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If an abelian subgroup $A$ has index $p^a$, the centralizer of any $1\ne g\in A$ contains $A$, so the class size of $g$ divides $p^a$. The prime-power conjugacy-class obstruction to simplicity rules this out. If $A=1$, the group is a p-group and has nontrivial center of a group, also ruling out nonabelian simplicity.

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.

For a finite group $G$, choose one representative $x_i$ from each conjugacy class with more than one element. The class equation iswhere $Z(G)$ is the center of a group and $C_G(x_i)$ is the centralizer of $x_i$. Each central element contributes its own one-element conjugacy class. Every remaining conjugacy class contributes its size, given by the orbit-stabilizer theorem. The sum counts each element exactly once. See McKernan, Group actions and automorphisms, lemma 23.6.

For conjugacy classes of the symmetric group on three letters, take $x_1=(12)$ and $x_2=(123)$. Their centralizers have sizes two and three, and the center of a group has size one, givingFor an abelian group, every centralizer is the whole group, every conjugacy class has size one, and the sum over noncentral representatives is empty.

The power of this counting identity is that the sizes of conjugacy classes constrain group structure. For a nontrivial finite p-group with group order $p^n$, every noncentral conjugacy class has size a positive power of $p$, by Lagrange's theorem. The class equation therefore gives $|Z(G)|\equiv 0(\mathrm{mod}p)$. Since the identity element lies in the center of a group, the center of a group must have at least $p$ elements. This obtains an element commuting with the entire group without constructing one. See McKernan, lemma 23.7.

Historically, these ideas belong to the nineteenth-century development of permutation groups and normal subgroups. Milne's historical opening records Galois's introduction of normal subgroups in 1832 and Jordan's systematic treatment of permutation groups in 1870. This is context for the conjugation action and the class equation, rather than a claim that either author first stated the modern formula.^[ref]

Induct on the order of the finite p-group. Its center of a group is nontrivial unless the group itself is trivial. Quotient by the center to obtain a smaller finite p-group and lift a central series supplied by induction. Prepending the center gives a central series for the original group: the first term is central, and the remaining commutator-containment conditions are exactly the lifted quotient conditions. Hence every finite p-group is a nilpotent group. Since central factors are Abelian, every nilpotent group is also a soluble group.

For an odd prime number $p$, a finite p-group that is a product of $d(G)$ cyclic subgroups is a powerful p-group. Indeed, $H=G/G^p$ has exponent of a finite group $p$ and the same minimal generator number $d$. Its cyclic factorization gives $|H|\le p^d$, while its Frattini quotient has order $p^d$. Thus $\Phi (H)=1$, $H$ is an elementary abelian group, and $[G,G]\subseteq G^p$. The condition on the number of factors is essential: the nonabelian Heisenberg group over ${\mathbb{F}}_p$ has exponent $p$ and is a product of three cyclic subgroups, although its minimal generator number is two.

Every complex irreducible representation of a finite p-group is induced from a one-dimensional representation of a subgroup. Induct on the group order. Nonfaithful representations reduce to a smaller quotient. For a faithful representation of a nonabelian group, choose an abelian normal subgroup not contained in the centre. Restriction cannot be isotypic, since it would make that subgroup scalar and faithfulness would then force it central. The Clifford correspondence induces the representation from a proper subgroup; induction there and transitivity of induction complete the proof. Abelian groups have only one-dimensional complex irreducible representations.

The class sizes sum to $|G|=36$. A degree-one character is a homomorphism into ${\mathbb{C}}^{\times }$. The supplied ${\chi }_2$ has order four and ${\chi }_2^2={\chi }_1$, so $1,{\chi }_2,{\chi }_1,{\bar{\chi }}_2$ are four distinct irreducible characters. A finite group has as many complex irreducible characters as conjugacy classes, so there are exactly two more. Their degrees $d,e$ satisfy $d^2+e^2=36-4=32$, whose only positive-integer solution is $d=e=4$.

The column orthogonality give ${\sum }_{\chi }|\chi (g){|}^2=|C_G(g)|$. Each of the last three classes has centralizer order $36/9=4$, already exhausted by the four degree-one rows. Thus both new rows vanish on $C_4,C_5,C_6$.

To justify that their remaining values are real, put $P=\ker {\chi }_2$. The supplied values show $P=C_1\cup C_2\cup C_3$, a normal subgroup of order nine, with $G/P\cong C_4$. A group of order nine is Abelian, by the standard p-group center argument and the fact that a cyclic quotient by the center forces the group to be Abelian. If $P$ were cyclic of order nine, conjugation through $G/P$ would act by an automorphism of order dividing both four and $|\mathrm{Aut}(C_9)|=6$, hence dividing two. Its nontrivial conjugacy classes could not have size four. Therefore $P\cong C_3^2$.

Let $h$ map to a generator of $G/P$, and let $A$ be its conjugation action on the two-dimensional vector space $P$ over ${\mathbb{F}}_3$. The two nonzero orbits have length four, so $A^2$ fixes no nonzero vector, while $A^4=I$. Since $t^2-1$ has distinct roots in characteristic three, the involution $A^2$ is diagonalizable; the absence of a $+1$ eigenvector gives $A^2=-I$. Thus conjugation by $h^2$ inverts every element of $P$. Every element in $C_2,C_3$ is therefore conjugate to its inverse, so every character value there is real.

Write one new row as $(4,a,b,0,0,0)$. Its orthogonality with the trivial row gives $4+4a+4b=0$, so $a+b=-1$. Its squared character inner product one gives $16+4a^2+4b^2=36$, so $a^2+b^2=5$ and $ab=-2$. Thus $a,b$ are $1,-2$ in some order. Column orthogonality with the identity column forces the other row's entries to be interchanged. Up to relabelling these two new characters, the completed character table isThe reality argument prevents an unjustified assumption when solving the orthogonality equations. This is character table from a cyclic quotient and two elementary-abelian orbits.

An Abelian group has the central series $1<G$, since $Z(G)=G$. In any central series, each factor is a subgroup of a center and hence Abelian. A finite normal series with Abelian factors makes the group soluble: inductively, if $N$ is soluble and $G/N$ is Abelian, then $G^{\prime }\subseteq N$, and successive derived groups eventually vanish. Hence every nilpotent group is soluble.

The group $S_3$ has the normal series $1<A_3<S_3$ with cyclic factors of orders three and two, so it is soluble. Its center is trivial, since a permutation commuting with both $(12)$ and $(23)$ must fix all three letters. Every nontrivial nilpotent group has nontrivial center: its first nonidentity central-series term is contained in $Z(G)$. Consequently $S_3$ is soluble but not nilpotent.

For a finite p-group, use induction on its order. The identity group is the base case. A nontrivial $G$ has nontrivial $Z=Z(G)$, so $G/Z$ has smaller order, still a power of $p$. By induction it has a central series. Lift that series to subgroups of $G$ and prepend $1<Z$. The first factor is central by definition of $Z$, and centrality of every lifted factor is exactly centrality of the corresponding quotient factor. This gives a central series for $G$. This proves finite p-groups are nilpotent:

For the Abelian case of Cauchy theorem for groups, use induction on $|G|$. Choose $a\ne e$. If $p$ divides $m=|a|$, then $a^{m/p}$ has order $p$. Otherwise $⟨a⟩$ is a nontrivial normal subgroup of order prime to $p$, and $G/⟨a⟩$ has smaller order still divisible by $p$. Induction gives a coset of order $p$; any lift $b$ has order divisible by $p$, because the order of its image divides its own order. A suitable power of $b$ then has order $p$. This uses no classification of finite Abelian groups.

For a finite p-group, the class equation iswhere the sum is over noncentral conjugacy classes. Each index is a positive power of $p$, hence divisible by $p$. Therefore $p$ divides $|Z(G)|$, and the center of a group is nontrivial.

If $|G|=p^n$ with $n>1$ and $Z(G)\ne G$, this is already a proper nontrivial normal subgroup. If $Z(G)=G$, the group is Abelian; the established case of Cauchy theorem for groups gives a subgroup of order $p$, which is normal and proper. Thus

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.

A subnormal series is a finite chain $1=G_r⊴G_{r-1}⊴\cdots ⊴G_0=G$, in which each term is normal in the preceding term. Its factors are the quotient groups $G_i/G_{i+1}$. A normal series of a group imposes the stronger condition that every term be normal in $G$. These conditions should not be conflated: inserting a subgroup normal in an intermediate term need not preserve normality in the whole group. A refinement inserts additional terms, and two series are equivalent when their nontrivial factors can be paired by group isomorphisms.

The basic refinement mechanism is the Zassenhaus butterfly lemma. For $A^{\prime }⊴A$ and $B^{\prime }⊴B$, let $C=A\cap B$ and $D=(A^{\prime }\cap B)(A\cap B^{\prime })$. Both factors of $D$ are normal in $C$, so $D⊴C$. The map from $C$ onto $A^{\prime }C/A^{\prime }(A\cap B^{\prime })$ has kernel $D$: if an element of $C$ equals $a^{\prime }{b}^{\prime }$ with $a^{\prime }\in A^{\prime }$ and $b^{\prime }\in A\cap B^{\prime }$, then $a^{\prime }=c{b}^{\prime -1}\in A^{\prime }\cap B$. Conversely each product in $D$ is in the kernel. The first isomorphism theorem therefore givesThe denominators are normal in their numerators: conjugation by $C$ preserves them, while conjugation by $A^{\prime }$ or $B^{\prime }$ changes their elements only by a factor already in that normal subgroup.

Given two subnormal series $(G_i)$ and $(H_j)$, refine the first by $G_{i+1}(G_i\cap H_j)$, and the second by $H_{j+1}(H_j\cap G_i)$. The endpoints are $G_i,G_{i+1}$ and $H_j,H_{j+1}$ respectively, and the Zassenhaus butterfly lemma pairs the successive factors indexed by $(i,j)$ and $(j,i)$. This proves the Schreier refinement theorem.

A composition series is a strict subnormal series all of whose factors are simple groups. Every finite group has one: choose a maximal proper normal subgroup, repeat inside it, and use decreasing orders to terminate. A simple factor cannot admit a nontrivial refinement. Applying the preceding refinement theorem to two composition series proves the Jordan–Hölder theorem: their lengths and multisets of simple factors agree, up to group isomorphism. The factors are the Jordan–Hölder factors, and their number is the composition length. If $N⊴G$, lifting a composition series of $G/N$ and adjoining one for $N$ gives

These factors do not determine the whole group, because they do not determine how its group extensions are assembled. For example, $C_4$ and the Klein four-group both have two factors $C_2$, but one contains an element of order four and the other does not. The symmetric group $S_3$ has a composition series $1<C_3<S_3$, with factors $C_3,C_2$, whereas the simple group $A_5$ has the single factor $A_5$.

A chief series is a maximal strict normal series of a group; each chief factor is a minimal nontrivial normal subgroup in the appropriate quotient of $G$. Finite chief factors are characteristically simple. To justify their structure, let $N$ be any nontrivial finite characteristically simple group and choose a minimal nontrivial normal subgroup $M$ of $N$. Its distinct images under automorphisms of $N$ are minimal normal subgroups, so distinct ones intersect trivially and commute. Choose a maximal family giving an internal direct product of groups. If any other image were not contained in that product, its intersection with the product would be trivial by minimality, and it could be added, a contradiction. Hence this product is characteristic and equals $N$. Each factor $M$ is simple: a proper nontrivial normal subgroup of $M$ would be normal in the whole direct product $N$, contrary to the choice of $M$. Thus $N$ is a direct product of groups of isomorphic finite simple groups. This applies to chief factors by conjugation from $G$.

Chief and composition factors need not coincide. In $A_4$, the chain $1<V_4<A_4$ is a chief series, since conjugation cycles the three order-two subgroups of $V_4$. A composition series instead inserts any one of those subgroups, giving factors $C_2,C_2,C_3$. Its inserted order-two subgroup is normal in $V_4$ but not in $A_4$. Chief series retain the ambient group's conjugation information that composition series discard.

The derived series gives a canonical criterion for a group to be soluble: put $G^{(0)}=G$ and $G^{(i+1)}=[G^{(i)},G^{(i)}]$. Its terms are characteristic, and each factor is Abelian. A finite group is soluble exactly when this series reaches the identity, equivalently when all its composition factors are cyclic of prime order. For one direction, the derived series supplies Abelian factors, whose simple refinements are cyclic of prime order. Conversely an extension of two soluble groups is soluble: if $(G/N{)}^{(r)}=1$, then $G^{(r)}\subseteq N$; if $N^{(s)}=1$, then $G^{(r+s)}=1$. Induction up a composition series with cyclic prime factors proves the other direction. In a finite soluble group, each minimal normal subgroup is elementary Abelian: its characteristic commutator subgroup must be trivial, its primary components force a single prime, and its characteristic subgroup of $p$th powers must be trivial. Thus every chief factor is a vector space over some ${\mathbb{F}}_p$. In a general finite group the Jordan–Hölder factors can instead be nonabelian simple groups.

The central series demand more than Abelian factors: each factor must be central in the corresponding quotient of the whole group. Define the upper central series by $Z_0=1$ and $Z_{i+1}/Z_i=Z(G/Z_i)$, and the lower central series by ${\gamma }_1=G$ and ${\gamma }_{i+1}=[{\gamma }_i,G]$. A group is nilpotent exactly when the upper series reaches $G$, equivalently when the lower series reaches the identity. Indeed, for any ascending central series $1=C_0\le \cdots \le C_c=G$, induction gives $C_i\subseteq Z_i$ and ${\gamma }_{c+1-i}\subseteq C_i$; reversing the lower series gives an ascending central series when it terminates. The least such $c$ is the nilpotency class. A central series has Abelian factors, so every nilpotent group is soluble; the converse fails, as $S_3$ is soluble but has trivial group center.

For a finite p-group, the class equation shows that the group center is nontrivial, since every noncentral conjugacy class has size divisible by $p$. Applying the same argument to successive central quotients makes the upper central series reach the group. More generally a finite group is nilpotent exactly when it is the direct product of groups of its Sylow subgroups. Here is the nontrivial direction. A central series gives the normalizer condition for nilpotent groups: for a proper subgroup $H$, choose the first $Z_i$ not contained in $H$. Then $[Z_i,H]\subseteq Z_{i-1}\subseteq H$, so $Z_i\subseteq N_G(H)$ and $N_G(H)>H$. If a Sylow subgroup $P$ had proper normalizer $N$, then $N_G(N)>N$; but $P$ is the unique Sylow subgroup of $N$, hence characteristic there, and $N_G(N)\subseteq N_G(P)=N$, a contradiction. Thus all Sylow subgroups are normal, different ones commute because their intersection is trivial, and they form an internal direct product. Conversely each finite Sylow subgroup is nilpotent, and products of their central series give a central series for their direct product.

Composition series record simple building blocks, chief series record ambient normal structure, derived series detect solubility, and central series detect nilpotence. Their associated factors and extension data explain why the same group can have very different useful series.

Let $\mathrm{Irr}(G)$ denote the irreducible characters of finite-dimensional complex group representations of $G$. A generalized character, or virtual character, is an integral linear combination of these irreducible characters. A class function is a function constant on conjugacy classes. For class functions use the character inner productA p-elementary group is a direct product of groups $P\times C$, where $P$ is a p-group and $C$ is a cyclic group of order coprime to $p$. An elementary subgroup is a subgroup of this form for some prime $p$. Either factor may be trivial. This notion is different from an elementary abelian group.

Brauer's characterization of characters says that a complex class function $f$ on $G$ is a generalized character exactly when its restriction to every elementary subgroup is a generalized character there. Equivalently,The character orthogonality relations make these inner products the coefficients of the restrictions in their irreducible bases. The theorem detects the integral character lattice. An ordinary character of a representation additionally requires every global irreducible coefficient $⟨f,\chi {⟩}_G$ to be nonnegative; integrality alone characterizes virtual characters, not necessarily actual representations.

Conversely, the same induction-norm formula shows that $I_G(\phi )=N$ for every nonprincipal $\phi \in \mathrm{Irr}(N)$. Conjugation by each $g\notin N$ fixes only $1_N$, so Brauer's permutation lemma says that it fixes only the identity conjugacy class of $N$. A nonidentity element centralizing $g$ would provide another fixed class. Therefore $C_N(g)=1$ for every $g\notin N$.

We first prove that $|N|$ and $[G:N]$ are coprime. If a prime $p$ divided both, a Sylow p-subgroup $S$ of $G$ would have $1<S\cap N<S$. The nontrivial normal subgroup $S\cap N$ meets $Z(S)$ nontrivially: the class equation for the conjugation action of the p-group $S$ on $S\cap N$ makes the number of fixed elements divisible by $p$, and one of them is the identity. Choose $1\ne z\in N\cap Z(S)$ and $s\in S\setminus N$. Then $z\in C_N(s)$, contradicting the previous conclusion.

Thus $N$ is a normal subgroup that is also a Hall subgroup. The Schur-Zassenhaus theorem supplies a group complement $H$, meaning $G=NH$ and $N\cap H=1$. The theorem's existence assertion requires only normality and coprimality of order and index. Here $H$ is nontrivial and proper by the assumptions on $N$.

For $n\in N\setminus \{1\}$, suppose $1\ne h\in H\cap nH{n}^{-1}$. Write $h=n{h}^{\prime }{n}^{-1}$ with $h^{\prime }\in H$. Projection to $G/N$ gives $hN=h^{\prime }N$, hence $h=h^{\prime }$ because $H\to G/N$ is injective. This makes $n$ centralize $h\notin N$, which is impossible. Hence $H\cap nH{n}^{-1}=1$. Writing an arbitrary element outside $H$ as $n{h}_0$ reduces its conjugate intersection to this case. It follows that $H$ is a Frobenius complement and $G$ is a Frobenius group.

Moreover, for each $1\ne h\in H$, the map $v\mapsto vh{v}^{-1}$ from $N$ into the coset $Nh$ is injective, since $C_N(h)=1$, and hence bijective. Thus every element outside $N$ belongs to a conjugate of $H$, while no nonidentity element of $N$ does. The Frobenius kernel is exactly $N$. This completes both directions of the Frobenius kernel criterion by irreducible induction.

Write $o(\theta )$ for the order of the determinant character of an irreducible character $\theta$. An extension has the same degree and restricts exactly to $\theta$, rather than just containing it as a constituent.

Extension theorem. Thompson's coprime character extension theorem says that if $N◃G$, $\theta \in \mathrm{Irr}(N)$ is $G$-invariant, and $\gcd (\theta (1),[G:N])=1$, then $\theta$ extends to $G$ exactly when its determinant character extends to a linear character of $G$. In particular, if also $\gcd (o(\theta ),[G:N])=1$, there is a unique extension with determinant order coprime to the index. For a p-group quotient, the version used below gives a canonical extension whenever both the degree and the determinant order are prime to $p$. This theorem is an assertion of extendibility, not the false statement that every invariant irreducible character extends.

The p-nilpotency lemma. Definewhere $m_p$ is the largest power of $p$ dividing an integer $m$. Isaacs' character-degree criterion for p-nilpotency assertsEquivalently, $s_p(G{)}_p=h_p(G)$. We prove both the divisibility and the criterion.

Let $N=O^p(G)$ be the p-residual of a finite group, the least normal subgroup with p-group quotient. It can also be described as the subgroup generated by all p-prime-order elements: every homomorphism to a p-group kills those elements, and after quotienting by their generated subgroup the image of every element has p-power order, so the quotient is a p-group by Cauchy's theorem.

The residual is p-perfect, meaning $O^p(N)=N$. Indeed $O^p(N)$ is characteristic in $N$, hence normal in $G$, and $G/O^p(N)$ is an extension of the two p-groups $N/O^p(N)$ and $G/N$. Minimality of $N$ forces equality. Consequently $N/N^{\prime }$ has order prime to $p$, since otherwise it would have a quotient of order $p$. Every linear character of $N$, in particular every determinant character of $N$, therefore has order prime to $p$.

Let $\mathcal{T}$ consist of the $G$-invariant irreducible characters $\theta$ of $N$ of degree prime to $p$. Each extends to an irreducible character $\hat{\theta }$ of $G$ by the extension theorem. The irreducible characters of $G$ lying over $\theta$ are exactlyHere is the representation argument for this description. If a representation lies over the invariant $\theta$, its $N$-module is copies of the module for $\theta$. Identify it with that module tensored with its multiplicity space. Dividing the $G$-action by the fixed extending action on the first factor leaves an action on the multiplicity space, trivial on $N$. Irreducibility is equivalent to irreducibility of that quotient action, proving the displayed parametrization.

Every irreducible degree of the p-group $G/N$ is a power of $p$, since it divides the group order. Thus the p-prime-degree characters lying over $\theta$ are exactly those with $\nu$ linear, and there areof them. For the last equality, $G/N{G}^{\prime }$ is the largest abelian p-group quotient of $G$, hence the p-primary part of its abelianization. In particular all these characters have degree $\theta (1)$.

Conversely, for any $\chi \in \mathrm{Irr}(G)$ of p-prime degree, Clifford theory writes its restriction to $N$ as $e$ times the sum of a single orbit of irreducible characters. The orbit size divides $[G:N]$ and also divides $\chi (1)$, so it is one; the constituent $\theta$ is invariant and has p-prime degree. Hence $\theta \in \mathcal{T}$. The preceding extension parametrization now forces the quotient character to be linear. We have therefore obtained the exact identityThis proves the required divisibility without assuming p-nilpotency.

The p-group $G/N$ acts on $\mathrm{Irr}(N)$ by conjugation. Every nonfixed orbit has size divisible by $p$, and degrees are constant on each orbit. Fixed characters whose degrees are divisible by $p$ also contribute zero modulo $p$. The sum of squares of irreducible degrees consequently givesThe quotient is prime to $p$ exactly when $N$ has order prime to $p$. Since $G/N$ is a p-group, that is exactly the assertion that $N$ is a normal p-complement. Conversely, any normal p-complement equals $O^p(G)$ because it contains all p-prime-order elements and has p-group quotient. This finishes the lemma.

Detection from the group algebra. For finite groups $G$ and $H$ with isomorphic complex group algebras, the Artin–Wedderburn theorem givesThe simple matrix-block sizes and their multiplicities are determined by the algebra isomorphism. Hence $G$ and $H$ have identical irreducible degree multisets. They have the same order, from the degree-square sum, and the same number of linear characters, which equals $|G:G^{\prime }|$. Thus $s_p$ and $h_p$ coincide for every prime $p$. The lemma shows that one group is p-nilpotent exactly when the other is.

A finite group is nilpotent exactly when it has a normal p-complement for every prime dividing its order. For completeness, if all those complements exist, set $L_q={\bigcap }_{p\ne q}{K}_p$, where $K_p$ is the normal p-complement. The diagonal map $G/L_q\to {\prod }_{p\ne q}G/K_p$ is injective, so $[G:L_q]$ is prime to $q$. At the same time, containment in each $K_p$ removes every prime other than $q$ from $|L_q|$. Thus $L_q$ is a normal Sylow q-subgroup. Distinct normal Sylow subgroups commute, giving the direct-product decomposition. Conversely, the direct-product decomposition of a finite nilpotent group provides all the complements. We concludeThis is the assertion that the complex group algebra detects p-nilpotency and hence nilpotency. It needs the multiplicities of the degrees, not just their distinct values.

Normal-complement consequence. If $p$ divides every nonlinear irreducible degree, the p-prime-degree characters are exactly the linear characters. ThereforeThe lemma yields a normal p-complement immediately:This proves Thompson's character-degree normal complement theorem, including the abelian case where the hypothesis on nonlinear characters is vacuous.

An inverse system of topological spaces over a directed set $I$ consists of spaces $X_i$ and continuous maps $f_{ij}:X_j\to X_i$ for $i\le j$, with $f_{ii}=1$ and $f_{ik}=f_{ij}{f}_{jk}$. Its inverse limit is a space $X$ with compatible continuous projections ${\pi }_i$ such that every compatible family of continuous maps $u_i:Y\to X_i$ factors uniquely through a continuous map $u:Y\to X$. To construct it, takewith the subspace topology inherited from the product topology. The map $u(y)=(u_i(y))$ has values in $X$ and is continuous by the definition of the product topology; its coordinates force uniqueness. If $X^{\prime }$ is another inverse limit, the two universal properties give continuous maps $X\to X^{\prime }$ and $X^{\prime }\to X$ whose composites have all the same projections as the identity. Uniqueness makes these composites the identity. The inverse limit exists and is unique up to a unique projection-preserving homeomorphism. An arbitrary inverse limit may be empty; nonemptiness requires additional hypotheses.

Use the usual Hausdorff convention for a topological group. An inverse limit of finite discrete groups is a closed subgroup of their product: each compatibility condition is closed because the finite factors are Hausdorff spaces. The Tychonoff theorem gives compactness, and distinct tuples are separated by a clopen coordinate cylinder, so the limit is totally disconnected. Coordinatewise multiplication and inversion are continuous, giving a profinite group.

Conversely, let $G$ be a compact Hausdorff space and a totally disconnected topological group. We use the standard topological fact that a compact Hausdorff space that is totally disconnected has a basis of clopen sets. Given an identity neighbourhood, choose a clopen identity neighbourhood $U$ inside it. Compactness of $U$ and continuity of multiplication give a symmetric identity neighbourhood $V$ with $VU\subseteq U$: cover $U$ by finitely many neighbourhoods on which multiplication by a sufficiently small identity neighbourhood stays inside $U$, then intersect those identity neighbourhoods. Symmetry gives $vU=U$ for $v\in V$. Consequently $H=\{g:gU=U\}$ is a subgroup containing $V$, hence an open subgroup, and $H\subseteq U$ since $1\in U$. Its open cosets cover the compact group, so there are finitely many. The normal core $N={\bigcap }_{g\in G}gH{g}^{-1}$ is a finite intersection of open subgroups, hence an open normal subgroup contained in $U$. Thus the open normal subgroups form an identity-neighbourhood basis and have trivial intersection.

The canonical group homomorphism $G\to {\underleftarrow{\lim }}_{N}G/N$ is continuous and injective. For a compatible tuple of quotient elements, its inverse-image cosets in $G$ are closed and have the finite intersection property: a finite family is refined by the quotient for the intersection of its normal subgroups. Compactness gives an element in all those cosets, proving surjectivity. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The compact totally disconnected and inverse-limit definitions of a profinite group agree.

Index the finite nonempty sets $\mathcal{P}(N)$ by open normal subgroups, with the transition from $M$ to $N$ when $M\subseteq N$ given by image under $G/M\to G/N$. Images of Sylow subgroups under surjections of finite groups are Sylow subgroups. These transitions are surjective: if $S\le G/N$ is a Sylow subgroup, choose a Sylow subgroup in its full inverse image in $G/M$. It maps onto $S$, and its order has the full $p$-part of $|G/M|$, so it is a Sylow subgroup of $G/M$ too. The transitions respect composition, giving an inverse system of finite sets.

Its inverse limit is nonempty. Indeed, finitely many compatibility conditions can be satisfied by choosing one Sylow subgroup in a quotient refining all the involved quotients and taking its images; compactness of the product of the finite sets then gives a compatible family $(P_N{)}_N$. DefineThis is a closed subgroup of $G$. Its image in $G/N$ is exactly $P_N$: for a prescribed $s\in P_N$, each finite collection of conditions defining $P$ together with ${\pi }_N(g)=s$ is satisfiable in a common refining quotient, because the transitions on the chosen Sylow subgroups are surjective. The finite intersection property again supplies $g\in P$. Hence $P\cong {\underleftarrow{\lim }}_N{P}_N$ is a pro-p group and $PN/N=P_N$ for every $N$.

Finally, if a pro-p subgroup $Q$ contains $P$, its image in every finite quotient is a p-group containing the Sylow subgroup $P_N$, so that image equals $P_N$. Thus every $q\in Q$ belongs to every ${\pi }_N^{-1}(P_N)$, whence $Q\subseteq P$. The constructed $P$ is a maximal pro-p subgroup.

Initially $G_1=G$ is powerfully embedded in itself. If $G_i$ is powerfully embedded in $G$, then $[G_i,G]\subseteq G_i^p$ and consequently $G_{i+1}=G_i^p$. The preliminary result says that $G_i^p$ is again powerfully embedded in $G$, completing the induction.

For a finite p-group $A$, the Frattini subgroup is $\Phi (A)=A^p[A,A]$. Here $[G_i,G_i]\subseteq [G_i,G]\subseteq G_i^p$, soSince the Frattini subgroup of a nontrivial finite p-group is proper, this series eventually reaches $1$. Thus $P_j=G^{p^j}$, with iterated power subgroups, terminates.

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.

Use induction on the order of the finite p-group. The trivial group is nilpotent. For $G\ne 1$, the permitted center theorem gives $Z=Z(G)\ne 1$, so the quotient group $G/Z$ is a smaller p-group. By induction it has a central seriesLet $H_i$ be the inverse image of ${\overline{H}}_i$ under the quotient map, so $H_0=Z$ and $H_s=G$. Centrality in the quotient gives $[H_i,G]\le H_{i-1}$ for $i\ge 1$: the image of every such group commutator lies in ${\overline{H}}_{i-1}$. Also $[Z,G]=1$. Thereforeis a central series of $G$. This proves every finite p-group is nilpotent, including the case $Z=G$, when the quotient is trivial and the series is simply $1\le G$.

Induct on the order of the p-group. Every complex irreducible representation of an abelian group is one-dimensional: commuting group operators preserve a common eigenspace, and irreducibility makes that eigenspace the whole space. This handles the abelian cases.

If $V$ is not faithful, it is an irreducible representation of a smaller quotient $G/K$, where $K$ is its kernel. Induction expresses it as induced from a one-dimensional representation of a subgroup of $G/K$. Pulling back the subgroup and inflating the character gives the same induction statement for $G$, because the coset spaces of the quotient subgroup and its inverse image agree.

Now suppose $V$ is faithful and $G$ is nonabelian. Use the allowed abelian normal subgroup $A$ not contained in the centre. If restriction to $A$ were isotypic, its irreducible type would be one-dimensional, so every $a\in A$ would act as a scalar. It would therefore commute with every representing matrix of $G$. Faithfulness would imply $ga=ag$ in $G$ for every $g$, contradicting the choice of $A$. By part (a), $V={\mathrm{Ind}}_H^{G}W$ for an irreducible representation of a proper subgroup $H$. Induction on $H$ gives $W={\mathrm{Ind}}_K^H\chi$ with $\chi$ one-dimensional. Transitivity of induction yieldsThis proves the monomiality of irreducible representations of finite p-groups.

The Hall theorem for soluble groups says that for every prime set $\pi$, a finite soluble group has Hall $\pi$-subgroups; they are all conjugate, and every $\pi$-subgroup is contained in one. A Hall $\pi$-subgroup has only primes in $\pi$ in its order and no primes in $\pi$ in its index.

For a proof outline, induct on the group order and choose a minimal normal subgroup $N$. In a finite soluble group this is elementary abelian of some characteristic $p$: its derived subgroup is characteristic and solvability forces it to be trivial; primary components and the subgroup of elements of order dividing $p$ are then characteristic, so minimality gives an elementary abelian p-group. Choose a Hall $\pi$-subgroup of $G/N$ and let $U$ be its preimage. If $p\in \pi$, $U$ is itself a Hall $\pi$-subgroup. If $p\notin \pi$, $U/N$ has order prime to $p$ and coprime splitting over an elementary abelian normal subgroup supplies a complement to $N$ in $U$, of the required Hall order.

Here the complement fact has a short averaging proof. Choose a section of $U/N=H$ and write its multiplication defect as $f(h,k)\in N$, in additive notation. Associativity givesSince $|H|$ is invertible on $N$, averaging over $t$ expresses the defect as $f(h,k)=b(h)+hb(k)-b(hk)$; adjusting the section by $-b$ gives a homomorphic section. Two such complements differ by $d(hk)=d(h)+hd(k)$. Averaging this identity gives $d(h)=b-hb$, so they are conjugate by $b\in N$. Inductive conjugacy in the quotient followed by this complement conjugacy proves Hall conjugacy. To embed an arbitrary $\pi$-subgroup $Q$, first put its quotient inside the chosen quotient Hall subgroup. In the coprime case $Q$ and the corresponding subgroup of a complement are both complements to $N$ inside $NQ$, hence are $N$-conjugate. In the other case $Q\le U$ directly. This proves Hall conjugacy and embedding in finite soluble groups as well as existence.

Now prove the assertion about maximal subgroups by induction on $|G|$. Suppose a nontrivial normal subgroup $K$ is contained in $A$. If $K\nleqq B$, maximality gives $KB=G$, and since $K\le A$ this gives $AB=G$. If $K\le B$, both $A/K$ and $B/K$ are maximal in $G/K$; induction yields either their product equal to $G/K$ or their conjugacy, lifting respectively to $AB=G$ or to conjugacy of $A,B$. The same applies with their roles reversed.

It remains to suppose that no nontrivial normal subgroup of $G$ lies in either maximum. Let $K$ be minimal normal, an elementary abelian p-group. Since $K$ lies in neither maximum, $KA=KB=G$. The intersection $K\cap A$ is normalized by $A$ and by the abelian group $K$, hence by $G$; it is therefore trivial. Likewise $K\cap B=1$. Thus both $A$ and $B$ complement $K$. Their action on $K$ is irreducible, since an invariant subspace would be a proper nontrivial normal subgroup of $G$. It is also faithful: $C_A(K)$ is normalized by $A$ and centralized by $K$, so is a normal subgroup of $G$ contained in $A$ and must be trivial.

If $G/K=1$, these assumptions force $A=B=1$ and $G$ to have prime order. Otherwise choose $L>K$ normal in $G$ with $L/K$ minimal normal in $G/K$. It is an elementary abelian r-group. Under $A\cong G/K$, its copy $R=A\cap L$ is a nontrivial normal subgroup of $A$. The prime $r$ differs from $p$: a p-group acting on the characteristic-p vector space $K$ has nonzero fixed points, by counting its orbits on $K$ modulo $p$. Its fixed subspace would be $A$-invariant, hence all of $K$ by irreducibility, contrary to faithfulness of $R$.

Therefore $A\cap L$ and $B\cap L$ are Sylow r-subgroups of $L=K(A\cap L)=K(B\cap L)$. They are conjugate in $L$. Conjugate $B$ so that these intersections coincide as $R$. Since $R⊲A$, $N_G(R)\ge A$. It cannot equal $G$, since $R$ would then be a nontrivial normal subgroup contained in $A$. Maximality gives $N_G(R)=A$, and the same argument gives $N_G(R)=B$. The conjugated maxima are equal. HenceThis is the factorization-or-conjugacy property of maximal subgroups of a finite soluble group.