I'm trying to prove the theorem that states that
Every simple group of order $60$ is isomorphic to $A_5$.
I'm trying to do it by assuming by contradiction that it doesn't hold and reach a contradiction.
So I was able to prove the following:
- $G$ doesn't have any subgroup with index less than $5.$
- If $G$ has a subgroup with index $5$ then it must be isomorphic to $A_5$.
- The number of $5$-sylow groups is $6.$
- The number of $2$-sylow groups is $15$ (if it's $5$ then $G$ is isomorphic to $A_5$ by previous claim - contradiction).
Now, in order to reach the contradiction, I want to show that the aforementioned conclusions result in $G$ having more than $60$ elements, but I need to show that each $P,Q$ $2$-sylow groups have trivial intersection, which I wasn't able to do (of-course if the intersection is trivial then we get the contradiction). Any help would be helpful.
You are almost there. If any two distinct $2$-Sylows intersect trivially, then you are done. If $P$ and $Q$ are different $2$-Sylows which intersect non-trivially, we have $|P \cap Q| = 2$.
Observe that, since $|P| = |Q| = 4$, we obtain that $P$ and $Q$ are Abelian groups. Hence $P$ and $Q$ are both contained in the centralizer $C$ of $P \cap Q$,
$$C = \{g \in G \ | \ gh = hg \ \ \forall h \in P \cap Q\}.$$
Hence, $C$ contains the subgroup $H$ generated by $P$ and $Q$. Note that $H \neq G$, since else the group $P \cap Q$ of order $2$ would be contained in the center of $G$, a contradiction to the simplicity.
So $4$ divides the order of $H$, and $H \neq G$. Moreover, since $P$ and $Q$ are distinct, $|H| > 4$. Hence the index $(G:H)$ is either $3$ or $5$.