Aluffi IV.2.25 (in the chapter on Sylow theorems) suggests the following exercise:
Assume $G$ is a simple group of order $60$.
- Use Sylow's theorems and simple numerology to prove that $G$ has either five or fifteen $2$-Sylow subgroups, accounting for fifteen elements of order $2$ or $4$.
- If there are fifteen $2$-Sylow subgroups, prove that there exists an element $g \in G$ of order $2$ contained in at least two of them. Prove that the centralizer of $g$ has index $5$.
Conclude that every simple group of order $60$ contains a subgroup of index $5$.
I'm stuck at the first step. Sure, I can see why there are $5$ or $15$ $2$-Sylow subgroups, but what's next?
Let's say there are fifteen subgroups. Each contains $3$ non-identity elements, so the union contains up to $45$ non-identity elements. It can't contain $45$ elements for sure — in fact, it can't contain more than $27$ elements (since there must be $6$ $5$-Sylow subgroups which all have a trivial intersection, so that results in $24$ non-identity elements, and similarly at least $4$ $3$-Sylow subgroups are responsible for $8$ non-identity elements), but how do I further bound that number?
Ok, after looking at this some more I guess I managed to prove the first point.
It's sufficient to note that there must be $10$ $3$-Sylow subgroups since, for a simple group $G$, $|G|$ divides $N_p!$ where $N_p$ is the number of $p$-Sylow groups for $p$ a divisor of $G$. Since $|G|$ has $5$ as its factor, this implies $N_3 \geq 5$, and further Sylow considerations show that it must be 10.
Then, repeating the same considerations shows that $3$-Sylow and $5$-Sylow groups combined take away $44$ non-identity elements, leaving at most $15$ to $2$-Sylow groups. It's then straightforward to show no other subgroups can bite any elements off this number.
But how do I then prove the second point — that the centralizer of $g$ must have index $5$? I managed to prove that such $g$ exists (by pigeon-hole principle basically), but I'm not sure about its index. I only managed to prove the centralizer itself has order more than $6$ (that is, at least $10$), but how do I refine that?
Best Answer
If there are $15$ $2$-Sylow subgroups, then you showed that not every pair has trivial intersection (otherwise there are too many $2$-elements). So take Sylow $2$-subgroups $H_1$, $H_2$ with intersection $\{1,g\}$. The centralizer $K=C_G(H_1\cap H_2)$ contains both $H_1$ and $H_2$ (as they are abelian), so $|K|=20$ or $12$ (since $4$ divides $|K|$ and $|K|\ge 6$). However $|G:K|$ can't be 3 since we would get an embedding of $G$ into $S_3$.
Note that this exercise gives you an embedding of $G$ in $S_5$, and since $G$ is simple, the embedding must land in $A_5$, and therefore $G\cong A_5$.