Number of $5$-Sylows of a simple group of order $660$.

Question

Let $G$ be a simple group of order $660$. I am trying to find $n_5$ - the number of Sylow $5$-subgroups of $G$.
I have easily proved that $n_5 \in \{1, 6, 11, 66\}$ using Sylow Theorems. Besides, I have concluded that if cannot be $1$ or $6$, but I don't know how to exclude the other option (so the result will follow).

Answer 1

$G$ must have $12$ Sylow $11$-subgroups. Consider its action by conjugation on this set of $12$ subgroups.

The normalizer of a Sylow $11$-subgroup has order $55$ and an element $g$ of order $5$ must consist of two $5$-cycles.

To see this, suppose that the $11$-cycle is $h=(2,3,4,5,6,7,8,9,10,11,12)$, with fixed point $1$. Then $\langle h \rangle$ is normalized by the $g$, and so $g$ most also fix $1$ and at least one other point, say $2$

If $g^{-1}hg=h$, then $gh$ has order $55$, which is clearly impossible. So $g^{-1}hg = h^i$ for some $i$ with $2 \le i \le 10$ (and in fact we must have $i^5 \equiv 1 \bmod 11$). Now we can see from the conjugation rule for permutations that $g$ must consist of two $5$-cycles. For example, if $i=4$, then $h^4=(2,6,10,3,7,11,4,8,12,5,9)$ and $g = (3,6,7,11,5)(4,10,12,9,8)$.

Alternatively, you could argue that if $g$ fixes another point, say $k$, then, since $h^{k-1}$ maps $2$ to $k$, so $g$ must centralize $h^{k-1}$, which also generates $\langle h \rangle$, so $g$ centralizes $h$, which is impossible.

Now $G$ is doubly transitive of order $660 = 12 \times 11 \times 5$, so $\langle g \rangle$ is the stabilizer of the two points $1$ and $2$. The normalizer of $\langle g \rangle$ must stabilize the set of fixed points of $g$, which is $\{1,2\}$, and so this normalizer has order $5$ or $10$. Hence it must be $10$, and there are $66$ Sylow $5$-subgroups.

Note that there is a group with all of these properties, namely ${\rm PSL}(2,11)$, and we are looking at its action on the projective line.

It would also be possible to rule out the case $|{\rm Syl}_5(G)| = 11$ with a similar argument, by considering the induced action on $11$ points and getting a contradiction. Interestingly, ${\rm PSL}(2,11)$ does act on $11$ points, also doubly transitively, but its point stabilizer is $A_5$ rather than the normalizer of a Sylow $5$-subgroup.