$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?
 A: 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.
A: 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)
