Simple group of order $p^2 q r$, where $p, q, r$ are distinct primes, is isomorphic to $A_5$ As stated, I need to prove that, up to isomorphism, the only simple group of order $p^2 q r$, where $p, q, r$ are distinct primes, is $A_5$ (the alternating group of degree 5).
Now I know the following: if $G$ is a simple group and $|G| = 60$, then $G$ is isomorphic to $A_5$. However, I don't even know how to begin the proof that $|G| = 60$, or anything similar.
 A: Here is a sketch solution. I can give more detail, but it depends on which results you are familiar with.
Let $G$ be simple of order $p^2qr$.
By Burnside's Transfer Theorem, $p$ must be the smallest of the three primes because, if for example $q$ was smallest then $G$ would have a normal $q$-complement so would not be simple.
Let $P \in {\rm Syl}_p(G)$. Then $P$ must be properly contained in its normalizer, since otherwise there would be a normal $p$-complement by Burnside's Transfer Theorem. So we can assume that $|N_G(P)| = p^2q$. Let $Q \in {\rm Syl}_q(N_G(P))$, so $|Q|=q$.
We cannot have $Q < C_G(P)$ or again there would be a normal $p$-complement, so $|Q|$ must divide $|{\rm Aut}(P)|$, which is equal to $p(p-1)$ if $P$ is cyclic and $p(p^2-1)$ if it is $C_p \times C_p$.
But since $q$ is prime and $p<q$, the only possibility is $p=2$, $q=3$, and $PQ \cong A_4$.
But now $|{\rm Syl}_r(G)|$ must divide $|G:R|=12$ and also be congruent to  $1$ mod $r$. We cannot have $|{\rm Syl}_r(G)|=12$, or $G$ would have a normal $r$-complement, so the only possibility is $|{\rm Syl}_r(G)| = 6$ and $r=5$.
A: The groups of order $p^2qr$ for distinct primes $p,q,r$ have been classified here by Oliver G. Glenn in $1906$.
With the exception of the group of order $2^2\cdot 3\cdot 5$,  simply isomorphic with the icosahedron-group $A_5$, all groups of order $p^2qr$ are solvable.
A: Since you asked for an "elementary" proof, I'll try to write one that doesn't use Burnside's transfer theorem.
Let $G$ be a simple group of order $p^2qr$, WLOG $q<r$. We call $n_p,n_q,n_r$ the numbers of the respective Sylow subgroups, which must be all greater than $1$: then, $n_p \in \{q,r,qr\}$ and $n_r \in \{p,p^2,pq,p^2q\}$. We also call $m \in \{1,p\}$ the largest cardinality of the intersection of two distinct Sylow $p$-subgroups.
First of all, we observe that $G$ has no subgroups of index $q$, otherwise it would be isomorphic to a subgroup of $A_q$, which has no elements of order $r$: in particular, $n_p \neq q$.
Case 1: $n_r \in \{p,p^2\}$.
This forces $p>r$: indeed, in both cases we must have $r|p^2-1$, i.e. $r|p\pm1$ and then $r \le p+1$, and the inequality $p>r$ only fails when $(p,r)=(2,3)$, which leaves no room for $q$. Then, since $n_p \ge p+1$, the only possibility for $n_p$ is $qr$.
Case 1.1: $n_r \in \{p,p^2\}$, $n_p=qr$ and $m=p$.
Let $P$ and $Q$ be two Sylow $p$-subgroups such that $|P \cap Q|=p$: then, $N_G(P \cap Q)$ has order a multiple of $p^2$ and it has more than one Sylow $p$-subgroup, so that it is forced to be the whole $G$, and $P \cap Q \triangleleft G$.
Case 1.2: $n_r \in \{p,p^2\}$, $n_p=qr$ and $m=1$.
There are exactly $qr(p^2-1)$ elements of order $p$ or $p^2$; among the remaining $qr$ elements, there are exactly $n_q(q-1)$ of order $q$ and $n_r(r-1)$ of order $r$, so that
$1+n_q(q-1)+n_r(r-1) \le qr \Rightarrow n_r(r-1) \le qr-1-n_q(q-1) < q(r-1) \Rightarrow n_r<q$,
which contradicts $n_r \ge r+1$.
Case 2: $n_r=p^2q$.
Arguing as in Case 1.2, there are exactly $p^2q$ elements of order $\neq r$, and if $m=1$ we would have $1+n_q(q-1)+n_p(p^2-1) \le p^2q \Rightarrow n_p<q$, contradiction. Therefore $m=p$, and taking $P,Q$ as in Case 1.1, $N_G(P \cap Q)$ has order $p^2q$ or $p^2r$: in the former case it is exactly the set of elements of order $\neq r$, so that it is a characteristic subgroup (and then normal), while in the latter it has index $q$ in $G$, but we had already ruled out this case at the beginning.
Case 3: $n_r=pq$ and $n_p=r$.
Write $r=kp+1$ and $pq=hr+1$, with $h,k \ge 1$. We have $pq=h(kp+1)+1=hkp+h+1$, so that $h=tp-1$ for some $t \ge 1$ and $q=hk+t$. Since $q<r$, we also have $(tp-1)k+t<kp+1 \Rightarrow (tp-p-1)k<1-t$, which is impossible if $t>1$, so that $t=1$ and $q=(p-1)k+1$. Finally, we have $k>1$ (otherwise $p=q$) and then $q>p$. Now, $n_p=r$ implies that any Sylow $p$-subgroup has a normalizer of order $p^2q$, so that the latter has a normal Sylow $q$-subgroup unless $(p,q)=(2,3)$ (which leads to $|G|=60$, in which case I take for granted that $G \cong A_5$), and then $n_q=r$. However, we should have $q|r-1=kp \Rightarrow q|k \Rightarrow q|\text{gcd}(k,q)=1$, impossible.
Case 4: $n_r=pq$ and $n_p=qr$.
If $m=1$, we get the same contradiction as in Case 1.2, therefore $m=p$, and similarly to Case 2, taking $P$ and $Q$ as usual, we must have $|N_G(P \cap Q)|=p^2q$. Since $N_G(P \cap Q)$ has more than one Sylow $p$-subgroup, it has a normal Sylow $q$-subgroup (by the classification of groups of order $p^2q$), so that the only possibility for $n_q$ is $r$, and then $r \equiv 1 (\bmod q)$. Also, since any Sylow $r$-subgroup $R$ has a normalizer of order $pr$, a subgroup $H<R$ of order $p$ cannot be normal in $R$, otherwise the index of $N_G(H)$ in $G$ would be $1$ or $q$, leading to a contradiction in both cases. Therefore $r \equiv 1 (\bmod p)$, and then $r \equiv 1 (\bmod pq)$, which contradicts $pq \equiv 1 (\bmod r)$.
