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The center of a group consists of the elements that commute with every element of that group:where $C_G(g)$ is the centralizer of $g$. These are operations whose position in any product can be changed freely. The center is an abelian group and a characteristic subgroup, hence a normal subgroup. A group is an abelian group precisely when $Z(G)=G$.

For the symmetric group $S_3$, the conjugacy classes areAn element belongs to the center precisely when its conjugacy class has one element, so $Z(S_3)=\{e\}$. Yet commutator subgroup of the symmetric group on three letters gives $S_3^{\prime }=A_3$, with three elements. Thus belonging to the commutator subgroup does not mean commuting with everything. The alternating group $A_3$ is an abelian group as a subgroup, but its nonidentity elements do not commute with the transpositions in the surrounding $S_3$.

For the dihedral group $D_8$ of group order eight in abelianization of the square symmetry group, the center is $\{e,r^2\}$. A rotation $r^i$ commutes with $s$ precisely when $r^i=r^{-i}$, which requires $i$ even. No reflection $r^{i}s$ commutes with $r$, since $(r^{i}s)r=r^{i-1}s$ whereas $r(r^{i}s)=r^{i+1}s$. ConsequentlyThe half-turn can be performed before or after any square symmetry with the same result. Its appearance both here and in the commutator subgroup is a feature of this example, rather than a general equality. See Conrad, Dihedral groups I, theorem 3.4 and example 3.5.

The center and the commutator subgroup answer different useful questions: which operations already commute with everything, and which operations must disappear for the entire group to commute. In an abelian group these answers are $Z(G)=G$ and $G^{\prime }=\{e\}$. Another way to use the center is through conjugation: the group homomorphism $g\mapsto (h\mapsto gh{g}^{-1})$ has group kernel $Z(G)$, so the first isomorphism theorem gives $G/Z(G)\cong \mathrm{Inn}(G)$, the inner automorphism group. The quotient group by the center thus records the distinct conjugation operations; it need not be an abelian quotient.

A nontrivial finite p-group always has a nontrivial center. In the class equation, each noncentral conjugacy class has size divisible by $p$, so $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$. The identity element ensures $Z(G)$ is nonempty, hence it has at least $p$ elements. See McKernan, MIT 18.703, lemmas 23.6–23.7 and definition-lemma 23.10.

The commutator subgroup of a group $G$ is the subgroup generated by all its group commutators:Here and throughout, $[x,y]=x^{-1}{y}^{-1}xy$. A group commutator is the identity element precisely when $x$ and $y$ commute. The generated subgroup includes finite products of group commutators and their inverse elements; an element of $G^{\prime }$ need not itself be a single group commutator. Thus $G^{\prime }=\{e\}$ precisely when $G$ is an abelian group.

The useful idea is to record exactly what must disappear to make multiplication commutative. The identity $xy=yx[x,y]$ shows how a group commutator corrects a change in multiplication order. Setting every such correction equal to $e$ produces the abelianization $G/G^{\prime }$, while preserving every piece of information that a group homomorphism to an abelian group can detect. This connects a local comparison of two operations to a canonical quotient group of the whole group.

The commutator subgroup is a characteristic subgroup, hence a normal subgroup: a group automorphism $\alpha$ sends $[x,y]$ to $[\alpha (x),\alpha (y)]$. For any normal subgroup $N$,Indeed, the group commutator of $xN$ and $yN$ is $[x,y]N$. This makes $G^{\prime }$ the smallest normal subgroup whose quotient group is an abelian group. See Conrad, Subgroup series I, section 4: commutators and abelian quotients.

Every virtually soluble group $K$ has a soluble characteristic subgroup of finite index. First take a soluble normal subgroup $K_0$ of finite index by the subgroup core construction. Choose a soluble normal subgroup $R\supseteq K_0$ maximizing its image size in the finite quotient $K/K_0$. For every soluble normal subgroup $S$, the product $RS$ is soluble, because its quotient by $R$ is a quotient of $S$. Maximality forces $S\le R$. Hence $R$ is the unique largest soluble normal subgroup and is invariant under all automorphisms. This avoids incorrectly intersecting infinitely many conjugates when proving closure under group extensions.

If $G$ has $d$ generators, there are at most $(n!{)}^d$ group homomorphisms $G\to S_n$. Each subgroup of index $n$ is a point stabilizer in a transitive coset group action, so there are at most $n(n!{)}^d$ such subgroups. Consequently finitely many subgroups have index at most any fixed bound. Intersecting them gives a finite-index characteristic subgroup, useful in proving residual finiteness of semidirect products.

The largest nilpotent normal subgroup of a finite group $G$. If $O_p(G)$ denotes its p-core, distinct p-cores commute, and $F(G)={\prod }_p{O}_p(G)$. Every normal nilpotent group has Sylow subgroups which are characteristic subgroups, so it is contained in this product.

A nontrivial minimal normal subgroup $N$ of a finite solvable group is abelian: its commutator subgroup is characteristic, and minimality plus solvability excludes $N^{\prime }=N$. A nontrivial Sylow subgroup of this abelian group is characteristic, so minimality makes $N$ a p-group. The characteristic subgroup of pth powers is proper, hence trivial by minimality. Therefore $N$ is an elementary abelian group.

A normal subgroup which is a nilpotent group. In a finite group its unique Sylow subgroups are characteristic subgroups and hence normal in the ambient group. Products of nilpotent normal subgroups are nilpotent groups: their normal Sylow subgroups of different prime characteristics commute.

For a finite group with Sylow $p$-subgroup $P$, a normal $p$-complement is a normal subgroup of order $|G|/|P|$. It consists exactly of the elements of order prime to $p$: every such element has trivial image in the $p$-group quotient, and every element of the complement has prime-to-$p$ order. Thus a normal $p$-complement, if it exists, is unique and a characteristic subgroup.

The Frattini subgroup is the intersection of the maximal subgroups of $G$, with $\Phi (1)=1$. A useful equivalent property isIndeed, a proper subgroup $H$ lies in a maximal subgroup, which also contains $\Phi (G)$. The Frattini subgroup is a characteristic subgroup, since automorphisms permute the maximal subgroups.

The Fitting subgroup is the largest nilpotent normal subgroup. To prove its existence, first form the p-core $O_p(G)$: the product of all normal p-subgroups. For two such subgroups, their product is normal and has order $|AB|=|A||B|/|A\cap B|$, a power of $p$. There are only finitely many subgroups, so $O_p(G)$ exists. For distinct primes $p,r$,Thus the product of the p-cores is a nilpotent normal subgroup, a direct product of groups of its Sylow subgroups. Conversely, a finite nilpotent normal subgroup $K$ has unique Sylow subgroups, each a characteristic subgroup of $K$. Each is normal in $G$ and lies in the corresponding p-core. Consequently the required largest subgroup exists, and

Write $F=F(G)$, $C=C_G(F)$ and $D=C\cap F=Z(F)$. Both $C$ and $D$ are normal in $G$. Suppose $C>D$. Choose a minimal normal subgroup $K/D$ of $G/D$ contained in $C/D$. In a finite soluble group, such a minimal normal subgroup is elementary abelian: the last nontrivial term of its derived series is an abelian characteristic subgroup, so minimality makes the whole subgroup abelian; its primary components and subgroup of elements killed by $p$ then force one prime and exponent $p$.

It follows that $K^{\prime }\le D$. Moreover $K\le C$ centralizes $D\le F$, so $K^{\prime }\le Z(K)$. Therefore $K$ is a nilpotent group of class at most two. Since $K$ is normal in $G$, the definition of the Fitting subgroup gives $K\le F$. But then $K\le C\cap F=D$, a contradiction. This proves the Fitting subgroup is self-centralizing in soluble groups:

It suffices to centralize each minimal normal subgroup $M$. The subgroup $M\cap F(G)$ is normal in $G$, so it is either $1$ or $M$. In the first caseIn the second case $M$ is a nilpotent group. Its nontrivial group center is a characteristic subgroup of $M$, hence normal in $G$, so minimality makes $M$ abelian. The primary components, which are characteristic subgroups, and the subgroup killed by a prime show that $M$ is elementary abelian of exponent $p$. Thus $M\le O_p(G)$.

The finite p-group $O_p(G)$ acts by conjugation on $M$. Its orbits have prime-power sizes; hence the number of fixed elements is congruent to $|M|=0$ modulo $p$. The identity is fixed, so there is a nonidentity fixed element. Therefore $M\cap Z(O_p(G))\ne 1$. This intersection is normal in $G$, and minimality makes it $M$. The other p-cores commute with $O_p(G)$, and hence with $M$. Thus every factor of $F(G)$ centralizes $M$. Since the group socle is generated by these minimal normal subgroups,

Put $\Phi =\Phi (G)$ and let $H$ be the full preimage of $F(G/\Phi )$ in $G$. The preceding part gives $\Phi \le F(G)$, so the normal nilpotent image $F(G)/\Phi$ lies in $F(G/\Phi )$.

For the reverse inclusion, choose a Sylow subgroup $P$ of $H$. Its image $P\Phi /\Phi$ is the unique Sylow subgroup of the nilpotent group $H/\Phi$, hence a characteristic subgroup there and normal in $G/\Phi$. Thus $P\Phi ⊴G$. The subgroup $P$ is also a Sylow subgroup of $P\Phi$: the p-part of $\Phi$ lies in $P$, because $P\cap \Phi$ is a Sylow subgroup of the normal subgroup $\Phi$ of $H$. Apply the Frattini argument to $P\Phi$:The Frattini subgroup property forces $N_G(P)=G$. Hence every Sylow subgroup of $H$ is normal, so $H$ is a nilpotent group and $H\le F(G)$. Together the inclusions proveThis is Frattini lifting of nilpotence; simply calling an extension of nilpotent groups nilpotent would not establish it.

Since the two-transitive group action is primitive, the orbits of any nontrivial normal subgroup form only the single block $\Omega$. Thus $N$ is transitive. If $N$ is regular, identify $\Omega$ with $N$ by $x\mapsto x\alpha$. Then $G=N{G}_{\alpha }$, $N\cap G_{\alpha }=1$, and conjugation by $G_{\alpha }$ gives automorphisms of $N$. The two-transitive group action makes these automorphisms transitive on $N\setminus \{1\}$. All nonidentity elements therefore have the same order, which must be a prime $p$: a suitable power of an element of composite order would have a smaller nontrivial order. Cauchy's theorem for finite groups now shows $|N|=p^d$.

The center of a group of prime-power order is nontrivial and is a characteristic subgroup. The transitive conjugation action on nonidentity elements therefore forces $Z(N)=N$. Hence $N$ is an elementary abelian group, which we identify with the additive vector space ${\mathbb{F}}_p^d$. Conjugation by $G_{\alpha }$ is linear on this vector space. It is faithful: an element fixing $\alpha$ and centralizing $N$ fixes every $x\alpha$, hence is the identity permutation. ConsequentlyHere the general affine group acts by translations followed by invertible linear maps.

A finite minimal normal subgroup is a characteristically simple group, and has the direct-product structure of a finite minimal normal subgroupwith isomorphic simple groups $T_i$. For completeness, choose a minimal nontrivial normal subgroup $T$ of $N$. Its distinct $G$-conjugates are minimal normal subgroups of $N$, so they centralize one another. Their product is normal in $G$, hence equals $N$. If $T$ is abelian the product is abelian and minimal primary characteristic subgroups make $N$ elementary abelian. Otherwise each $T$ has trivial center: its center is a characteristic subgroup of $T$, hence normal in $N$, and minimality would make $T$ abelian. Each new conjugate then meets the product of previous conjugates trivially, because it centralizes that product and is centerless. The product is direct. Every normal subgroup of a direct factor is normal in $N$, so minimality of that factor makes it simple.

Assume now that $N$ acts primitively and nonregularly. Any nontrivial normal subgroup of this faithful primitive action is transitive. Thus each $T_i$ is transitive. If $k\ge 2$, the transitive commuting groups $T_1$ and $T_2$ are regular: an element of one fixing $\alpha$ commutes with the other and hence fixes its entire orbit. But the centralizer of a regular group has order $n$, as its elements are determined by the image of $\alpha$. Therefore all remaining factors lie in the order-n centralizer of $T_1$, and the transitive group $T_2$ already fills it. There can be only two factors, with $|N|=n^2$. Now $|N_{\alpha }|=n$, whereas its common nontrivial subdegree $r$ divides both $n$ and $n-1$. Hence $r=1$, making $N_{\alpha }$ trivial, a contradiction. Thus $k=1$, and $N$ is simple. An abelian transitive group is regular, so $N$ is nonabelian.

Finally $C_G(N)$ is normal in the primitive group $G$. If nontrivial it would be transitive, and commuting with transitive $N$ would force $N$ regular. Thus $C_G(N)=1$, and conjugation embeds $G$ in $\mathrm{Aut}(N)$. The centerless simple group $N$ is identified with its inner automorphism group. Therefore

For $H\le G$ of finite subgroup index, let $H^{\prime }=[H,H]$ and choose a right-coset transversal $T$ for the cosets $Ht$. For each $g\in G$ writeThe transfer homomorphism isThe product is taken in the abelianization, so its order is irrelevant. If another transversal has representative $a_{t}t$ for $Ht$, with $a_t\in H$, its factor is $a_{t}h(t,g)a_{t_g}^{-1}$. The map $t\mapsto t_g$ permutes $T$, so the extra factors cancel in $H/H^{\prime }$. This proves independence of the transversal. AlsoMultiplying over $t$ and again permuting the second indices proves $V(g_1{g}_2)=V(g_1)V(g_2)$. Thus it is indeed a group homomorphism.

The Burnside transfer theorem states that if a Sylow $p$-subgroup $P$ of a finite group satisfies $P\subseteq Z(N_G(P))$, then $G$ has a normal $p$-complement: a normal subgroup of order $|G|/|P|$. The hypothesis first makes $P$ abelian. It also makes fusion inside $P$ trivial. To prove this, suppose $u,v\in P$ and $u^g=v$, using $u^g=g^{-1}ug$. Both $P$ and $P^g$ lie in $C_G(v)$: the former because $P$ is abelian, and the latter because $v\in P^g$. They are Sylow subgroups of that centralizer. Choose $c\in C_G(v)$ with $(P^g{)}^c=P$. Then $gc\in N_G(P)$ and $u^{gc}=v^c=v$, while the central-normalizer hypothesis gives $u^{gc}=u$. Thus $u=v$.

Evaluate transfer to the abelian group $P$ on $u\in P$. In the action of $u$ on right cosets, a cycle of length $r$ contributeswith $t_r=t_0$. This is a conjugate of $u^r$ inside $P$ and hence equals $u^r$ by the fusion argument. Summing the cycle lengths therefore givesThe exponent is coprime to $p$, so this power map is an automorphism of the finite abelian $p$-group $P$. Transfer is consequently onto $P$, and its kernel is normal of subgroup index $|P|$ and of order prime to $p$. This proves the theorem. Notice also that a normal $p$-complement consists exactly of the elements of prime-to-$p$ order: they map trivially to the $p$-group quotient, while all elements of the complement have such order. It is therefore unique and a characteristic subgroup.

Now let $G\le A_p$ be primitive of degree $p=2q+1$, with $q$ prime. Every nontrivial normal subgroup of a faithful primitive group action is transitive: its orbits form a $G$-invariant block partition, and singleton orbits would make it trivial. Such a subgroup has order divisible by $p$. A Sylow $p$-subgroup $P$ has order $p$, since $p!$ contains only one factor $p$. The normalizer of a regular $p$-cycle in $S_p$ is the affine group $x\mapsto ax+b$ on ${\mathbb{F}}_p$, of order $p(p-1)$. Translations are even, whereas a generating multiplier gives a $(p-1)$-cycle and is odd. Its normalizer in $A_p$ therefore has order $p(p-1)/2=pq$.

Suppose $G$ is not simple and choose $1<N◃G$ proper. It is transitive, so we can choose $P\le N$. The Frattini argument gives $G=N{N}_G(P)$: for each $g$, the subgroup $P^g$ is a Sylow subgroup of $N$ and can be conjugated back to $P$ within $N$. Since $N_G(P)$ contains $P$ and lies in the order-$pq$ normalizer, it has order $p$ or $pq$. If its order were $p$, the Burnside theorem applied to $G$ would give a normal $p$-complement. A nontrivial such complement cannot be transitive because its order is prime to $p$, contradicting primitivity. It must be trivial, giving $G=P$, contrary to the choice of $N$. Thus $|N_G(P)|=pq$.

The intersection $N_N(P)$ again has order $p$ or $pq$. If it had order $pq$, the whole normalizer would lie in $N$ and the Frattini equality would give $G=N$. Hence $N_N(P)=P$. Applying Burnside transfer within $N$ gives its normal $p$-complement $K$. By the uniqueness argument, $K$ is a characteristic subgroup of $N$, and so normal in $G$. Primitivity again forces this prime-to-$p$ subgroup to be trivial. Thus $N=P$ is normal in $G$, and $G=N_G(P)$ has order $pq$. The even primitive groups of safe-prime degree conclusion is

A subnormal series of a finite group is a chain $1=G_0◃G_1◃\cdots ◃G_r=G$. A normal series of a group requires every $G_i$ normal in the whole group. This distinction matters: a composition series is a subnormal series with nontrivial simple factors, whereas a chief series is a strict, unrefinable normal series of a group, whose factors are minimal normal subgroups of the appropriate ambient quotients. Both exist in a finite group, by successively choosing maximal proper normal subgroups or a maximal chain of ambient-normal subgroups. The direct-product result proved above shows that chief factors are powers of isomorphic simple groups, rather than necessarily simple themselves.

The Jordan–Hölder theorem says that any two composition series have the same multiset of simple factors, up to isomorphism and order. Here is an induction proof. The assertion is trivial for the identity group. Let $A,B$ be the penultimate terms of two series for $G$; they are maximal proper normal subgroups. If $A=B$, apply induction there. Otherwise $AB=G$, since its image is a nontrivial normal subgroup in each simple quotient. Put $C=A\cap B$. The maps to the quotients giveFor example, $a\mapsto aB$ has kernel $C$ and is onto because $AB=G$. Take a composition series of $C$ and append $A$, obtaining a composition series of $A$ because $A/C$ is simple; do the same with $B$. Induction inside these smaller groups shows that the factors in the two original series are, in both cases, the factors of $C$ together with $G/A$ and $G/B$. This proves the theorem and justifies the phrase Jordan–Hölder factors independently of a chosen series.

The derived series is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$. Its terms are characteristic subgroups, hence normal. A group is soluble if this series reaches the identity. Equivalently it admits a subnormal series with abelian factors. One direction uses the derived series itself. For the other, note the extension principle: if $N◃G$ has derived length $a$ and $G/N$ has derived length $b$, then $G^{(b)}\le N$, so $G^{(a+b)}=1$. Applying this principle along an abelian-factor series proves solubility.

Subgroups and quotients of a soluble group are soluble: $H^{(i)}\le G^{(i)}$ for a subgroup, and the commutator subgroup of a quotient is the image of the commutator subgroup. A finite soluble simple group is abelian, since its commutator subgroup is normal and cannot equal the whole group if the derived series terminates. An abelian simple finite group is cyclic of prime order. Therefore a finite group is soluble exactly when all its Jordan–Hölder factors are cyclic of prime order: necessity follows from the subgroup/quotient property, and sufficiency follows from the extension principle. Its chief factors are elementary abelian, by the minimal-normal-subgroup result.

For nilpotence, the lower central series is ${\gamma }_1(G)=G$, ${\gamma }_{i+1}(G)=[{\gamma }_i(G),G]$. A group is nilpotent if ${\gamma }_{c+1}=1$ for some $c$. An ascending central series $1=N_0\le \cdots \le N_c=G$ satisfies $[N_{i+1},G]\le N_i$. Reversing a terminating lower central series gives such a central series. Conversely, induction gives ${\gamma }_i(G)\le N_{c-i+1}$, so any such central series forces ${\gamma }_{c+1}=1$. The upper central series $Z_0=1$, $Z_{i+1}/Z_i=Z(G/Z_i)$ contains every ascending central series term by term, and thus reaches $G$ exactly in the nilpotent case. Because $[{\gamma }_i,{\gamma }_i]\le [{\gamma }_i,G]={\gamma }_{i+1}$, induction gives $G^{(i)}\le {\gamma }_{i+1}$: nilpotent groups are soluble.

For finite groups there is the stronger structural characterization: a finite group is nilpotent exactly when it is the direct product of groups formed from its Sylow subgroups. To prove the forward direction, the upper central series gives the normalizer condition. If $H<G$, choose its first term $Z_i$ not contained in $H$, with $Z_{i-1}\le H$. An element of $Z_i\setminus H$ has all group commutators with $H$ in $Z_{i-1}$, so normalizes $H$. Hence $H<N_G(H)$.

For a Sylow subgroup $P$, its normalizer is self-normalizing. Indeed, if $x$ normalizes $N_G(P)$, the two Sylow subgroups $P^x,P$ of that normalizer are conjugate there; adjusting $x$ by an element of $N_G(P)$ puts it in $N_G(P)$ already. Applying the strict normalizer condition to $N_G(P)$ shows that it cannot be proper, so $P$ is normal. Normal Sylow subgroups of distinct primes commute, since their group commutators lie in their trivial intersection. Their product has cardinality $|G|$ and is direct.

Conversely, a nontrivial finite $p$-group has nontrivial centre: in the class equation every noncentral conjugacy class has size divisible by $p$, so $p$ divides the centre's order. Induct on its order through the quotient by its centre to obtain an upper central series. Thus every finite $p$-group is nilpotent, and a finite direct product of groups of them is nilpotent because its upper central series is obtained componentwise.

A finite supersolvable group has a normal series of a group with cyclic prime-order factors. It is soluble, but the ambient-normal condition is stronger than the condition on a composition series. Finite nilpotent groups are supersolvable: a $p$-group has a central subgroup of order $p$, and induction in its quotient builds an ambient-normal prime-factor series; direct products of groups of the Sylow factors combine these series. The converses fail. The group $S_3$ is supersolvable via $1<C_3<S_3$, but is not nilpotent because its Sylow $2$-subgroups are not normal. The group $A_4$ is soluble via $1<V_4<A_4$, but is not supersolvable: no subgroup of order two is normal, since the three involutions are conjugate, and no subgroup of order three is normal, since there are four such subgroups. In this example $1<C_2<V_4<A_4$ is a composition series whose first $C_2$ is not normal in $A_4$, while $1<V_4<A_4$ is a chief series with a nonsimple chief factor $V_4$.

Finally, a perfect group satisfies $G^{\prime }=G$. In any finite group, the derived series eventually stabilizes at a perfect characteristic subgroup, because a descending sequence of finite subgroup orders eventually stops decreasing. It is trivial exactly when the group is soluble. A nonabelian simple group is perfect: its nontrivial commutator subgroup is normal and must be the whole group. Thus series distinguish successively stronger notions, summarize the building blocks through Jordan–Hölder factors, and separate soluble behaviour from a persistent perfect part.

For a set of primes $\pi$, a Hall $\pi$-subgroup $H$ of a finite group is a subgroup whose order uses only primes in $\pi$ and whose subgroup index uses only primes outside $\pi$. Equivalently, $|H|$ is the full $\pi$-part of $|G|$. A Hall subgroup is a Hall $\pi$-subgroup for some $\pi$. The theorem of P. Hall for finite soluble groups states: Hall $\pi$-subgroups exist for every $\pi$, any two are conjugate, and every $\pi$-subgroup is contained in a Hall $\pi$-subgroup.

For the existence proof, first give the explicit coprime splitting calculation needed in the induction. Let $N$ be elementary abelian of field characteristic $r$, normal in $E$, with $r\nmid |E/N|$. Write $Q=E/N$ and use additive notation on $N$. A normalized section $s:Q\to E$ gives an action $x\cdot n=s(x)ns(x{)}^{-1}$ independent of its choice, since $N$ is abelian. Write $s(x)s(y)=f(x,y)s(xy)$. Associativity givesSum over $z\in Q$, put $S(x)={\sum }_{z}f(x,z)$, and divide by $|Q|$ in the vector space $N$. This is legitimate because $r\nmid |Q|$. With $b(x)=|Q{|}^{-1}S(x)$ one obtainsThe new section $s^{\prime }(x)=(-b(x))s(x)$ therefore has zero multiplication defect and is a group homomorphism. Its image is a complement to $N$.

We will also need a single conjugation carrying one complement to the other. If two homomorphic sections differ by $d(x)$, their difference satisfies $d(xy)=d(x)+x\cdot d(y)$. Summing over $y$ gives $d(x)=c-x\cdot c$, where $c=|Q{|}^{-1}{\sum }_{y}d(y)$. Thus the second section is $cs(x)c^{-1}$. This proves, by explicit averaging, both coprime splitting over an elementary abelian normal subgroup and the existence of such a conjugation by an element of $N$.

Now induct on $|G|$ for Hall subgroup existence in soluble groups. A minimal normal subgroup $N$ of a finite soluble group is elementary abelian of some field characteristic $r$. Indeed, $N^{\prime }$ is a characteristic subgroup of $N$ and normal in $G$; it cannot equal $N$ because $N$ is soluble, so it is trivial. In this abelian group, a nontrivial primary component is a characteristic subgroup, forcing a single prime, and the characteristic subgroup annihilated by that prime forces exponent $r$.

By induction $G/N$ has a Hall $\pi$-subgroup $\overline{H}$, with preimage $M$. If $M<G$, induction in $M$ supplies a Hall $\pi$-subgroup of $M$, which is also Hall in $G$ because $[G:M]$ is prime to all primes in $\pi$. If $M=G$, the quotient is a $\pi$-group. When $r\in \pi$, $G$ itself is a $\pi$-group. When $r\notin \pi$, the averaging construction splits off $N$, and its complement has order $|G/N|$ and is the required Hall $\pi$-subgroup. This completes the existence proof, including the identity group and empty prime set.

A Sylow basis is a family $\{P_p:p\mid |G|\}$ containing one Sylow subgroup for each prime, with $P_p{P}_q=P_q{P}_p$ for each pair. This is permutability of subgroups, not commutation of all their elements. The product of any subfamily is a subgroup: commuting the factors as sets proves closure and closure under inversion. Inductively its order is the product of the prime-power orders, since the newly adjoined factor has trivial intersection with the earlier product. It is thus the corresponding Hall subgroup.

Prove existence of a Sylow basis by induction on $|G|$, again using elementary abelian $N$ of field characteristic $r$. Choose a basis $\{{\overline{P}}_p\}$ of $G/N$, taking ${\overline{P}}_r=1$ if $r$ does not divide that quotient's order. The product $\overline{H}={\prod }_{p\ne r}{\overline{P}}_p$ is its Hall $r^{\prime }$-subgroup. In its full preimage $T$, the averaging lemma supplies a complement $H$ to $N$, mapping isomorphically to $\overline{H}$. For $p\ne r$, let $P_p$ be the inverse image of ${\overline{P}}_p$ under this isomorphism within $H$. Let $P_r$ be the full preimage of ${\overline{P}}_r$ in $G$. These are Sylow subgroups of $G$. The other-prime pairs permute because they do so in $\overline{H}$. For $P_r$ and $P_p$, their product is contained in the full preimage of the subgroup ${\overline{P}}_r{\overline{P}}_p$ and has exactly that preimage's order $|N||{\overline{P}}_r||{\overline{P}}_p|$. Equality follows, so the product is a subgroup and these two factors permute as well. This constructs a basis.

To obtain a simultaneous conjugation of two bases, take two bases. Their images in $G/N$ are bases: $N$ lies in every Sylow $r$-subgroup, and the other-prime Sylow subgroups project injectively. By induction, conjugate one basis so that all corresponding images agree. The products of their non-$r$ factors are then two complements to $N$ in the same preimage $T$ of their common Hall $r^{\prime }$-subgroup. The averaging lemma for complements supplies a single element of $N$ conjugating these complements. In the resulting common complement each corresponding Sylow subgroup is the unique lift of its agreed quotient subgroup, so all non-$r$ factors agree. The $r$-factors already equal the full preimage of their common quotient factor and are preserved by conjugation from $N$. Thus all Sylow bases of a finite soluble group are conjugate.

In $S_4$ there are three Sylow $2$-subgroups of order eight: the action on the three pair-partitions of four points has kernel $V_4$, and these subgroups are the preimages of the three order-two subgroups of $S_3$. There are four Sylow $3$-subgroups, one for each three-point support. Every choice $P_2,P_3$ has trivial intersection and product of cardinality $8\cdot 3=24$, so $P_2{P}_3=S_4=P_3{P}_2$. ThereforeFor $E=V⋊S_4$, $V={\mathbb{F}}_5^4$, the Sylow $5$-subgroup is uniquely $V$. In a Sylow basis the product $P_2{P}_3$ is a Hall $5^{\prime }$-subgroup, hence a complement to $V$. Conversely, a basis of any complement together with $V$ is a basis of $E$, because $V$ is normal. The averaging lemma makes all complements conjugate by $V$. The stabilizer of the standard complement $S_4$ under this conjugation is $C_V(S_4)$: if a vector normalizes it, its group commutator with each complement element lies in both $V$ and that complement, hence is trivial. The fixed vectors under the natural coordinate permutation action are exactly $(a,a,a,a)$, so $|C_V(S_4)|=5$. There are consequently $5^4/5=125$ complements, and each has twelve bases. The complement is uniquely recovered from its basis as $P_2{P}_3$, so no basis is counted twice. The counting Sylow bases in a coprime elementary abelian extension formula yields

For a pro-p group, the lower p-series is the descending sequence of closed characteristic subgroupsThe bar means closure of the subgroup generated by the displayed powers and group commutators. Throughout this question $p$ is odd. For a finitely generated powerful pro-p group, group commutator collection gives $P_i(G)=G^{p^{i-1}}$, each $P_i$ is powerful, and $\Phi (P_i)=P_{i+1}$. All these subgroups are open. A uniform pro-p group is such a group with all subgroup indices $[P_i:P_{i+1}]$ equal to $[G:P_2]$. We use the standard powerful-group torsion criterion: a finitely generated powerful pro-p group is uniform exactly when it is a torsion-free group. For a uniform group of generator number $d$, the indices are $p^d$, and $x\mapsto x^{p^a}$ is a homeomorphism from $G$ onto $G^{p^a}$.

We record explicitly the collection facts used for the operations below. With $U_b=G^{p^b}=P_{b+1}$, one hasThe second statement is the power-congruence property: the $p$th-power map is an isomorphism on adjacent uniform layers, and collection gives $(yu{)}^p\equiv y^p{u}^p(\mathrm{mod}{U}_{b+2})$ for $u\in U_b$. Thus the first nonzero layer of a difference moves down exactly one step under a $p$th power. Iterating gives the asserted equivalence and also justifies taking roots of congruences. These facts follow from powerful-group group commutator collection, using divisibility by $p$ of $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ and the successive-layer power isomorphisms; they are the uniform-group structure results being used here.

The transported addition on a uniform pro-p group isThe product is in $U_n$ and has a unique $p^n$th root in $G$. For the comparison at $n=1$ define $x{+}_{0}y=xy$. The numbered proofs below establish associativity, approximate commutativity and compatibility of these operations.

The intrinsic Lie algebra of a uniform pro-p group has underlying set $G$, zero equal to the group identity, additive inverse $x^{-1}$, and operationsHere the group commutator is $[x,y]=x^{-1}{y}^{-1}xy$. For $\lambda \in {\mathbb{Z}}_p$, $x^{\lambda }$ is the limit of $x^{m_j}$ for integers $m_j\to \lambda$; the pro-$p$ topology makes it well defined. The addition sequence is Cauchy by the compatibility proved below. The group commutator estimate puts $[x^{p^n},y^{p^n}]$ in $U_{2n+1}$, so its displayed root exists and lies in $U_1$. Collection of group commutators shows that successive bracket approximants differ in layers tending to the identity, giving the other limit. The group commutator identities give additivity of the limiting bracket, and the Hall-Witt identity gives its Jacobi identity. Approximate commutativity makes limiting addition commutative. The resulting lattice $L_G$ is a free module over ${\mathbb{Z}}_p$ of rank $d(G)$, and $[L_G,L_G]\le p{L}_G$.

A powerful Lie algebra over p-adic integers, in the convention relevant here, is a Lie algebra $L$ whose underlying module is finite free over ${\mathbb{Z}}_p$ and whose bracket satisfies $[L,L]\le pL$. For $p=2$ the condition is instead $[L,L]\le 4L$. For odd $p$, define multiplication on its underlying set by the convergent Baker--Campbell--Hausdorff formulaAn iterated bracket of degree $m$ lies in $p^{m-1}L$. The BCH coefficient bound loses at most $\lfloor (m-1)/(p-1)\rfloor$ powers of $p$, so all higher-degree terms are integral, and their valuations tend to infinity. This explains both convergence and why multiplying a lattice with small bracket stays inside the lattice. Formal associativity of BCH passes to the convergent series, the identity is $0$, the inverse is $-x$, and every group power satisfies $x^{\ast m}=mx$. The quotients by $p^{a}L$ are finite p-groups. Hence the resulting compact group is a finitely generated torsion-free group and a pro-p group, and its group commutator lies in $pL$. Thus $(L,\ast )$ is a uniform pro-p group, with $P_i(L,\ast )=p^{i-1}L$ and successive index $p^{{\mathrm{rank}}_{{\mathbb{Z}}_p}L}$.

The standard uniform-group/Lie-lattice correspondence identifies these constructions as mutually inverse: $L_{(L,\ast )}\cong L$ and $(L_G,\ast )\cong G$. To see why the inverse limits recover the lattice operations, the addition approximant for a BCH group is $p^{-n}\mathrm{BCH}(p^{n}x,p^{n}y)$: its linear term is $x+y$ and every higher term tends to zero. Similarly its group commutator approximant has limiting term $[x,y]$. In the other direction, a local logarithm in the complete normed group algebra identifies the intrinsic operations with algebra addition and commutator and identifies group multiplication with BCH; the uniform power maps extend the identification from a sufficiently small power subgroup to all of $G$. Continuous group homomorphisms respect powers, unique roots and limits, while Lie algebra homomorphisms respect every BCH term, so the equivalence also preserves morphisms.

The objective is a continuous faithful representation of a uniform pro-p group $G$ in some ${\mathrm{GL}}_N({\mathbb{Z}}_p)$. The proof proceeds from the intrinsic Lie lattice to a small matrix group, then extends the resulting representation to all of $G$. This last step matters: a representation defined only on an open subgroup does not yet establish linearity of the original group.

Let $L=L_G$ be the Lie algebra of a uniform pro-p group and put $\mathfrak{g}={\mathbb{Q}}_p{\otimes }_{{\mathbb{Z}}_p}L$. The uniform-group/Lie-lattice correspondence identifies the group law on $L$ with BCH. The substantial finite-dimensional Lie-theoretic input is the Ado theorem: every finite-dimensional Lie algebra over a characteristic-zero field admits a faithful finite-dimensional representation. Apply it to obtainAn Adjoint representation would not suffice here, since it kills the centre; Ado supplies injectivity even in central directions.

Choose a p-adic lattice $\Lambda$ spanning $V$. Since $L$ has finitely many lattice generators, their representing matrices have bounded denominators. There is therefore an integer $a\ge 0$ such thatOn this matrix lattice the series $\exp X={\sum }_{n\ge 0}{X}^n/n!$ and $\log (1+Y)={\sum }_{n\ge 1}(-1{)}^{n+1}{Y}^n/n$ converge and are inverse. For example $v_p(n!)\le n/(p-1)$, so the valuations of the exponential terms tend to infinity with this choice of $ϵ$; the logarithm terms do as well. They take values in $1+p^{ϵ}\mathrm{End}(\Lambda )$ and in $p^{ϵ}\mathrm{End}(\Lambda )$ respectively. The formal identity $\exp (\mathrm{BCH}(X,Y))=\exp X\exp Y$ holds for the convergent series.

Under the correspondence, $U=G^{p^a}$ has Lie lattice $p^{a}L$ and is an open characteristic subgroup. Definewhere ${\log }_G$ denotes the identification of $U$ with $p^{a}L$. A Lie homomorphism preserves BCH, so the exponential identity proves $\eta (uv)=\eta (u)\eta (v)$. If $\eta (u)=I$, the matrix logarithm gives $\rho ({\log }_{G}u)=0$; faithfulness of $\rho$ gives $u=1$. Thus $\eta$ is a continuous faithful integral representation of $U$.

To extend it, take the moduleValues on a transversal of $U\backslash G$ determine $f$, so $W$ is a free module over ${\mathbb{Z}}_p$ of rank $[G:U]\dim V$. These functions are continuous, since on each open coset they are given by the continuous representation $\eta$. Define $(T_{g}f)(x)=f(xg)$. The equivariance condition is preserved, and $T_g{T}_h=T_{gh}$. In a transversal basis each $T_g$ is a permutation of blocks followed by matrices from $\eta (U)$, so it belongs to $\mathrm{GL}(W)$ and depends continuously on $g$.

Faithfulness can be checked directly. If $g\notin U$, right multiplication moves the coset $U$ to a different coset, so $T_g$ moves a function supported on one block and cannot be the identity. If $g\in U$ and $T_g=1$, evaluation at $1$ gives $f(g)=f(1)$ for every $f$, whereas equivariance gives $f(g)=\eta (g)f(1)$. Since $f(1)$ is arbitrary, $\eta (g)=I$, and hence $g=1$. ThereforeCompactness of $G$ makes this continuous injection a homeomorphism onto its closed image. This proves the linearity of a uniform pro-p group for the full group, including its centre, and explains why taking a sufficiently small power lattice permits matrix integration before finite-index induction restores the whole group.

For a finite group, define the Frattini subgroup bythe intersection is $G$ when there are no maximal subgroups, which for a finite group happens only for $G=1$. A second definition is the set of non-generators: an element $x$ such that $⟨S,x⟩=G$ always implies $⟨S⟩=G$.

If $x\in \Phi (G)$ and $H=⟨S⟩<G$, put $H$ in a maximal subgroup $M$. Then $x\in M$, so $⟨S,x⟩\le M<G$. Conversely, if $x\notin \Phi (G)$, some maximal subgroup $M$ omits it; maximality gives $⟨M,x⟩=G$, although $M\ne G$. This proves equivalence.

Every group automorphism permutes maximal subgroups, so $\Phi (G)$ is a characteristic subgroup, in particular a normal subgroup. Alsoa proper $H$ would lie in a maximal subgroup containing both factors.

Set $F=\Phi (G)$ and take a Sylow subgroup $P$ of $F$. For $g\in G$, normality of $F$ makes $P^g$ a Sylow subgroup of $F$. Conjugating inside $F$ therefore gives the Frattini argument factorization $G=F{N}_G(P)$. The preceding generating property forces $N_G(P)=G$. Thus every Sylow subgroup of $F$ is normal, and part (i) shows that $F$ is a nilpotent group.

Now suppose $G$ is a finite p-group. The normalizer condition makes every maximal subgroup $M$ normal: $M<N_G(M)$ and maximality force $N_G(M)=G$. The quotient $G/M$ has no proper nontrivial subgroup, hence is cyclic of prime order; its p-power order makes that prime $p$. Consequently every such $M$ containswhere the two factors are the commutator subgroup and the subgroup generated by pth powers. Thus $D\le \Phi (G)$.

The quotient $G/D$ is an elementary abelian group, hence a vector space over ${\mathbb{F}}_p$. If $x\notin D$, extend the nonzero vector $xD$ to a basis and choose a linear functional nonzero on it. Its kernel pulls back to a subgroup of index $p$, a maximal subgroup omitting $x$. Therefore $x\notin \Phi (G)$, provingThe two factors in the last formula are multiplied; the superscripted expression in the TeX aid is not the formula printed in the PDF.