Let \(p<q<r\) be prime integers. Show that a group of order \(pqr\) cannot be simple.
Consider groups of orders \(2^2\cdot 3\cdot p\) where \(p\) has the values 5, 7, and 11. For each of those values of \(p\), either display a simple group of order \(2^2\cdot 3\cdot p\), or show that there cannot be a simple group of that order.
Solution
1. Let \(G\) have order \(pqr\) with \(p<q<r\) prime. Then \(G\) is not simple.
Suppose for contradiction that \(G\) is simple. Let \(n_r\) be the number of Sylow \(r\)-subgroups. The Sylow theorems give \[ n_r\equiv1\pmod r, \qquad n_r\mid pq. \] Since \(G\) is simple, \(n_r\ne1\). The only divisors of \(pq\) are \(1,p,q,pq\), and both \(p\) and \(q\) are strictly less than \(r\), so neither can be congruent to \(1\) modulo \(r\). Hence \[ n_r=pq. \] Distinct Sylow \(r\)-subgroups intersect trivially, so they contribute \[ pq(r-1) \] nonidentity elements of order \(r\).
Now let \(n_q\) be the number of Sylow \(q\)-subgroups. Again simplicity gives \(n_q>1\), while \[ n_q\equiv1\pmod q, \qquad n_q\mid pr. \] Since \(p<q\), we cannot have \(n_q=p\). Therefore \(n_q\ge r\), because the remaining nontrivial divisors of \(pr\) are \(r\) and \(pr\). Distinct Sylow \(q\)-subgroups intersect trivially, so there are at least \[ r(q-1) \] nonidentity elements of order \(q\).
The sets of nonidentity elements of order \(r\) and of order \(q\) are disjoint. Hence \(G\) would contain at least \[ pq(r-1)+r(q-1) \] nonidentity elements. But \[ pq(r-1)+r(q-1)-(pqr-1) =r(q-1)-pq+1. \] Since \(q>p\), we have \(q-1\ge p\), and since \(r>q\), \[ r(q-1)\ge rp>pq. \] Thus \[ pq(r-1)+r(q-1)>pqr-1, \] more nonidentity elements than \(G\) possesses. This contradiction proves that \(G\) is not simple.
2. There is a simple group of order \(2^2\cdot3\cdot5=60\).
The alternating group \(A_5\) has order \[ |A_5|=\frac{5!}{2}=60, \] and \(A_5\) is simple.
3. No group of order \(2^2\cdot3\cdot7=84\) is simple.
Let \(G\) have order \(84\). If \(n_7\) denotes the number of Sylow \(7\)-subgroups, then \[ n_7\equiv1\pmod7, \qquad n_7\mid12. \] Among the divisors of \(12\), only \(1\) is congruent to \(1\) modulo \(7\). Hence \[ n_7=1. \] Thus the Sylow \(7\)-subgroup is normal, so \(G\) is not simple.
4. No group of order \(2^2\cdot3\cdot11=132\) is simple.
Suppose for contradiction that \(G\) is simple of order \(132\). For the Sylow \(11\)-subgroups, \[ n_{11}\equiv1\pmod{11}, \qquad n_{11}\mid12. \] Thus \(n_{11}=1\) or \(12\). Simplicity excludes \(1\), so \[ n_{11}=12. \] These Sylow subgroups contribute \[ 12(11-1)=120 \] nonidentity elements of order \(11\).
Now \[ n_3\equiv1\pmod3, \qquad n_3\mid44. \] The possibilities are \(n_3=1,4,22\). Simplicity excludes \(1\). If \(n_3=22\), the Sylow \(3\)-subgroups would contribute \[ 22(3-1)=44 \] nonidentity elements of order \(3\), disjoint from the \(120\) nonidentity elements of order \(11\). That would give at least \(164\) nonidentity elements in a group with only \(131\), impossible. Hence \[ n_3=4. \]
Conjugation gives an action of \(G\) on the set of its four Sylow \(3\)-subgroups, hence a homomorphism \[ \varphi:G\longrightarrow S_4. \] The action is transitive by Sylow conjugacy and has more than one point, so \(\varphi\) is nontrivial. Its kernel is normal in \(G\). Since \(G\) is assumed simple, a nontrivial homomorphism must have trivial kernel, so \(\varphi\) would be injective. Then \(|G|\) would divide \(|S_4|=24\), contradicting \(|G|=132\). Therefore no group of order \(132\) is simple.