Nonexistence of a simple group of order 576

Let $G$ be simple of order $576 = 64 \times 9$. The number of Sylow $2$-subgroups is 1,3 or 9 by Sylow's Theorem, but $G$ simple implies that it cannot be 1 or 3, so it must be 9. Then the conjugation action of $G$ on the set $\Omega$ of Sylow $2$-subgroups of $G$ induces an embedding $G \to A_9$.

Let $S \in {\rm Syl_2(G)}$. Now a Sylow $2$-sugroup of $A_9$ has order $64 = |S|$ so in its action on $\Omega$, $S$ is a Sylow $2$-subgroup of ${\rm Alt}(\Omega)$ and, in particular $S$ is transitive on $\Omega \setminus \{ S \}$, and hence the action of $G$ on $\Omega$ is 2-transitive.

A 2-point stabilizer in this action is a subgroup $T$ of $S$ of order $8$. Since $T$ fixes more than one point of $\Omega$, it is contained in more than one Sylow $2$-subgroup of $G$, so its normalizer $N_G(T)$ in $G$ has more than one Sylow $2$-subgroup.

Since $N_G(T) \ne G$, $N_G(T)$ cannot have 9 Sylow $2$-subgroups, so it must have 3. Since $T$ is properly contained in $N_S(T)$, we have $|N_S(T) \ge 16$, so $|N_G(T)| = 2^k \times 3$ where $4 \le k \le 6$, and hence $|G:N_G(T)| = 2^{6-k} \times 3 = 3, 6$, or $12$. From its order $G$ is not contained in$A_6$, so we must have $k=4$ and $|N_G(T)|=48$.

To proceed further we can use the fact that $T$ is isomorphic to a Sylow $2$-subgroup of $A_6$. We can identify $T$ with $\langle (4,5)(6,7), (6,7)(8,9), (4,8)(7,9)\rangle$, which is dihedral of order $8$ (the group $D_8$) and has centre $U = \langle (4,5)(8,9) \rangle$ of order $2$.

Now the normalizer of $U$ in $S_8$ contains $D_8 \times D_8$ of order $64$, so its normalizer in $A_8$ and hence in $S$ has order (at least) 32.

Since $N_G(U)$ contains $N_G(T)$ and $|N_S(U)| \ge 32$, we have $|N_G(U)| \ge 32 \times 3$, so $|G: N_G(U)| \le 6$, which is impossible because $G$ does not embed in $A_6$.

By Sylow, the number of Sylow $3$-subgroups of $G$ is either $16$ or $64$, as $G$ is simple. Let a Sylow $3$-subgroup $S$ of $G$ act on the set $Syl_3(G)$ of Sylow $3$-subgroups by conjugation. If $g\in S$ fixes a point $T\in Syl_3(G)$, then $\langle g, T\rangle$ is a $3$-group, so $g\in T$. So $S$ has a unique orbit $\{S\}$ of length $1$. The other orbits have length $3$ or $9$.

If all orbits have length $9$, then $S$ intersects all other Sylow $3$-subgroups $T$ trivially ($S\cap T = 1$), and as the Sylow $3$-subgroups are conjugated, any two Sylow $3$-subgroups have trivial intersection. The number of Sylow $3$-subgroups being $1 \bmod 9$ is $64$, and the union of all Sylow $3$-subgroups consists of $8\cdot 64 = 512$ elements. The set of the remaining $64$ elements of $G$ is then the (unique) Sylow $2$-subgroup of $G$, contradicting the simplicity of $G$.

Hence $S$ has an orbit of length $3$, and for $T$ in this orbit $U := S\cap T$ has order $3$. As the Sylow $3$-subgroups are abelian (being of order prime squared), the centralizer $C := C_G(U)$ contains both Sylow $3$-subgroups $S$ and $T$, so by Sylow, the number of Sylow $3$-subgroups of $C$ is a multiple of $4$. As $G$ is simple, the order of $C$ is either $36$ or $72$.

If the order of $C$ is $36$, then $C/U$ is isomorphic to $A_4$ (the only group of order $12$ without normal Sylow $3$-subgroup). A Sylow $2$-subgroup $V$ of $C$ centralizes $U$ and its image $\bmod U$ is normal in $C/U$, so $V$ is normal in $C$. Its normalizer $N:=N_G(V)$ contains $C$, but also elements in $S\setminus V$, as a proper subgroup of a (finite) $p$-group is a proper subgroups of its normalizer. As $G$ is normal, $N$ has order $72$.

In any case, there exists a subgroup $H$ of $G$ of order $72$. As $G$ is simple, its normalizer $N_G(H)$ equals $H$, i.e., $H$ is self-normalizing and $H$ has $8$ conjugates, which are also self-normalizing. Let $H$ act on the set $\Omega := \{H^g\mid g\in G\}$ of its conjugates by conjugation. If $h\in H$ fixes a point $H^g\in\Omega$, then $h\in H\cap H^g$, as $H^g$ is self-normalizing. So $H$ has a unique orbit $\{H\}$ of length $1$, and the other orbits have lengths $2, 3$ or $4$.

If $H^g$ lies in an orbit of length $2$, then $H\cap H^g$ has index $2$ in both $H$ and $H^g$, and therefore has to be normal in both. As $U$ is the maximal normal $3$-subgroup of $H\cap H^g$, it is characteristic in a normal subgroup of $H^g$, and hence normal in $H^g$. So the normalizer of $U$ contains $H$ and $H^g$ contradicting the simplicity of $G$.

As $G$ is simple, its action by conjugation on $\Omega$ is faithful, and $G$ embeds into $A_8$ such that its subgroup $H$ has a fixed point and two orbits of length $3$ and $4$. As $H$ has order $72$, $H = (S_4\times S_3)\cap A_8$, where $S_4$ and $S_3$ are the symmetric groups on $\{1,2,3,4\}$ rsp. $\{5,6,7\}$. So $H$ contains the $3$-cycles $(123), (124), (134), (234)$ and $(567)$.

As the subgroup generated by all $3$-cycles of $G$ is normal in $G$, and as $G$ is simple, $G$ has to contain another $3$-cycle $(abc)$ (not contained in $H$). The subgroup generated by two non-disjoint $3$-cycles acts transitively on the union of their orbits. As $5$ does not divide the order of $G$, $(abc)$ cannot intersect any $3$-cycle of $H$ in exactly one point, so $\{a,b,c\}\subset\{5,6,7,8\}$. But then $(567)$ and $(abc)$ generate $A_4$ on the points $\{5,6,7,8\}$, so $G$ contains $A_4\times A_4$, a subgroup of order $144$ contradicting its simplicity.