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Let $p<q$ be distinct prime numbers. Every group of group order $p^2{q}^2$ is Abelian if and only if $p$ does not divide $q^2-1$.

For sufficiency, the Sylow theorems make the number of Sylow p-subgroups one of $1,q,q^2$, congruent to one modulo $p$. The divisibility hypothesis excludes both nontrivial choices, so that subgroup is normal. The number of Sylow subgroups for $q$ is one of $1,p,p^2$, congruent to one modulo $q$. The choice $p$ is impossible because $p<q$. If $q$ divided $p^2-1=(p-1)(p+1)$, primality and $q>p$ would force $q=p+1$, hence $(p,q)=(2,3)$; this pair violates the hypothesis. Thus the other Sylow subgroup is normal too. Their intersection is trivial, and their group commutators lie in that intersection, so the group is their direct product of groups. A group of prime-square order is abelian: its nontrivial center of a group is either the whole group or has cyclic quotient, and a cyclic central quotient also forces commutativity. Both factors are therefore abelian.

For necessity, suppose $p\mid q^2-1$. The multiplicative group of a finite field is cyclic, so ${\mathbb{F}}_{q^2}^{\times }$ contains $\alpha$ of order $p$. Let the cyclic group $C_{p^2}$ act on the additive group of ${\mathbb{F}}_{q^2}$ by mapping its generator to multiplication by $\alpha$. This is a well-defined nontrivial group action, since ${\alpha }^{p^2}=1$. The resulting semidirect product has order $p^2{q}^2$ and is nonabelian, because the generator does not commute with a nonzero field element. Thus a nonabelian group of this order exists exactly when $p\mid q^2-1$.

The Sylow theorems give either one or eight Sylow subgroups of order seven. In the first case that subgroup is normal. In the second, their six nonidentity elements are pairwise disjoint, accounting for 48 elements. Only eight elements remain, including the identity, and every Sylow subgroup of order eight lies entirely among them. Thus there is exactly one, and it is a normal subgroup.

For a finite simple group with a subgroup of index $n\ge 3$, the coset action is nontrivial and its kernel is a normal subgroup, hence trivial. Its sign homomorphism must also be trivial: otherwise simplicity would give an isomorphism to $C_2$, which cannot act transitively on at least three points. Thus the image lies in the alternating group. In particular Lagrange's theorem gives $|G|\mid n!/2$. This can exclude simplicity with very little counting: a group of order $112$ has a subgroup of index seven by Sylow theorems, but $112$ does not divide $7!/2$.

Every finite group of odd order is a solvable group. In particular, a nonabelian simple group has even order and therefore has an involution by Cauchy's theorem for finite groups. This is a deep theorem, not an elementary consequence of the Sylow theorems.

For distinct primes $p,q$, the Sylow theorems force a proper nontrivial normal subgroup in any group of order $p^{2}q$. If $p>q$, the Sylow $p$-subgroup is unique. If $p<q$ and both Sylow counts were nontrivial, the count congruences force $(p,q)=(2,3)$. In order twelve, four Sylow $3$-subgroups exhaust eight nonidentity elements; only three remain for a Sylow $2$-subgroup, making that subgroup unique. Thus the group is not a simple group.

These are all the groups of group order $70$ up to group isomorphism, with $D_n$ denoting a dihedral group of order $n$. The Sylow theorems make the Sylow subgroups of orders five and seven unique: the possible counts divide fourteen and ten, respectively, and only one has the required residue. These normal subgroups of coprime order commute, giving a normal subgroup $N\cong C_{35}$. Cauchy's theorem for finite groups provides an element of order two outside $N$, so $G=N⋊C_2$.

The automorphism of a finite cyclic group defining this semidirect product is a unit $a$ modulo $35$ with $a^2\equiv 1$. The Chinese remainder theorem makes its restrictions modulo five and seven independently $+1$ or $-1$. The resulting exponents are $1,34,6,29$, giving the four displayed groups in order. Their center of a group orders are $70,1,5,7$, so no two are isomorphic. Every case exists as the specified semidirect product.

A Sylow subgroup for the prime $p$ is a maximal $p$-subgroup; in a finite group of order $p^{a}m$ with $p\nmid m$ it has order $p^a$. Sylow theorems assert existence, containment of every $p$-subgroup in a Sylow $p$-subgroup, conjugacy of all Sylow $p$-subgroups, and the counting conditions $n_p\mid m$, $n_p\equiv 1(\mathrm{mod}p)$.

Since $N$ is a normal subgroup, $NP$ is a subgroup andThe final index is prime to $p$. Since $N\cap P$ is a $p$-subgroup, its index in $N$ being prime to $p$ means it contains the entire $p$-part of $|N|$. Hence $N\cap P$ is Sylow in $N$.

For a finite group with $|G|=p^{a}m$ and $p\nmid m$, a Sylow p-subgroup has order $p^a$. The Sylow theorems assert existence, containment of every p-subgroup in a Sylow subgroup, conjugacy of all Sylow subgroups, and $n_p\mid m$, $n_p\equiv 1(\mathrm{mod}p)$.

For Sylow subgroups of A6, $|A_6|=6!/2=360=2^3\cdot 3^2\cdot 5$. Explicit choices areTheir elements are even permutations, and their orders are the largest powers of five and three dividing $360$.

Every nonidentity element of $P_5$ is a five-cycle. There are $\left(\genfrac{}{}{0ex}{}{6}{5}\right)4!=144$ five-cycles, and distinct subgroups of order five meet only at the identity. Each contributes four such elements, givingFor $P_3$, each unordered partition of the six letters into two triples gives the subgroup generated by the two triple-cycles. It is independent of the chosen orientation of either cycle. There are $\frac{1}{2}\left(\genfrac{}{}{0ex}{}{6}{3}\right)=10$ such partitions. Different partitions give different subgroups, since the four single three-cycles in a subgroup identify the two triples. Every Sylow subgroup of order nine is conjugate to the displayed $P_3$, so every one arises from a partition in this way. HenceThe normalizer is $N_G(H)=\{g\in G:gH{g}^{-1}=H\}$. Its definition makes $H$ normal in $N_G(H)$. If $H$ is a Sylow subgroup of $G$, it is also a Sylow subgroup of $N_G(H)$, since $H\le N_G(H)\le G$ leaves no room for a larger power of $p$. The Sylow theorems make all Sylow subgroups of the normalizer conjugate there, and normality of $H$ makes every such conjugate equal to $H$. Thus $H$ is the unique Sylow subgroup of its normalizer.

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

Suppose $|G|=p^{2}q$ with distinct primes and $G$ were simple. A Sylow subgroup of order $q$ has order $q$, and $n_q>1$ divides $p^2$, so $n_q=p$ or $p^2$. If $n_q=p$, conjugation gives a nontrivial homomorphism $G\to S_p$. Its kernel is normal and, by simplicity, trivial. Hence $p^{2}q$ would divide $p!$ by Lagrange's theorem, impossible because $p!$ has only one factor of $p$.

If $n_q=p^2$, distinct Sylow subgroups of order $q$ meet only in the identity and together contain $p^2(q-1)$ nonidentity elements. Only $p^2$ elements of $G$ remain, including the identity. Every Sylow p-subgroup lies entirely in that remaining set and itself has $p^2$ elements. They must therefore all be the same subgroup, making it normal, again a contradiction. Thus no group of order $p^{2}q$ is simple. If $p=q$, the preceding p-group argument applies.

For distinct $p,q$, a finite Abelian group is the direct product of groups of its primary subgroups. A group of order $p^2$ is either cyclic or has every nonidentity element of order $p$, in which case choosing two independent elements gives $C_p\times C_p$. Consequently the possibilities areIf the primes coincide, the Fundamental theorem of finitely generated abelian groups gives instead $C_{p^3}$, $C_{p^2}\times C_p$, and $C_p^3$. This classification is used only for the requested list, not for the proof above.

For the transfer homomorphism, choose representatives $T$ of the right cosets $H\backslash G$. For $t\in T$ and $g\in G$, write $tg=h(t,g)t_g$, where $h(t,g)\in H$ and $t_g\in T$. The permutation $t\mapsto t_g$ represents right multiplication on the cosets. DefineThe product has no order ambiguity in the abelianization $H/H^{\prime }$. Replace each representative $t$ by $a_{t}t$, with $a_t\in H$. The new factor is $a_{t}h(t,g)a_{t_g}^{-1}$. In the Abelian quotient, the extra products cancel because $t\mapsto t_g$ permutes $T$. Thus the transfer is independent of the transversal. Moreover,Multiplying and using the same permutation of representatives gives $V(g_1{g}_2)=V(g_1)V(g_2)$. This proves that group transfer is a well-defined group homomorphism.

For the Burnside transfer theorem, let $P\subseteq Z(N_G(P))$ be a Sylow $p$-subgroup. In particular $P$ is Abelian. We first prove the fusion fact that two elements of $P$ which are conjugate in $G$ must be equal. Suppose $u\in P$ and $v=g^{-1}ug\in P$. Both $P$ and $g^{-1}Pg$ centralize $v$, and they are Sylow subgroups of $C_G(v)$: they have the maximal $p$-order possible even in $G$. By Sylow theorems inside this centralizer, choose $h\in C_G(v)$ with $h^{-1}{g}^{-1}Pgh=P$. Thus $gh\in N_G(P)$ andBut $gh$ centralizes $P$ by hypothesis, so $v=u$.

Evaluate transfer to $P$ on $u\in P$. Partition the right cosets of $P$ into cycles under multiplication by $u$. For a cycle of length $r$ beginning at $Pt$, the product of its transfer factors telescopes to $t{u}^r{t}^{-1}\in P$. It is conjugate to $u^r\in P$, so the fusion fact makes it equal to $u^r$. The cycle lengths sum to $m=[G:P]$. Since $P$ is Abelian, all factors commute andThe integer $m$ is prime to $p$, and this power map is an automorphism of the finite Abelian $p$-group $P$: choose an inverse exponent modulo the exponent of $P$. Therefore $V_{G,P}:G\to P$ is surjective. Its kernel $K$ is normal and the first isomorphism theorem givesThus $K$ is the required normal p-complement; it also has trivial intersection with $P$.

Now suppose $P$ is cyclic of order $p^a$, where $p$ is the smallest prime divisor of $|G|$. Conjugation embeds $N_G(P)/C_G(P)$ into $\mathrm{Aut}(P)$, whose order is $p^{a-1}(p-1)$. Since $P\subseteq C_G(P)$ and $P$ is Sylow in $N_G(P)$, the quotient has order prime to $p$. Any remaining prime factor would divide $p-1$ and hence be smaller than $p$, but would also divide $|G|$, an impossibility. Thus $N_G(P)=C_G(P)$ and $P\subseteq Z(N_G(P))$. The Burnside transfer theorem proves that a cyclic Sylow subgroup at the least prime divisor gives a normal complement.

Apply this to a finite nonabelian simple group. If $p^3\nmid |G|$, its Sylow $p$-subgroup has order $p$ or $p^2$. A cyclic Sylow subgroup would give a normal complement; simplicity would then force either a trivial complement and a $p$-group, or an improper complement, neither possible for a nonabelian simple group. Groups of order $p^2$ are Abelian: their nontrivial group center, supplied by the class equation, is either the whole group or has cyclic quotient, which also forces the group Abelian. Thus the only remaining Sylow possibility is $P\cong C_p^2$.

The conjugation quotient $N_G(P)/C_G(P)$ now embeds in $G{L}_2(p)$ and is still prime to $p$. Its possible prime divisors come fromIf $p$ is odd, every prime divisor of $p-1$ is smaller than $p$, and every prime divisor of the even number $p+1$ is at most $(p+1)/2<p$. Since $p$ was the least prime divisor of $|G|$, the quotient must be trivial. Then the Burnside transfer theorem again contradicts simplicity. If $p=2$, the only nontrivial odd divisor of $|G{L}_2(2)|=6$ is three. Trivial conjugation would again give a normal complement, so three must divide $|G|$; the Sylow subgroup has order four. We have proved the least-prime divisibility constraint for a finite simple group:

For infinitely many instances of the latter alternative only, takeThese are simple, as permitted in the question (and also covered by the next solution). For $q=5^{2m+1}$, the group order is $q(q^2-1)/2$. Since $q\equiv 5(\mathrm{mod}8)$, $q-1$ is divisible by four but not eight and $q+1$ by two but not four. Hence the order is divisible by four but not eight. Also $q\not\equiv 0(\mathrm{mod}3)$, so $3\mid q^2-1$. Their smallest prime divisor is two andTheir orders strictly increase with $m$, so they give infinitely many distinct examples.

A soluble group is a group whose derived series $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$ terminates at $1$. Equivalently, it has a finite subnormal series with abelian factors. For the reverse implication, if $G/N$ has derived length $r$, then $G^{(r)}\le N$; if $N$ has derived length $s$, then $G^{(r+s)}=1$. Iterating this observation through the series proves the equivalence. For a finite group, refining the abelian factors to composition factors says that solubility is equivalent to all composition factors being cyclic of prime order.

Subgroups and quotient groups of soluble groups are soluble, since their derived series lie in, or are images of, the original series. The same calculation proves closure under group extensions. Finite p-groups are nilpotent groups, hence soluble: their nontrivial centers allow induction to construct a central series. Every finite nilpotent group is a direct product of groups of its Sylow subgroups. The converse fails: $S_3$ has the abelian-factor series $1<C_3<S_3$, but its order-two Sylow subgroups are not normal. The symmetric group $S_4$ is also soluble, with series $1<V_4<A_4<S_4$ and abelian factors. A finite soluble group has elementary abelian minimal normal subgroups, by the argument in 1(c). In contrast, a nonabelian simple group is a perfect group and is not soluble.

The Fitting subgroup collects the normal nilpotent structure. For a nontrivial finite soluble group it is nontrivial, since it contains an elementary abelian minimal normal subgroup. Part 1(c) proves $C_G(F)\le F$, so conjugation embeds $G/C_G(F)$ in $\mathrm{Aut}(F)$. This shows why controlling a soluble group through normal abelian or nilpotent layers is effective, even when the whole group is not a nilpotent group.

For a set of primes $\pi$, a Hall subgroup of type $\pi$ has order divisible only by primes in $\pi$ and index divisible only by primes outside $\pi$. The Hall theorem for soluble groups says that a finite soluble group has such subgroups, any two are conjugate, and every subgroup of $\pi$-order is contained in one. These assertions generalize the Sylow theorems, which treat a single prime. Here is a proof of all three assertions.

First we need coprime splitting over an elementary abelian normal subgroup. Suppose $V⊴E$ is an elementary abelian p-group and $Q=E/V$ has order $h$ prime to $p$. A section $s:Q\to E$ defines an action on $V$ and an additive factor set $f(x,y)$ by $s(x)s(y)=f(x,y)s(xy)$. Associativity saysSumming over $z\in Q$ and dividing by $h$ in $V$ gives $f(x,y)=b(x)+xb(y)-b(xy)$, where $b(x)=h^{-1}{\sum }_{z}f(x,z)$. Replacing $s(x)$ by $-b(x)s(x)$ removes the factor set, giving a complement. Two complements differ by a map $d$ satisfying $d(xy)=d(x)+xd(y)$. Averaging over $y$ gives $d(x)=b-xb$, so the complements are conjugate by an element of $V$. This proves both splitting and conjugacy, without assuming that an arbitrary extension splits.

Induct on $|G|$, taking an elementary abelian minimal normal subgroup $V$ of exponent $p$. Choose a quotient Hall subgroup $\bar{H}\le G/V$ and let $K$ be its preimage. If $p\in \pi$, then $K$ itself is a Hall subgroup. If $p\notin \pi$, the preceding splitting argument gives a complement $H$ to $V$ in $K$, and that complement is a Hall subgroup of $G$.

For conjugacy, the images of two Hall subgroups are quotient Hall subgroups, so induction allows them to have the same image and lie in the same $K$. If $p\in \pi$, every Hall subgroup contains $V$: its product with $V$ is a $\pi$-subgroup and cannot be larger by the subgroup-order formula. Both are then $K$. If $p\notin \pi$, both are complements to $V$ in $K$, so the averaging argument conjugates them.

Finally let $U\le G$ be any $\pi$-subgroup. By induction its image lies in a quotient Hall subgroup, so $U\le K$. If $p\in \pi$, $K$ is already a Hall subgroup containing $U$. Otherwise write $K=V⋊H$. In $VU$, both $U$ and the inverse image in $H$ of the image of $U$ are complements to $V$. Complement conjugacy therefore places $U$ in a conjugate of $H$. This proves embedding and completes the induction.

Thus existence, conjugacy and containment all hold for every prime set in a finite soluble group. The solubility hypothesis matters: $A_5$ has no Hall subgroup for $\pi =\{2,5\}$. Such a subgroup would have index three; its coset action would give a nontrivial group homomorphism $A_5\to S_3$, impossible because $A_5$ is a simple group and has order $60$. This example also distinguishes a Hall subgroup from a merely arbitrary subgroup whose order uses the chosen primes.

Work with finite groups, as required for these Sylow subgroup counts. Let $P$ act by conjugation on the set of Sylow $p$-subgroups. The stabilizer of $Q$ is $P\cap N_G(Q)$. Since $Q$ is the normal Sylow subgroup of its own normalizer, every $p$-subgroup of that normalizer lies in $Q$. Consequently the stabilizer is exactly $P\cap Q$, and the orbit-stabilizer theorem gives orbit size $[P:P\cap Q]$. The orbit of $P$ has size one. Every other orbit size is a power of $p$ at least $p^a$, hence is divisible by $p^a$. Adding orbit sizes proves the strengthened Sylow congruence from intersections:Now suppose $G$ is simple of order $2^e\cdot 15$. If $e=0$, its Sylow $5$-subgroup is normal because its number divides three and is one modulo five. If $e=1$, an involution in the regular permutation action of $G$ swaps fifteen pairs, so its sign is negative. The sign of a permutation would give a nontrivial group homomorphism $G\to C_2$ with a proper nontrivial normal kernel. Both cases contradict simplicity. Thus $e\ge 2$.

The Sylow theorems make $n_2$ a divisor of fifteen. It is not one, since a nontrivial Sylow subgroup would then be normal. It is not three, since the conjugation action on three Sylow subgroups would give a faithful group homomorphism $G\hookrightarrow S_3$, impossible by order. If $n_2=5$, take a Sylow subgroup $P$ itself: its normalizer has subgroup index five. If $n_2=15$, not all distinct Sylow subgroups can intersect $P$ in subgroup index at least four, since that would give $15\equiv 1(\mathrm{mod}4)$. Choose $Q\ne P$ with $[P:P\cap Q]=2$. Put $R=P\cap Q$, of order $2^{e-1}$. It has subgroup index two in both $P,Q$, so is normal in each. Hence $N_G(R)$ contains both distinct Sylow subgroups and has order greater than $2^e$. It is proper: otherwise $R$ would be a nontrivial proper normal subgroup of $G$. Its order is therefore $2^e\cdot 3$ or $2^e\cdot 5$. The latter would give an index-three subgroup and again an impossible faithful action on three cosets. Thus $[G:N_G(R)]=5$.

In either case we have the requested subgroup of order $2^e$ or $2^{e-1}$ with index-five normalizer. The action on the five cosets of that normalizer is nontrivial and faithful by simplicity. Its image lies in $A_5$, since composing with sign cannot give a nontrivial map from this simple group of order greater than two to $C_2$. Thus $2^e\cdot 15$ divides $60$, forcing $e=2$ and equality of orders. This proves the simple groups with order a power of two times fifteen conclusion:For $S{L}_2(5)$, choose the first column: there are $5^2-1=24$ nonzero choices. For each, determinant one is a nonzero linear equation in the second column and has five solutions. Therefore $|S{L}_2(5)|=120$. A central matrix commuting with $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ has zero lower-left entry and equal diagonal entries; commuting also with $\left(\begin{array}{cc}1& 0\\ 1& 1\end{array}\right)$ makes the upper-right entry zero. It is scalar. Determinant one then givesTo establish the projective special linear group over the field with five elements isomorphism without assuming its simplicity, compute its classes. For determinant-one matrices, the characteristic polynomial is $X^2-tX+1$ with trace $t$. The possibilities over ${\mathbb{F}}_5$ exhaust the group:For trace zero the eigenvalues are $2,3$, so the determinant-one centralizer is the split diagonal torus of order four. For traces $\pm 1$, the discriminant is the nonsquare $2$: the centralizer is the norm-one subgroup of ${\mathbb{F}}_{25}^{\times }$, of order six. In both semisimple cases the determinant map on the full linear centralizer is surjective onto ${\mathbb{F}}_5^{\times }$; consequently each trace type is a single special-linear class. For the noncentral traces $\pm 2$, a Jordan block has full linear centralizer $\left\{\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right):a\ne 0\right\}$ with determinant $a^2$. The determinant image consists of the two squares, so each general-linear class splits into two special-linear classes; the determinant-one centralizer has order $2\cdot 5=10$. These sizes sum to $120$.

Quotienting by $Z$ identifies a matrix with its negative. The trace-zero class gives one projective class of size fifteen; the two traces $\pm 1$ give one class of size twenty; the four noncentral unipotent classes give two classes of size twelve. Their respective projective orders are $2,3,5,5$, by the characteristic polynomials. A normal subgroup is a union of classes containing the identity. The nontrivial proper possible union sizes are $13,16,21,25,28,33,36,40,45,48$, none of which divides $60$. Thus the quotient is simple. Applying the order-$60$ result above givesFinally, if $A^2=I$ over an odd-characteristic field, its minimal polynomial divides the square-free polynomial $(X-1)(X+1)$, so it is diagonalizable with eigenvalues $\pm 1$. In dimension two, determinant one requires both eigenvalues equal. Hence $A=I$ or $A=-I$. This proves the unique involution in SL2 over an odd field assertion: the only element of order two in $S{L}_2(5)$ is $-I$. If an index-two subgroup $H$ existed, it would be normal and have order sixty. By Cauchy theorem it would contain an involution, hence contain $Z$. Then $H/Z$ would be a normal subgroup of order thirty in the simple quotient $PS{L}_2(5)$, a contradiction. Thus $S{L}_2(5)$ has no index-two subgroup.

Choose a minimal nontrivial normal subgroup $L$ of $K$ itself. Every conjugate $L^g$ for $g\in G$ is also a minimal normal subgroup of $K$, since $K$ is normal in $G$. Distinct such subgroups intersect trivially: their intersection is normal in $K$ and minimality forces either equality or trivial intersection. Moreover $[L_i,L_j]\subseteq L_i\cap L_j=1$, so distinct ones commute.

Choose a maximal family of these conjugates whose product $M=L_1\times \cdots \times L_r$ is direct. For any further conjugate $L^g$, its intersection with $M$ is normal in $K$ and is either all of $L^g$ or trivial. In the latter case it commutes with all the existing factors and could be adjoined to the direct product of groups, contrary to maximality. Thus every conjugate is contained in $M$. The product of all conjugates is consequently $M$, which is nontrivial and normal in $G$. Minimality of $K$ as a $G$-normal subgroup gives $M=K$.

Any normal subgroup of a factor $L_i$ is normal in $K$: the other direct factors commute with it, and $L_i$ normalizes it. Minimality of $L_i$ as a $K$-normal subgroup makes it simple. All factors are isomorphic because they are conjugate to $L$. This proves the direct-product structure of a finite minimal normal subgroup:An abelian simple group is cyclic of prime order, so the abelian case is elementary abelian. Now let $H$ be a maximal proper subgroup of a nonabelian simple group $X$. It is nontrivial, since $X$ has nontrivial proper cyclic subgroups. Choose a minimal normal subgroup $K$ of $H$. Then $H\le N_X(K)$. This normalizer cannot be $X$, since that would make $1<K\le H<X$ normal in $X$. Maximality givesFor each of the two groups in this question, the allowed absence of proper nonabelian simple subgroups makes the factors of such a $K$ cyclic of prime order. Hence it suffices to examine normalizers of nontrivial elementary abelian subgroups.

In $A_5$, the possible primes are two, three and five. The Sylow subgroups for the last two primes are cyclic of orders three and five. Every involution is a double transposition; for $(12)(34)$ the centralizer in $A_5$ isthe Klein four-group fixing point five. Thus an elementary abelian $2$-subgroup is cyclic of order two or a Klein four-group. The normalizer of a cyclic order-two subgroup equals its centralizer and has order four, so is contained in a point stabilizer of order twelve. A Klein four-group's common fixed point is intrinsic, and the full point stabilizer $A_4$ normalizes it; its normalizer is therefore exactly that stabilizer.

An order-three subgroup has support a triple. Its normalizer preserves the triple and the remaining pair, and is $(S_3\times S_2)\cap A_5$, of order six. An order-five subgroup acts regularly on all five points. Label the points by ${\mathbb{F}}_5$, with the subgroup acting by translations. Its normalizer in $S_5$ is the twenty affine maps $x\mapsto ax+b$. The translations are even, whereas multiplication by a generator of ${\mathbb{F}}_5^{\times }$ is a four-cycle and is odd. Exactly half these maps are even, so the normalizer in $A_5$ has order ten.

The only candidates for maximal subgroups are thus the normalizers of orders twelve, ten and six. They are all maximal: none can lie in a larger candidate, except potentially an order-six subgroup in an order-twelve $A_4$. But an index-two subgroup of $A_4$ would contain every order-three element; there are eight such elements, plus the identity, already more than six. Hence that containment is impossible. Point stabilizers are conjugate; order-three and order-five subgroup normalizers each form one class by the Sylow theorems. Therefore the maximal subgroups of A5 form preciselyTheir numbers are five, six and ten respectively, since these proper maximal subgroups are self-normalizing in the simple group.

For $X=G{L}_3(2)$, choosing independent columns gives $|X|=(8-1)(8-2)(8-4)=168$. If $t$ is an involution, write $t=I+N$. Field characteristic two gives $N^2=0$, and $\mathrm{im}N\subseteq \ker N$ in dimension three forces its nonzero rank to be one. Thus $t=I+vf$, with nonzero vector $v$, nonzero linear functional $f$ and $f(v)=0$. This transvection fixes the plane $\ker f$ and has image line $⟨v⟩$. Every flag consisting of an incident line and plane occurs, giving $7\cdot 3=21$ involutions, all conjugate. Its centralizer, and hence the normalizer of its cyclic subgroup, is the flag stabilizer of order $168/21=8$. It lies in both the line and plane stabilizers of order $24$, so cannot be maximal.

For the elementary abelian subgroups in GL3 over F2, consider commuting transvections $I+vf$, $I+wh$. Their rank-one parts satisfyIf $v,w$ are independent, equality forces both sides zero; thus $f,h$ vanish on $⟨v,w⟩$ and are the same nonzero functional. If $v=w$, they have a common image line. Consequently the two types of order-four elementary abelian subgroup areNo extra transvection can commute with all elements of $K_L$: if its image were different, its functional would have to equal each of the two independent functionals occurring in $K_L$. Dually, the same holds for $K_W$. Thus these exhaust the elementary abelian $2$-subgroups of rank at least two. The normalizer of $K_L$ is the line stabilizer and the normalizer of $K_W$ is the plane stabilizer, each of order $168/7=24$. They give different conjugacy classes because their common fixed spaces have dimensions one and two. All line stabilizers are conjugate, as are all plane stabilizers.

For an order-three element, its minimal polynomial is $(X-1)(X^2+X+1)$: it has a fixed line and an invariant complementary irreducible plane. Its normalizer preserves both and is the full $G{L}_2(2)$ on that plane, of order six, since the order-three subgroup is normal in $G{L}_2(2)$. This normalizer lies in the corresponding line and plane stabilizers, so is not maximal.

An order-seven element is irreducible. Over the finite field ${\mathbb{F}}_2$ one has $X^7-1=(X-1)(X^3+X+1)(X^3+X^2+1)$, and both cubic factors are irreducible because neither has a root in ${\mathbb{F}}_2$. Its minimal polynomial divides this square-free product and cannot be just $X-1$, since the element is not the identity. It therefore contains a cubic factor; the three-dimensional space forces that irreducible cubic to be the whole minimal polynomial. Identify the space with ${\mathbb{F}}_8$ and the element with multiplication by a generator $\alpha$ of ${\mathbb{F}}_8^{\times }$. Its centralizer consists of the seven nonzero multiplication maps. A normalizer induces an ${\mathbb{F}}_2$-automorphism of this field, so it sends $\alpha$ to $\alpha$, ${\alpha }^2$ or ${\alpha }^4$. All three possibilities occur via the field automorphisms $x\mapsto x^{2^j}$. Therefore the normalizer of the Singer cycle has order $7\cdot 3=21$, and all such normalizers are conjugate by the Sylow theorems.

The maximal-normalizer reduction leaves exactly the two order-twenty-four stabilizer types and the order-twenty-one type. Each is maximal because it cannot be contained in any larger remaining candidate; equal-order subgroups cannot properly contain one another and $21$ does not divide $24$. Thus

Relabel the two primes so that $p<q$. By the Sylow theorems, the number $n_q$ of Sylow $q$-subgroups divides $p$, so it is either $1$ or $p$. But $n_q\equiv 1(\mathrm{mod}q)$ and $1<p<q$, ruling out $n_q=p$. Thus there is a unique Sylow subgroup of order $q$, which is a normal subgroup. It is nontrivial and proper because $1<q<pq$. Consequently no group of order $pq$ is simple. No additional assumption on divisibility of $q-1$ by $p$ is required.

The remaining Sylow theorems state that every $p$-subgroup of a finite group is contained in a Sylow subgroup, and that all Sylow $p$-subgroups are conjugate. If $|G|=p^{n}r$ with $p\nmid r$ and $n_p$ is the number of Sylow $p$-subgroups, thenIn particular $n_p=1$ if and only if that Sylow subgroup is a normal subgroup.

Let $\mathcal{S}$ be the set of all subsets of $G$ with $p^n$ elements. First show that $|\mathcal{S}|=\left(\genfrac{}{}{0ex}{}{p^{n}r}{p^n}\right)$ is not divisible by $p$. In the polynomial ring over the field of $p$ elements, $(1+X{)}^p=1+X^p$, since the intermediate binomial coefficients are divisible by $p$. Iterating givesComparison of coefficients of $X^{p^n}$ yields $\left(\genfrac{}{}{0ex}{}{p^{n}r}{p^n}\right)\equiv r\not\equiv 0(\mathrm{mod}p)$.

Let $G$ act on $\mathcal{S}$ by left translation. Because its orbit sizes sum to a number not divisible by $p$, some orbit has size prime to $p$. Choose a subset $S$ in that orbit, with stabilizer $H$. The orbit-stabilizer theorem gives the orbit size $[G:H]$, so $p\nmid [G:H]$. On the other hand $H$ acts freely on the elements of $S$: if $hs=s$, cancellation in the group gives $h=1$. Every $H$-orbit in $S$ therefore has $|H|$ elements, and $|H|$ divides $|S|=p^n$. Sinceand the second factor has no factor of $p$, the first must contain all $p^n$. Hence $\boxed{|H|=p^n}$, proving the first of the Sylow theorems. This is Sylow existence by subset action and includes the trivial case $n=0$.

Write the cardinality of the finite group as $|G|=p^{a}m$, where $p$ is prime and $p\nmid m$. The Sylow theorems say that a Sylow subgroup of order $p^a$ exists; every p-group is contained in a Sylow subgroup; and all Sylow subgroups are conjugate subgroups. If $n_p$ is their number, thenEquivalently, for a Sylow subgroup $P$, conjugation and the orbit-stabilizer theorem give $n_p=[G:N_G(P)]$, where $N_G(P)$ is the normalizer. The containment statement is often included in the conjugacy form: every p-group lies in a conjugate of any fixed Sylow subgroup.

Suppose an extension $H$ existed. Its degree would be $p$, and its order would be $2p(p-1)$ by the orbit-stabilizer theorem. A Sylow $p$-subgroup $P$ consequently has order $p$, and the Sylow theorems imply that its number $n_p$ divides $2(p-1)$ and is congruent to one modulo $p$. Since $2(p-1)<2p$, the only possibilities are $1$ and $p+1$. The latter would require $p+1\mid 2(p-1)$ and hence $p+1\mid 4$, impossible for $p\ge 5$. Thus $P$ is a normal subgroup of $H$.

A nonidentity element of $P$ is a $p$-cycle, so $P$ acts regularly on the $p$ points. Label them by ${\mathbb{F}}_p$ so that $P$ consists of translations. A permutation normalizing $P$ must have the form $x\mapsto ax+b$: its conjugation sends translation by one to translation by some $a\ne 0$, and its value at zero determines $b$. Therefore the normalizer of $P$ in $S_p$ has order $p(p-1)$. Normality would place $H$ inside this normalizer, contradicting $|H|=2p(p-1)$. Hence the specified dihedral permutation group has no one-point extension.

For a finite group of order $p^{a}m$ with $p$ prime and $p\nmid m$, the Sylow theorems assert that a subgroup of order $p^a$ exists; every subgroup whose order is a power of $p$ is contained in one such Sylow subgroup; all Sylow subgroups for that prime are conjugate; and their number $n_p$ divides $m$ and is congruent to one modulo $p$. A unique Sylow subgroup is a normal subgroup, since conjugation cannot move it to another one.

Here $56=7\cdot 8$, so $n_7$ divides eight and is congruent to one modulo seven. Hence $n_7=1$ or $8$. In the first case the Sylow subgroup of order seven is normal. In the second case, any two distinct order-seven subgroups intersect only at the identity, because a nontrivial intersection would generate both prime-order subgroups. Their nonidentity elements therefore account for $8\cdot 6=48$ elements.

Exactly eight elements of the group remain. Every Sylow subgroup of order eight consists entirely of these remaining elements: a nonidentity element in it has order a power of two and cannot be one of the order-seven elements already counted. Each such Sylow subgroup must consequently equal the remaining eight-element set. It is unique and hence normal. Thus a group of order 56 has a normal subgroup of order seven or of order eight.

For $|G|=p^{a}m$ with $p\nmid m$, Sylow theorems assert existence of a subgroup of order $p^a$, containment of every p-subgroup in one, conjugacy of all these Sylow subgroups, andA group of order $20$ has $n_5\mid 4$ and $n_5\equiv 1(\mathrm{mod}5)$, so its Sylow 5-subgroup $P$ is unique. In $S_6$ it is generated by a five-cycle and has one fixed point. Its normalizer preserves that point and, on the other five points identified with ${\mathbb{F}}_5$, is the affine group $x\mapsto ax+b$. Every automorphism of the cyclic translation group is realized by a multiplier. Thus $|N_{S_6}(P)|=5\cdot 4=20$, forcing

Inside $S_5$ there are $120/20=6$ Sylow 5-subgroups. Conjugation gives a transitive action on them and a homomorphism into $S_6$. Its group homomorphism kernel is a normal subgroup of $S_5$, whose only possibilities are $1,A_5,S_5$. The latter two would leave an image of order at most two, incompatible with transitivity on six points. Hence it is faithful. Denote the resulting transitive copy of $S_5$ by $J\le S_6$.

The action of $S_6$ on the six cosets of $J$ is also faithful: its group homomorphism kernel cannot contain the simple subgroup $A_6$, since an image of order at most two is not transitive of degree six, and a normal subgroup disjoint from $A_6$ centralizes $A_6$ and is trivial. It therefore gives an automorphism $\tau :S_6\to S_6$ carrying $J$ to a natural point stabilizer. This automorphism is outer, because conjugation cannot carry a transitive subgroup to an intransitive one.

There are precisely two classes of copies of $S_5$ in $S_6$. A faithful intransitive copy must have an orbit of length five: all partitions of six into smaller orbits embed it in groups of order below $120$. Thus it is a natural point stabilizer. A transitive copy is a degree-six coset action of an abstract $S_5$, with an order-$20$ stabilizer. Every order-$20$ subgroup of $S_5$ is a Sylow-five normalizer, so these stabilizers are conjugate and the degree-six actions are permutation-equivalent. Hence all transitive copies form one class. The outer automorphism constructed above swaps these two classes; it cannot preserve the point-stabilizer class, since any automorphism preserving that class induces a relabeling of its six members, and equivariance of their faithful conjugation action then makes the automorphism inner.

Identify $S_6$ with its inner automorphism group inside $\overline{G}$. Every Sylow 5-subgroup of $\overline{G}$ is contained in $S_6$, since that normal subgroup has index two. There are $720/20=36$ of them. ConsequentlyTo prove maximality, first classify the proper overgroups of $N$ in $S_6$. The affine multiplier in $N$ of order four is a four-cycle and is odd. Within $H\cong S_5$, a proper overgroup larger than $N$ could only have order $60$, but the index-two subgroup is $A_5$ and cannot contain that odd element. Thus $N$ is maximal in $H$. A nontransitive overgroup of $N$ must fix its unique fixed point, so is $H$ if it is larger than $N$. A transitive overgroup $M$ has $M\cap H\ge N$; this point stabilizer is either $N$ or $H$. In the latter case $M=S_6$, and in the former $|M|=6\cdot 20=120$, making it a transitive copy of $S_5$.

There is exactly one such transitive copy containing $N$. There are six conjugates of $J$: applying the preceding overgroup argument to its own Sylow-five normalizer shows that $J$ is maximal, and $J$ is not normal. Each contains six Sylow 5-subgroups. Counting pairs of a conjugate copy and a Sylow subgroup gives $6\cdot 6=36$, with the same number through every one of the 36 Sylow subgroups of $S_6$. Thus each Sylow subgroup, and hence its normalizer $N$, is contained in exactly one transitive copy. In particular the only proper strict overgroups of $N$ are its natural $H$ and this $J$.

The subgroup $\overline{N}$ has an outer element, because it has twice the order of $N$. Such an element normalizes $P$ and swaps the unique overgroups $H,J$. If $\overline{N}\le U\le \overline{G}$, then $U\cap S_6$ is normalized by $\overline{N}$, so it cannot be just $H$ or just $J$. It is therefore $N$ or $S_6$. In the latter case $U=\overline{G}$. In the former $U$ normalizes $N$ and hence its unique Sylow 5-subgroup $P$, giving $U\le \overline{N}$. Therefore