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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.

GBA
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1 Answers1

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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 $5$.

Marktmeister
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  • In this context, the index of H in G divides 15, so then we can say that it is either 3 or 5, but why does it divide 15? – sbws Nov 23 '22 at 11:19
  • @luis_orion The order of $H$ is divisible by $4$, since it contains $P$ as a subgroup. Hence the Index of $H$ in $G$ divides 15. – Marktmeister Nov 24 '22 at 05:17