a group $G$ on a set $X$, let $\star$ denote an action of a group $H$ on a set $Y$, let $(f, \sigma)$ be an isomorphism from the first action onto the second action. Then an element $g$ of $G$ fixes a point $x$ of $X$ (for the action $\cdot$) if and only $\sigma(g)$ fixes $f(x)$ (for the action $\star$).

Proof. Easy.

Definition. Let $\cdot$ denote an action of a group $G$ on a set $X$, let $\star$ denote an action of a group $H$ on a set $Y$. If there exists an isomorphism from $\cdot$ onto $\star$, we say that these actions are isomorphic. (Some authors say "equivalent". Aschbacher, Finite Group Theory, 2d edition, p. 9, says "quasiequivalent".)

Lemma 4. Let $G$ be a simple group, let $\cdot$ be a nontrivial action of $G$ on a set $E$, let $\varphi$ denote the homomorphism from $G$ to $S_{E}$ corresponding to this action. Then $\varphi$ is injective (in other words, the action is truthful) and the action $\cdot$ of $G$ on $E$ is isomorphic to the natural action of the permutation group $\varphi (G)$. If $E$ is finite and $\vert G \vert > 2$, then $\varphi$ takes its values in $A_{E}$.

Proof. For the injectivity of $\varphi$, note that the kernel of $\varphi$ is a normal subgroup of the simple group $G$ and this kernel is not the whole $G$, since $G$ acts nontrivially. For an isomorphism from the action $\cdot$ onto the natural action of $\varphi (G)$, take the ordered pair $(f, \psi)$, where $f$ is the identity bijection from $E$ onto itself and where $\psi$ is the group isomorphism $G \rightarrow \varphi(G) : g \mapsto \varphi(g)$ from $G$ onto $\varphi (G)$. For the last statement, use lemma 2.

Lemma 5. Let $G$ be a finite nonabelian simple group. (This amounts to say : let $G$ be a finite simple group whose order is not a prime number.) Let $p$ be a prime factor of $\vert G \vert$. Let $E$ denote the set of all Sylow $p$-subgroups of $G$, let $n$ denote the number $\vert E \vert$ of Sylow $p$-subgroups of $G$. Then the action of $G$ on $E$ by conjugation is isomorphic to the natural operation of a subgroup of $A_{n}$.

Proof. Since $G$ is a finite nonabelian simple group, it has more than one Sylow $p$-subgroup, thus the (transitive) action of $G$ on $E$ is nontrivial. In view of the preceding lemmas, the action of $G$ on $E$ by conjugation is isomorphic to the natural operation of a subgroup of $A_{E}$. Now, if $X$ and $Y$ are equipotent finite sets, the natural action of a subgroup of $A_{X}$ is isomorphic to the natural action of a subgroup of $A_{Y}$.

Lemma 6. Let $G$ be a finite group, let $p$ be a prime number. If $P$ is a Sylow $p$-subgroup of $G$, if $g$ is an element of $G$ whose order is a power of $p$ and which normalizes $P$, then $g$ is in $P$. If $P$ and $Q$ are Sylow $p$-subgroups of $G$, if $P$ normalizes $Q$, then $P = Q$.

Proof. Classical. (Since $g$ normalizes $P$, the order of the subgroup $<P, g>$ generated by $P$ and $g$ is a power of $p$, thus $<P, g>$ is equal to $P$ by maximality of Sylow $p$-subgroups.)

Lemma 7. Let $G$ be a finite group, let $p$ be a prime number. Let $E$ denote the set of all Sylow $p$-subgroups of $G$. The action of $G$ by conjugation on $E$ has the following properties :

1° for each Sylow $p$-subgroup $P$ of the operating group, there is one and only one point in the set $E$ that is fixed by every element of $P$;

2° for every point in the set $E$, there is one and only one Sylow $p$-subgroup $P$ of the operating group such that every element of $P$ fixes this point;

3° if $P$ is a Sylow $p$-subgroup of the operating group, if $x$ denotes the only point in $E$ that is fixed by each element of $P$, then the stabilizer of $x$ in $G$ is $N_{G}(P)$;

4° if it is moreover assumed that two different Sylow $p$-subgroups of $G$ always intersect trivially, then every nontrivial $p$-element of the operating group (I mean by "$p$- element" an element whose order is a power of $p$) fixes one and only one point of $E$.

Proof. Use Lemma 6. (In the statement of Lemma 7, I made a distinction between a Sylow $p$-subgroup of $G$ as a point of $E$ and as a subgroup of $G$, in order to forget what is not essential.)

Definition (nonstandard). Let us define a Cole group as a simple subgroup $G$ of order 360 of $A_{10}$ with the following properties :

1° for each Sylow $3$-subgroup $P$ of $G$, there is one and only one point in $\{1, \ldots , 10 \}$ that is fixed (for the natural operation) by every element of P$; 2° for every point in the set$\{1, \ldots , 10 \}$, there is one and only one Sylow$3$-subgroup$P$of$G$such that every element of$P$fixes this point; 3° if$P$is a Sylow$3$-subgroup of$G$, if$x$denotes the only point in$E$that is fixed by each element of$P$, then the stabilizer of$x$in$G$is$N_{G}(P)$; 4° every nontrivial$3$-element of the operating group (I mean by "$3$- element" an element whose order is a power of$3$) fixes one and only one point of$\{1, \ldots , 10 \}$. Lemma 8. Every simple group of order 360 is isomorphic to a Cole group. Proof. Use Lemmas 3, 5 and 7. (Recall that user641 has proved that a simple group$G$of order 360 has exactly$10$Sylow$3$-subgroups and that two distinct Sylow$3$-subgroups of$G$interset always trivially.) Now, I think that user641' statement : "Note that$N_G(P)$(...) is a point stabilizer in$G\$" should be clear for the average student. (Again, if this proof is too complicated, please say it.)

If nobody has objections, I will write other answers in order to make other arguments from the proof more explicit.

• No comments ? Is my work unnecessary ? Jun 2, 2017 at 5:43

There is a nice geometric proof: using only Sylow theorems one can show that G has 30 subgroups isomorphic to Z_2 x Z_2 which break into classes of 15 each. Denote by P one class and by L the other. Refer to elements of P as points and elements of L as lines. Say a "point", E, is on the "line", F, E and F intersect in a subgroup of order 2. This is a generalized quadrangle of order 2 which is easily shown to be unique and have isomorphic group S_6. In this way one gets an injective how of G into S_6 which has a unique subgroup of index 2 (since A_6) is simple.

A similar geometric proof can be used to prove a simple group of 168 is the automorphism of a Fano plane (projective plane of order 2)