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Use the Sylow theorems. The number $n_p$ divides $qr$ and is congruent to one modulo $p$. Since $q,r<p$, it is either one or $qr$. Suppose $n_p=qr$. Different subgroups of prime order intersect trivially, so their nonidentity elements occupy $qr(p-1)$ places, leaving only $qr-1$ nonidentity elements of other prime orders.

If neither the Sylow $q$-subgroup nor the Sylow $r$-subgroup is normal, then $n_q\ge p$ and $n_r\ge q$. Indeed $n_q$ divides $pr$, and its possible divisor $r<q$ cannot satisfy the Sylow congruence; the smallest remaining nontrivial possibility is at least $p$. Likewise any nontrivial divisor of $pq$ is at least $q$. Their elements would require at leastplaces, since the excess is $(p-1)(q-1)>0$. Hence some Sylow subgroup of order $s\in \{q,r\}$ is normal.

In the quotient by this normal subgroup, the largest prime $p$ has a normal Sylow subgroup: for a group of order $pt$ with $p>t$ its Sylow count divides $t<p$ and so equals one. Pulling back gives a normal subgroup of order $ps$. Inside it, the subgroup of order $p$ is again the unique Sylow $p$-subgroup. It is characteristic in that normal subgroup and therefore normal in $G$, contradicting $n_p=qr$. Thus $\boxed{n_p=1}$.

Let $P$ be this normal Sylow subgroup. In $G/P$, of order $qr$, its subgroup of order $q$ is normal by the same argument. Its preimage $K$ is a normal Hall $\{p,q\}$-subgroup. The serieshas factors of orders $p,q,r$, hence cyclic and abelian. Therefore $\boxed{G\text{ is soluble}.}$ This establishes solubility before using any Hall-existence conclusion that itself assumes solubility.

If $|G|=p^{a}m$ with $p\nmid m$, a Sylow subgroup for $p$ is a subgroup of order $p^a$. The Sylow theorems assert: such subgroups exist; every $p$-subgroup is contained in a Sylow subgroup; all Sylow $p$-subgroups are conjugate; and their number $n_p$ divides $m$ and satisfies $n_p\equiv 1(\mathrm{mod}p)$.

For groups of order p squared q are not simple, if $p>q$, then $n_p\mid q$ and $n_p\equiv 1(\mathrm{mod}p)$ force $n_p=1$. Its unique Sylow subgroup is a nontrivial proper normal subgroup. Suppose instead $p<q$. If either Sylow count is one, the conclusion already follows. Otherwise $n_p=q$, while $n_q$ is either $p$ or $p^2$. The value $p$ is impossible because $0<p-1<q$. Thus $n_q=p^2$, and $q\mid (p^2-1)=(p-1)(p+1)$. Primality and $q>p$ imply $q\mid p+1$, hence $q=p+1$. The only consecutive primes are $p=2,q=3$.

In that exceptional order-$12$ case, four distinct Sylow $3$-subgroups contribute eight distinct nonidentity elements: their intersections are trivial. Only three nonidentity elements remain. Every subgroup of order four must contain exactly those remaining three, so there can be only one Sylow $2$-subgroup, contradicting the assumption that both counts were nontrivial. Therefore no group of order $p^{2}q$ with distinct primes is simple.

For the final factorization, normality of $H$ ensures $g^{-1}Pg$ is a Sylow $p$-subgroup of $H$. By Sylow conjugacy within $H$, choose $h\in H$ such that $g^{-1}Pg=h^{-1}Ph$. Then $k=g{h}^{-1}$ satisfiesThus $k\in N_G(P)$ and $g=kh$. We have proved the Frattini argumentThe normaliser factorization includes the case $P=\{1\}$, when its normalizer is all of $G$.

Write $|G|=p^{r}m$ with $p$ prime and $p\nmid m$. The Sylow theorems assert that a subgroup of order $p^r$ exists; every p-group subgroup is contained in a conjugate of a Sylow subgroup; all Sylow subgroups are conjugate; and their number $n_p$ satisfies $n_p\equiv 1(\mathrm{mod}p)$ and $n_p\mid m$.

For the containment and conjugacy assertions, fix a Sylow subgroup $P$ and let any p-group subgroup $Q$ act by left multiplication on the $m$ left cosets $G/P$. By the orbit-stabilizer theorem, every orbit has size a power of $p$. Since $p\nmid m$, at least one orbit has size one. If $gP$ is such a fixed point, then $qgP=gP$ for all $q\in Q$, equivalently $g^{-1}Qg\subseteq P$. Thus $Q\subseteq gP{g}^{-1}$. If $Q$ too has order $p^r$, this inclusion is equality, proving conjugacy. This is Sylow containment from a coset fixed point.

Since $351=3^3\cdot 13$, the number of Sylow subgroups for $p=13$ divides $27$ and is $1$ modulo $13$, hence is $1$ or $27$. In the first case the unique subgroup is a nontrivial proper normal subgroup.

In the second case the $27$ subgroups of order $13$ have pairwise trivial intersection: two such subgroups sharing a nonidentity element would both be generated by it. They account for $27\cdot 12=324$ nonidentity elements of order $13$. Exactly $27$ elements, including the identity, remain. Every Sylow subgroup for $p=3$ has order $27$, and all its elements lie among these remaining elements, by Lagrange's theorem. It must therefore be precisely that set. The Sylow subgroup for $p=3$ is unique and hence normal. In either case $G$ is not a simple group.

For the general linear group over ${\mathbb{F}}_{31}$, choose a nonzero first column in $31^2-1$ ways and a second column outside its span in $31^2-31$ ways. ThusSince $5^3=125\equiv 1(\mathrm{mod}31)$ and $5\ne 1$, the residue $5$ has order three. The nine diagonal matrices $\mathrm{diag}(5^i,5^j)$, with $0\le i,j<3$, form a subgroup isomorphic to $C_3\times C_3$. It is an abelian group and a Sylow subgroup for $p=3$. Every other Sylow subgroup for $p=3$ is conjugate to it by the Sylow theorems, hence is also Abelian.

The Sylow theorems state that if a finite group has order $|G|=p^{a}m$ with $p$ prime and $p\nmid m$, it has a subgroup of order $p^a$, called a Sylow subgroup. Every p-group subgroup is contained in a Sylow subgroup, and all Sylow subgroups for the same prime are conjugate subgroups. Their number satisfiesMore precisely $n_p=[G:N_G(P)]$ for a Sylow subgroup $P$, where $N_G(P)$ is its normalizer.

A finite nonabelian simple group cannot be a p-group: a nontrivial finite p-group has nontrivial centre of a group, since its class equation makes every noncentral conjugacy class size divisible by $p$, and hence makes the centre size a positive multiple of $p$. That centre is a normal subgroup, so must be the whole group if $G$ is a simple group, making it abelian; an abelian simple group has prime order. Therefore every Sylow subgroup here is nontrivial and proper. It cannot be normal, so $n_p>1$.

To obtain a simple group embedding from Sylow conjugation, the conjugation action on Sylow subgroups gives a group homomorphism $\rho :G\to S_{n_p}$. This group action is transitive by the Sylow theorems and is nontrivial since $n_p>1$. Its kernel of a group homomorphism is a normal subgroup, so the defining property of a simple group makes the kernel of a group homomorphism trivial: $\rho$ is an embedding. Now compose with the sign of a permutation $S_{n_p}\to \{\pm 1\}$. A nontrivial composite would again be injective because $G$ is a simple group, embedding $G$ in a group of order two, impossible for a nonabelian group. Thus $\rho (G)$ lies in the alternating group $A_{n_p}$. By Lagrange's theorem,The argument only invokes the factorial formula once $n_p\ge 2$; the impossible case $n_p=2$ itself is also excluded by the embedding.

For order $48=2^4\cdot 3$, the Sylow theorems give $n_2\mid 3$ and $n_2\equiv 1(\mathrm{mod}2)$, hence $n_2=1$ or $3$. If $n_2=1$, the Sylow subgroup of order $16$ is a nontrivial proper normal subgroup. If $G$ were simple in the other case, it would be nonabelian (a finite abelian simple group has prime order), and the preceding result would force $48\mid 3!/2=3$, impossible. Thus every group of order 48 is not simple.

An example with no normal Sylow subgroup for $p=2$ isIn $S_4$, $P=⟨(1234),(13)⟩$ is a dihedral group of order $8$: conjugation by $(13)$ inverts $(1234)$. Thus $P\times C_2$ has order $16$ and is a Sylow subgroup of $G$. The only transpositions in $P$ are $(13)$ and $(24)$; conjugating $(13)$ by $(23)$ gives $(12)\notin P$. Therefore $P$ and $P\times C_2$ are not normal. By conjugacy, no other Sylow subgroup for $p=2$ is normal either.