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.
On
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 $r=5$.
On
The proof is aimed not to use the Burnside's Transfer Theorem.
Let $G$ be a simple group of order $|G|=p^2 qr$ and let $P∈Syl_p (G)$, $Q∈Syl_q (G)$, $R∈Syl_r (G)$.
Assume (w.l.o.g) that $q<r$. Our goal will be to show that $|G|=60$.
(1) $G$ has no subgroup of order $p^2 r$.
To see this, suppose the contrary: $|H|=p^2 r$. Then $|G∶H|=(p^2 qr)/(p^2 r)=q$. Let $G$ act by left multiplication on the set of all distinct left cosets of $H$ in $G$. This action affords a homomorphism $G→S_q$, so by the 1st Isomorphism Theorem, since $G$ is simple, $G≅L≤S_q$ and by the Lagrange’s Theorem $|G| \ | \ |S_q |$ or $p^2 qr \ |\ q!$ ⇔ $p^2 r \ | \ (q-1)!$, but this is not possible since $q<r$ are primes. We’ve reached a contradiction.
(2) Choose $P_1,P_2∈Syl_p (G)$ over all pairs of distinct Sylow $p$-subgroups of G with the order $|P_1 ⋂ P_2 |$ being maximal.
Suppose $|P_1 ⋂ P_2 |>1$. Then $|P_1 ⋂ P_2 |=p$. Hence $|P_1 ∶P_1 ⋂ P_2 |=|P_2 ∶P_1 ⋂ P_2 |=p^2/p=p$, therefore $P_1 ⋂ P_2⊲P_1$, $P_1 ⋂ P_2⊲P_2$ (as subgroups of index $p$ — the smallest prime dividing $p^2$) and so $P_1,P_2≤N_G (P_1 ⋂ P_2 )$, therefore, since $|P_1 |=|P_2 |=p^2$, we have that $p^2 \ | \ |N_G (P_1 ⋂ P_2 )|$ and $|N_G (P_1 ⋂ P_2 )|>p^2$ (otherwise there wouldn’t be enough elements in $N_G (P_1 ⋂ P_2 )$ to contain both $P_1,P_2$). Then by the Lagrange’s Theorem and (1) it is only possible that $|N_G (P_1 ⋂ P_2 )|=p^2 q$. Let’s denote $N_G (P_1 ⋂ P_2 )=M$. Consider the set $Syl_p (M)$. Because $|M|=p^2 q$, it follows that every Sylow $p$-subgroup of $M$ is also a Sylow $p$-subgroup of $G$, that is, $Syl_p (M)⊆Syl_p (G)$. Hence $G$ acts transitively by conjugation on $Syl_p (M)$. The kernel $K$ of this action which is also a kernel of the afforded homomorphism $G→S_{|Syl_p (M)}|$ is $K=\bigcap_{i=1}^{|Syl_p (M)|} N_G (P_i)$. Note that $P_1 ⋂ P_2$ lies in the intersection of all the Sylow $p$-subgroups of $M$. To see this, take a Sylow $p$-subgroup of $M$ containing $P_1 ⋂ P_2$, conjugate it to another Sylow $p$-subgroup of $M$ with some $x∈M$ and you get that $x(P_1 ⋂ P_2 ) x^{-1}=P_1 ⋂ P_2$ is contained in this another Sylow $p$-subgroup of $M$; since all the Sylow $p$-subgroups of $M$ are conjugate in $M$, it follows that $P_1 ⋂ P_2$ is contained in all Sylow $p$-subgroups $P_i$ of $M$. Therefore, $P_1 ⋂ P_2≤\bigcap_{i=1}^{|Syl_p (M)|} P_i$. Since $P_i≤N_G (P_i )$ for all $P_i∈Syl_p (M)$, it follows that $P_1 ⋂ P_2≤\bigcap_{i=1}^{|Syl_p (M)|} N_G (P_i)=K$, i.e., the kernel $K$ is nontrivial. Since $K⊲G$ and $G$ is simple, this is a contradiction.
Thus, $|P_1 ⋂ P_2 |=1$. Then $P_i ⋂ P_j=1$ for every pair $P_i,P_j ∈Syl_p (G)$. Hence the number of elements of $G$ contained in its Sylow $p$-subgroups is $(p^2-1) n_p$.
Consider $N_G (P)$. Obviously $P≤N_G (P)$, hence the Lagrange’s Theorem and (1) leave only two possibilities: $|N_G (P)|∈\{p^2,p^2 q\}$.
(3) Suppose $|N_G (P)|=p^2$. Then $n_p=|G∶N_G (P)|=(p^2 qr)/p^2 =qr$. Hence the number of elements of $G$ contained in its Sylow $p$-subgroups is $(p^2-1) n_p=(p^2-1)qr$. For $n_q$ and $n_r$ we have $n_q \ | \ p^2 r$ and $n_r \ | \ p^2 q$. Obviously $n_q≠1$ and $n_p≠1$ since $G$ is simple. Suppose $n_q=r$, $n_r=q$. Then the total number $A$ of elements of $G$ contained in its Sylow subgroups is $A=(p^2-1)qr+(q-1)r+(r-1)q=p^2 qr-qr+qr-r+qr-q=p^2 qr+qr-(q+r)>p^2 qr=|G|$ Note that actually $n_r≠q$ since $q<r$ and so $n_r=q≢1 (mod \ r)$. Hence to satisfy the congruence condition there must be $n_r>q$ and because $n_r \ | \ p^2 q$ the minimal possible $n_r$ is either $n_r=pq$ (if $p>q$) or $n_r=p^2$ (if $p<q$). But $A>|G|$ even for the lesser number of $n_r$, hence there must be $n_q<r$. Because $n_q \ | \ p^2 r$ the minimal possible $n_q$ is $n_q=p$ (assuming that $p<r$; if this assumption is not true, we reach a contradiction immediately).
If $n_r=pq$, then $$A=(p^2-1)qr+(q-1)p+(r-1)pq=p^2 qr-qr+pq-p+pqr-pq=p^2 qr+pqr-(qr+p)>p^2 qr=|G|$$
a contradiction.
If $n_r=p^2$, then $p<q$ as was noted (and hence $p<q<r$). But now we cannot have $n_q=p$ because $n_q=p≢1(mod \ q)$. Hence, since $n_q \ | \ p^2 r$ and $n_q<r$, the minimal possible $n_q$ is $n_q=p^2$. But since $n_r=p^2$, by Sylow’s Theorem, $n_r=p^2≡1(mod \ r)$, hence $p^2>r$ and so $n_q≠p^2$ because there must be $n_q<r$. So there is no $n_q<r$ satisfying $n_q \ | \ p^2 r$. We’ve reached a contradiction.
So, if $|N_G (P)|=p^2$, the total number $A$ of elements of $G$ contained in its Sylow subgroups exceeds $|G|$. So, $|N_G (P)|≠p^2$.
Thus, $|N_G (P)|=p^2 q$. Therefore, $n_p=|G∶N_G (P)|=(p^2 qr)/p^2 =r$. Also, because $P_i ⋂ P_j=1$ for every pair $P_i,P_j ∈Syl_p (G)$, it follows that $n_p≡1(mod \ p^2 )$ and hence $r≡1(mod \ p^2 )$.
(4) Because $|N_G (P)|=p^2 q$ there is $Q∈Syl_q (G)$ such that $Q∈Syl_q (N_G (P))$. Denote $N_G (P)=N$.
Suppose $Q⊲N_G (P)$. Then $N_G (Q)=N_G (P)=N$. Hence $n_p=n_q=|G∶N|=(p^2 qr)/(p^2 q)=r$. In (3) it was shown that $r≡1(mod \ p^2 )$ and hence $p^2 \ | \ r-1$. Also since $n_q≡1(mod \ q)$ we have $r≡1(mod \ q)$ and hence $q \ | \ r-1$. Therefore $p^2 q \ | \ r-1$ and so $r=1+kp^2 q$, where $k∈Z^+$. On the other hand, $n_r≡1(mod \ r)$, $n_r \ | \ p^2 q$. So the maximal possible $n_r$ is $n_r=p^2 q$. But even if $k=1$ and so $r=1+p^2 q$ we have that $n_r=p^2 q<(1+p^2 q)=r$, hence $n_r≢1(mod \ r)$, a contradiction. So, $Q⋪N_G (P)$.
Thus, $N_G (Q)≠N_G (P)$.
(5) Consider $N$ of order $|N|=p^2 q$. Since $P⊲N_G (P)=N$, it follows that $n_p (N)=1$. Since $Q$ is not normal in $N$, as it was just shown in (4), it follows that $n_q (N)>1$. For the number of Sylow $q$-subgroups of $N$ we have: $n_q (N)≡1(mod \ q)$, $n_q (N) \ | \ p^2$. Then it is either: $n_q (N)=p$ (and so $p≡1(mod \ q)$) or $n_q (N)=p^2$.
Either way, though, $p^2≡1(mod \ q)$.
⦁ Our subgoal from now on will be to establish the inequality relation between $p$ and $q$.
(6) Consider $|N_G (Q)|$. In (4) it was shown that $N_G (Q)≠N_G (P)$, hence $|N_G (Q)|≠p^2 q$. Denote $N_G (Q)=H$. Then, since $Q≤N_G (Q)=H$, by the Lagrange’s Theorem $q \ | \ |H|$ and therefore $|H|∈\{q,pq,qr,pqr\}$.
⦁ Suppose $|H|=q$. Then $n_q=|G∶N|=(p^2 qr)/q=p^2 r$. So we have $p^2 r≡1(mod \ q)$ and since $p^2≡1(mod \ q)$ (see (5)), it follows that $r≡1(mod \ q)$. On the other hand $r≡1(mod \ p^2)$ (see (3)). These conditions for $r$ taken together lead to the exactly the same contradiction as in (4). So, $|H|≠q$.
⦁ Suppose $|H|=qr$. Then by Sylow’s Theorem, there is $R∈Syl_r (H)$ with $|R|=r$ and so $R∈Syl_r (G)$. Then, since $Q⊲H$ and $Q ⋂ R=1$, $H≅Q⋊R$. Since $Q⊲H$, $H$ acts on $Q$ by conjugation. This action affords a homomorphism $H→Aut(Q)$ with the kernel $C_H (Q)$, hence by the 1st Isomorphism Theorem, $H⁄C_H (Q)≅L≤Aut(Q)$. Since $Q≅Z_q$, it follows that $Aut(Q)≅Z_{q-1}$. By the Lagrange’s Theorem, $|H⁄C_H (Q) | \ | \ |Aut(Q)|$ or $qr/|C_H (Q)| \ | \ (q-1)$. Since $Q$ is abelian, $Q≤C_H (Q)$. Because $r>q$, $r∤(q-1)$, what forces $|C_H (Q)|=qr=|H|$ and so $Q≤Z(H)$. Then $H≅Q×R$, therefore $R⊲H$ and so $H≤N_G (R)$. Then it is either $|N_G (R)|=qr$ or $|N_G (R)|=pqr$ and hence $n_r=|G∶N_G (R)|=(p^2 qr)/qr=p^2$ or $n_r=|G∶N_G (R)|=(p^2 qr)/pqr=p$ correspondingly. Both cases though are not possible since there must be $n_r≡1(mod \ r)$ but $r≡1(mod \ p^2)$ (see (3)), hence $p^2<r$ and so $p^2≢1(mod \ r)$ and $p≢1(mod \ r)$. We’ve reached a contradiction. So, $|H|≠qr$.
⦁ Suppose $|H|=pqr$. As in the previous case, there is $R∈Syl_r (H)$ with $|R|=r$ and so $R∈Syl_r (G)$. Since $Q⊲H$, it follows that $QR<H$ and since $Q ⋂ R=1$, its order is $|QR|=qr$. Considering the action of $QR$ on $Q$ by conjugation exactly analogously to the previous case gives that $QR≅Q×R$, therefore $R⊲QR$, so $QR≤N_G (R)$, hence either $|N_G (R)|=qr$ or $|N_G (R)|=pqr$, what leads to exactly the same contradiction as in the previous case. So, $|H|≠pqr$.
Thus, $|N_G (Q)|=pq$.
(7) Suppose $p>q$.
Thus, $p<q$.
⦁ So the inequality relation between $p,q,r$ is $p<q<r$.
(8) Going back to (5), we now have $|N|=p^2 q$ with $p<q$, therefore $p≢1(mod \ q)$. Because $n_q (N) \ | \ p^2$, $n_q (N)>1$ and $p≢1(mod \ q)$, it can only be that $n_q (N)=p^2$. Then, as in (7), we have $q \ | \ (p-1)(p+1)$, but now $p<q$. If $p$ is an odd prime, i.e. $p>2$, then $(p-1)(p+1)$ is even. Because $q>p$, $q$ is also an odd prime, hence $q \ ∤ \ (p-1)(p+1)$, a contradiction. So it can only be that $p=2$. Now $q \ | \ (2-1)(2+1) ⇔ q \ | \ 3$, therefore it can only be that $q=3$.
Thus, $p=2,q=3$.
(9) Now $|G|=p^2 qr=2^2⋅3⋅r=12r$. Then for the number of Sylow $r$-subgroups of G we have: $n_r≡1(mod \ r)$, $n_r \ | \ 12$. Since $G$ is simple, $n_r≠1$, hence $n_r∈{2,3,4,6,12}$. Note that $n_r≠2$, because $r≠1$. Also $n_r≠3$, because this would imply $r=2=p$, but $r≠p$. Also $n_r≠4$, because this would imply $r=3=q$, but $r≠q$. So it there are only two possibilities: $n_r=6$ and so $r=5$, or $n_r=12$ and so $r=11$.
On the other hand, in (4) it was shown that $r≡1(mod \ p^2)$, therefore $r≡1(mod \ 4)$. Since $11≡3≢1(mod \ 4)$, it can only be that $r=5$, which does satisfy this condition.
Thus, $r=5$.
(10) Finally, we have $p=2, q=3, r=5$, therefore $|G|=p^2 qr=2^2⋅3⋅5=60$. Since $G$ is a simple group of order $60$, it is isomorphic to $A_5$.
On
References to Dummit and Foote's Abstract Algebra, 3rd edition, 2004.
If $p^{2}qr$ is odd, $G$ is solvable ((Feit-Thompson) Theorem~6.1.11(3)) so there is a proper normal subgroup, a contradiction. It must be that $p^{2}qr$ is even and $q=2$ or $r=2$ implies a normal subgroup of index $2$ (Exercise~4.2.13), again a contradiction. Then $p=2$ so $\left|G\right|=4qr$ and $q<r$ implies $n_{r}$ is either $2q$ or $4q$. If $n_{r}=4q$ there is the minimal (the minimal index is $r$) element count $4q\left(r-1\right)+r\left(q-1\right)+\left(r\cdot2+1\right)+1>4qr$, a contradiction. It must be that $n_{r}=2q$ and $2r>2q$ so $n_{r}=r+1=2q$. Since $n_{q}$ must be at least $2q+1>r$, $n_{q}$ is either $2r$ or $4r$. Observe that $4r=8q-4\equiv 1\bmod 5$ and $r=2\cdot 5-1=9$ is not prime, a contradiction. It must be that $n_{q}=2r=4q-2\equiv 1\bmod 3$ so $q=3$ and $r=5$. Thus the result.
On
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 $pq \equiv 1 (\bmod r)$.
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.