Let $G$ be a simple group such that $|G|=p^2q r$ ($p$ and $q$ are distinct prime numbers) then how prove $|G|=60$? Let $G$ be a simple group such that $|G|=p^2q r$ ($p$ and $q$ are distinct prime numbers, $r$ some positive integer) then how to prove $G$ and $A_5$ are isomorphic or ($|G|=60$)?
Thanks in advance 

Here is a solution which makes repeated use of the Burnside Transfer Theorem (or BTT for short), which we can take as:

Theorem: If a Sylow subgroup is central in its normalizer, then the group $G$ is not simple.

Because groups of order $l$ or $l^2$ (here $l$ is a prime) are always abelian, we can use the following form of BTT for this question:

The normalizer of a Sylow subgroup in $G$ cannot be abelian.

OK, now let's get down to business.


*

*First, let's count the number of Sylow $p$-subgroups.  There cannot be $qr$ of them, since that contradicts BTT.  There also obviously cannot be $1$ of them.  So there are, let's say, $r$ of them. [At this point, $q$ and $r$ are indistinguishable.] Then if $P$ is a Sylow $p$-subgroup, $N_G(P)$ has index $r$ in $G$. The conjugation action of $G$ on the Sylow $p$-subgroups embeds $G$ into $A_r$, and thus $p^2q$ divides $(r-1)!$. In other words, $r$ is the largest prime dividing $G$.

*Now let's show $p<q$.  If not, then $q$ would be the smallest prime dividing $|G|$.  We would thus have, for a Sylow $q$-subgroup $Q$, $|N_G(Q)/C_G(Q)|$ dividing both $q-1$ and $p^2r$; this means $|N_G(Q)/C_G(Q)|=1$, or in other words, $Q$ is central in $N_G(Q)$.  Once again, a contradiction of BTT; so $p<q$.

*Now suppose $Q$ is a Sylow $q$-subgroup in $N_G(P)$.  If $Q$ was normal in $N_G(P)$, we would have $N_G(P)=P\times Q$, an abelian group, contradicting BTT. So instead, there are $p^2$ Sylow $q$-subgroups in $N_G(P)$, and thus $q$ divides $p^2-1$.  Since $p<q$, this means $q=p+1$.  This is only possible if $p=2$, $q=3$.

*Finally, let's count the Sylow $r$-subgroups.  By BTT, there cannot be $p^2q$ of them.  Since $r$ is bigger than $p$ and $q$, there cannot be $p$ or $q$ of them.  Thus, there are $pq$ Sylow $r$-subgroups. This means $r$ divides $pq-1=2\cdot3-1=5$, and so $r=5$.
We have then shown that $|G|=2^2\cdot3\cdot5=60$.
