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For the Integer Heisenberg group, choose $x=I+E_{12}$, $y=I+E_{23}$ and central $z=I+E_{13}$. Every element has a unique form $y^c{z}^b{x}^a$, and $xy=zyx$. The group algebra is therefore the displayed skew Laurent polynomial ring, with $\sigma (z)=z$. The base is a commutative Laurent polynomial ring and an integral domain; the skew Hilbert basis theorem gives both-sided Noetherianity and the leading coefficient of a product proves that there are no zero divisors.

In the group algebra $\mathbb{C}{S}_n$, setThese elements commute: for each triple $a<i<j$, the only overlapping contributions cancel as $[(ai),(ij)+(aj)]=0$. They generate the Gelfand–Tsetlin algebra, and on a standard Young tableau vector $X_r$ acts by the Content of a Young-diagram cell containing $r$.

On the set underlying a uniform pro-p group, define scalar multiplication by $\lambda x=x^{\lambda }$, and define addition and the Lie bracket byRoots are the unique roots in the appropriate power layers. The limits give a powerful Lie lattice. In a suitable complete normed group algebra, the p-adic logarithm identifies these operations with ordinary addition and the associative-algebra commutator.

Modular representation theory studies representations of a finite group $G$ over a field whose characteristic divides $|G|$. The group algebra is then generally nonsemisimple, and projective modules, radicals, blocks and vertices replace complete reducibility as central tools.

In characteristic $p$, the augmentation ideal $I$ of the group algebra of a finite $p$-group is nilpotent. Choose a central element $z$ of order $p$ and put $t=z-1$, so $t^p=0$. The quotient by $t$ is the group algebra of the smaller quotient group. Induction gives $I^m\subseteq tkP$ for some $m$, hence $I^{mp}=0$. For a normal $p$-subgroup $P$ of $G$, conjugation stability gives $(IkG{)}^m=I^{m}kG$, proving nilpotence of the kernel of $kG\to k(G/P)$.

Use $[g]$ for the basis element of the group algebra $A={\mathbb{F}}_p[{\mathbb{Z}}_p]$, so $[g][h]=[g+h]$ and $[0]=1$. The kernel of the augmentation map is the augmentation ideal $I$. This algebra must not initially be identified with a Laurent polynomial ring: the additive group ${\mathbb{Z}}_p$ is not abstractly generated by the integer $1$.

For $r\ge 1$, map $A$ onto ${\mathbb{F}}_p[{\mathbb{Z}}_p/p^r{\mathbb{Z}}_p]$ and call its kernel $J_r$. It is generated by $[g]-1$ for $g\in p^r{\mathbb{Z}}_p$: differences of basis elements in one coset have the form $[h]([g]-1)$. Since each such $g=p^{r}h$,by the characteristic-$p$ binomial identity. Thus $J_r\subseteq I^{p^r}$.

The quotient group is a cyclic group of order $p^r$, generated by the image of $1$. Its group algebra isThe image of $I$ is $(T)$ and its $p^r$th power vanishes, proving the reverse inclusion. HenceThis is cyclic finite quotients of the p-adic augmentation filtration.

For any $n\ge 0$, choose $p^r>n+1$. The finite quotient then identifies $I^n/I^{n+1}$ with $(T^n)/(T^{n+1})$, a one-dimensional space generated by the image of $([1]-1{)}^n$. The degree-one symbol $X$ of $[1]-1$ consequently gives a surjective graded map ${\mathbb{F}}_p[X]\to {\mathrm{gr}}_{I}A$, and no power of $X$ vanishes in its degree. ThusThe same finite quotients show that the I-adic completion is ${\mathbb{F}}_p[[T]]$, and its associated graded ring is identical. The proof therefore covers the completed group algebra interpretation too, while proving the literal ordinary-algebra statement without assuming completion. Applying part ii to a ring would require its actual completeness, not just this polynomial associated graded ring.

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.

Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra, choose a Cartan subalgebra $\mathfrak{t}$ and a set $R^{+}$ of positive roots, and write $W$ for the Weyl group. Let $\lambda$ be a dominant integral weight and $L(\lambda )$ the finite-dimensional irreducible highest-weight representation with that highest weight. If $L(\lambda {)}_{\mu }$ denotes its weight space of weight $\mu$, its formal character is $\mathrm{ch}L(\lambda )={\sum }_{\mu }\dim L(\lambda {)}_{\mu }{e}^{\mu }$. The symbols $e^{\mu }$ belong to the group algebra of the weight lattice and multiply by $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$; they are formal symbols, not matrix exponentials.

Put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the half-sum of positive roots, and let $\ell (w)$ be the Coxeter length of $w$, the least number of simple reflections expressing it. Then $(-1{)}^{\ell (w)}=\det (w)$ in the real reflection representation. The Weyl character formula statesBy the Weyl denominator formula, the denominator is $e^{\rho }{\prod }_{\alpha \in R^{+}}(1-e^{-\alpha })$, giving the equivalent expressionThe quotient can be interpreted in the fraction field of the formal group algebra; the theorem says it simplifies to the finite formal character above. In particular the apparent denominator does not mean that the representation has infinitely many weights.

Put $U_0=G$ and $U_{j+1}=U_j^p$, where powers mean closed generated subgroups. The finite-quotient arguments of Question 3 pass to a powerful pro-p group: in particular $U_j$ is powerfully embedded, the power map is onto $U_{j+1}$, andFor the root assertion, apply the finite result to every finite quotient and use the finite intersection property on the closed sets of possible roots in the compact group. Each $U_j$ is finitely generated by the appropriate powers of a finite generating tuple, and $U_j/U_{j+1}$ is finite and elementary abelian. Thus the $U_j$ are open. They form an identity-neighbourhood basis: in any finite continuous quotient their images form a terminating lower p-series, so some $U_j$ is contained in the kernel. In particular ${\bigcap }_j{U}_j=1$. Collection also givesOne can see the extra index directly by induction from $[U_a,G]\subseteq U_{a+1}$: collect $[v,w^p]$ for $w\in U_b$. Its leading factor $[v,w{]}^p$ advances one layer, and the remaining iterated commutators, with their binomial coefficients, lie at least as deep. Products and limits preserve the inclusion. This is the usual strongly central property of the lower p-series.

A uniform pro-p group is a finitely generated powerful pro-p group for which all the indices $|U_j:U_{j+1}|$ equal $|G:G^p|$. The preliminary congruence gives surjective linear mapsIf $G$ is uniform, these maps are bijective. If $x\ne y$, write $y=xu$ and choose the unique layer $j$ with $u\in U_j\setminus U_{j+1}$. Bijectivity of ${\theta }_j$ and the power congruence imply $y^p(x^p{)}^{-1}\in U_{j+1}\setminus U_{j+2}$, so $x^p\ne y^p$. Thus the power map is injective. A finite-order element of a pro-p group has $p$-power order; repeated injectivity of the power map excludes every nonidentity such element. Hence a uniform group is a torsion-free group.

Conversely, suppose some ${\theta }_j$ has a nonzero kernel. Choose $x\in U_j\setminus U_{j+1}$ with $x^p\in U_{j+2}$. At step $r\ge j+1$, if the current $x_r$ satisfies $x_r^p\in U_{r+1}$, surjectivity of ${\theta }_r$ supplies $u_r\in U_r$ withSet $x_{r+1}=x_r{u}_r$. The power congruence gives $x_{r+1}^p\in U_{r+2}$. Since the corrections tend to $1$ in the complete profinite group, the $x_r$ converge to $z$, with $z^p=1$ and $z\equiv x(\mathrm{mod}{U}_{j+1})$. In particular $z\ne 1$. A torsion-free group cannot have such an element, so every ${\theta }_j$ is injective and all layer indices are equal. This proves the root correction criterion for uniformity andFor a uniform group the power maps are continuous bijections $U_j\to U_{j+1}$, hence homeomorphisms. Their inverses define the unique roots used below. Comparison at the first differing layer also proves the power congruences in a uniform pro-p group:

The intrinsic Lie algebra of a uniform pro-p group has the same underlying set as $G$, zero equal to the group identity, and operationsHere $x^{\lambda }$ is the limit of integral powers approaching $\lambda$. The first root is taken in $U_n$ and the second in $U_{2n}$; the commutator actually lies in $U_{2n+1}$. The following faithful algebra realization verifies convergence and the Lie algebra axioms, rather than assuming them as part of the definition.

Choose a minimal tuple $a_1,\dots ,a_d$ of topological generators. The layer isomorphisms ${\theta }_j$ show that the images of $a_i^{p^j}$ are bases of $U_j/U_{j+1}$. The ordered-coordinate argument of Question 3 therefore gives a continuous bijectionExistence follows by successive digits and compactness; uniqueness follows by the first digit at which the coordinates differ, because its images are linearly independent in the corresponding power layer. Thus this map is a homeomorphism.

Here are details of the norm needed on the group algebra. In ${\mathbb{Z}}_p[[G]]$ put $b_i=a_i-1$. Its elements have unique ordered expansions ${\sum }_{\alpha }{c}_{\alpha }{b}_1^{{\alpha }_1}\cdots b_d^{{\alpha }_d}$ with $c_{\alpha }\in {\mathbb{Z}}_p$. To verify this description, identify the ordered coordinates of $G$ with ${\mathbb{Z}}_p^d$. Elements of the completed group algebra are integral measures on this space; their values on the binomial polynomials ${\prod }_i\left(\genfrac{}{}{0ex}{}{{\lambda }_i}{{\alpha }_i}\right)$ are precisely the coefficients $c_{\alpha }$. The finite-difference identity pairs the ordered monomial $b^{\beta }$ with this binomial polynomial as ${\delta }_{\alpha \beta }$. The Mahler expansion of a continuous function therefore proves both existence and uniqueness of the measure with the specified coefficients. This also proves that the ordinary rational group algebra embeds in these expansions.

Choose $1/(p-1)<s<1$ and setThis defines the weighted group algebra of a uniform pro-p group. Its multiplication is submultiplicative in the valuation convention, $v(AB)\ge v(A)+v(B)$. Indeed, $a_i^p-1=p{b}_i+\left(\genfrac{}{}{0ex}{}{p}{2}\right)b_i^2+\cdots +b_i^p$ has valuation $1+s$; collection of ordered coordinates gives the same lower bound for $g-1$ when $g\in U_1$. Interchanging two adjacent $b_i$ in a product introduces a correction coming from $[a_i,a_j]-1$, of valuation at least $1+s>2s$. Iterating the interchange rule never lowers the total weight; the corrections have strictly increasing weight, so infinite collection converges. The same argument with $p^j$-th powers showsThe equality follows from the first nonzero coordinate digit: its leading terms are $p^j{\lambda }_i^{\prime }{b}_i$, at least one with unit ${\lambda }_i^{\prime }$, and cannot cancel in the ordered expansion. Completing the rational group algebra in this norm produces a complete normed algebra $A$ in which $G$ embeds faithfully and with its original topology.

Since $v(g-1)\ge s>1/(p-1)$, the seriesconverge on the relevant balls in $A$ and are inverse there. Their convergence follows from $v_p(r!)\le (r-1)/(p-1)$ and $v_p(r)\le (r-1)/(p-1)$; formal one-variable identities prove that they are inverse. Put $X=\log x$ and $Y=\log y$. The Baker--Campbell--Hausdorff formula in $A$ now givesbecause every term of degree at least two tends to zero after this rescaling. Similarly expansion of $\exp (-p^{n}X)\exp (-p^{n}Y)\exp (p^{n}X)\exp (p^{n}Y)$ givesThe leading term is the indicated associative-algebra commutator and all terms of degree at least three tend to zero. These expressions are the logarithms of the unique roots in the intrinsic formulas. The image $\log G$ is compact and hence closed in $A$, so both limits belong to $\log G$, and the inverse logarithm takes the limits back to $G$. Also $\log (x^{\lambda })=\lambda \log x$. Consequently $\log$ identifies the intrinsic operations with addition, scalar multiplication and commutation in $A$. The Jacobi identity follows by expanding the associative products; thus $L_G$ is indeed a Lie algebra over ${\mathbb{Z}}_p$.

The commutator roots in the intrinsic formula belong to $U_1$, so their limit gives $[L_G,L_G]\subseteq \log U_1=p{L}_G$. Iterating the power congruence gives $x^{p^n}{y}^{p^n}\equiv (xy{)}^{p^n}(\mathrm{mod}{U}_{n+1})$. Taking the unique roots shows that intrinsic addition modulo $U_1$ is the original group product modulo $U_1$. Consequently $L_G/p{L}_G$ is the original elementary abelian quotient $G/U_1$, and the images of the $a_i$ form a basis. Successively removing digits expresses every element uniquely as an additive sum ${\sum }_i{\lambda }_i{a}_i$: existence uses completeness, while a nonzero relation would have a first nonzero digit contradicting the basis in $U_j/U_{j+1}$. Hence $L_G$ is a free module of rank $d$ over the p-adic integers and is a powerful Lie lattice.

Conversely a powerful Lie algebra over p-adic integers, for odd $p$, is a finite free ${\mathbb{Z}}_p$-module $L$ with a bilinear alternating Lie bracket satisfying the Jacobi identity and $[L,L]\subseteq pL$. DefineFor completeness, integrality and convergence can be verified without guessing which individual denominators are units. In ${\mathbb{Q}}_p\otimes L$ set $M=p^{-1}L$. Then $[M,M]\subseteq M$, so its integral universal enveloping algebra has an ordered basis by the Poincaré-Birkhoff-Witt theorem. Its coefficientwise $p$-adic norm is submultiplicative, and $L=pM$ consists of elements of valuation at least one. The homogeneous degree-$r$ terms in $\log (\exp X\exp Y)$ have coefficients of valuation at least $-\lfloor (r-1)/(p-1)\rfloor$: expand the logarithm, use $v_p(k!)\le (k-1)/(p-1)$ on each nonconstant exponential factor, and $v_p(\ell )\le (\ell -1)/(p-1)$ on a product of $\ell$ such factors. Therefore evaluating degree $r$ on $X,Y\in pM$ gives valuation at leastwhich tends to infinity and is at least two when $r\ge 2$. Each homogeneous term is a rational Lie polynomial by the formal Baker--Campbell--Hausdorff formula. The ordered enveloping-algebra basis shows that its intersection with ${\mathbb{Q}}_{p}M$ is exactly $M$, so the estimate places the sum in $L$, with the non-linear part in $pL$.

Associativity follows from the identity $\exp (X\ast Y)=\exp X\exp Y$ in the completed universal enveloping algebra. Its identity is $0$, its inverse is $-X$, and $X^{\ast m}=mX$, since the bracket of $X$ with itself is zero. The same coefficient estimate for the group commutator puts it in $pL$. Conjugation in the BCH group is $\exp (\mathrm{ad}X)$; each term $\mathrm{ad}(X{)}^r/r!$ preserves $p^{j}L$, because each bracket adds a factor of $p$ and $v_p(r!)<r$. Thus the submodules $p^{j}L$ are normal subgroups for $\ast$, their intersection is zero, and their finite quotients have $p$-power order. Thus $(L,\ast )$ is a powerful pro-p group, withA module basis topologically generates this group: it generates the quotient $L/pL$, and successive power digits lift that generation through all $p^{j}L$. The BCH construction therefore gives a uniform pro-p group.

Take finite-dimensional modules over $k$ and finite $R$-free lattices over the complete valuation ring, the usual integral modular representation theory convention. If a completion has several local factors, use the factor supporting the modular block. A block of a group algebra is an indecomposable two-sided ideal $B=RGb$, where $b$ is a primitive central idempotent. The identity decomposes as the sum of orthogonal block idempotents, and $M={⨁}_{b}bM$ for every module. Hence an indecomposable module belongs to $B$ exactly when $bM=M$.

A defect group of a block in characteristic $p$ is a maximal p-subgroup $D$ with ${\mathrm{Br}}_D(b)\ne 0$. Equivalently, it is minimal with $b\in {\mathrm{Tr}}_D^G((kG{)}^D)$; the trace criterion for defect groups and conjugacy are proved below. For a complete discrete valuation ring, use the corresponding reduced modular block; modular reduction of block idempotents preserves its defect groups. The equivalent bimodule description says that the module vertices of $B$, with $(x,y)$ acting by $a\mapsto xa{y}^{-1}$, are conjugates of $\Delta D=\{(d,d):d\in D\}$. This basic characterization follows from the relative-projectivity trace criterion for the bimodule action.

Fix a Sylow subgroup $P$ containing $D$. The regular group algebra is the permutation module $R[G]$ for $G\times G$, and multiplication by $b$ is its block projection. Restriction to a subgroup containing a chosen module vertex retains a summand with that module vertex: a module source at that module vertex is a summand of the module restriction, and the Mackey restriction formula and minimality prevent all the restricted summands from having smaller module vertices. Thus $B{\downarrow }_{P\times P}$ has an indecomposable summand with module vertex $\Delta D$.

For a finite p-group $Q$, the transitive permutation module $k[Q/L]$ is indecomposable: the group algebra of a p-group in characteristic p is local, and this module has a one-dimensional top, incompatible with a sum of two nonzero modules. Its module vertex is $L$, because it is induced from the trivial $L$-module and its Brauer quotient of a module at $L$ has the nonzero fixed coset $L$. The fixed-basis rule for this quotient follows by removing nonsingleton orbit sums, which are proper-subgroup traces. It excludes a smaller module vertex. Reduction gives the same statements for permutation lattices. Therefore Krull-Schmidt decomposition identifies the chosen block summand with an entire transitive $P\times P$ orbit in $G$.

Choose a point $c$ in that orbit whose stabilizer is exactly $\Delta D$. Being fixed by every $(d,d)$ gives $c\in C_G(D)$. Its full stabilizer isEquality with $\Delta D$ gives $P\cap c^{-1}Pc=D$. Conjugating, and using that $c$ centralizes $D$, provesThis is defect groups are centralizer-conjugate Sylow intersections.

If $Q◃G$ is a p-subgroup, then $QP$ is a p-subgroup for every Sylow $P$, so $QP=P$. It lies in both Sylow factors above. Thus every normal p-subgroup of $G$ is contained in $D$.

Finally put $N=N_G(D)$, choose $S\in {\mathrm{Syl}}_p(N)$, and extend it to a Sylow $P$ of $G$. Then $D\le S$ and $P\cap N=S$. Apply the proved intersection result to this $P$. Its element $c$ belongs to $C_G(D)\le N$, so $cP{c}^{-1}\cap N=cS{c}^{-1}$ is Sylow in $N$ too. Hence $D=S\cap cS{c}^{-1}$. The p-core $O_p(N)$ is contained in both factors, whereas normality of $D$ in $N$ gives the opposite inclusion. Therefore $D=O_p(N_G(D))$, so $D$ is a p-radical subgroup. The centralizer condition on the conjugating element is important for this deduction.

For a profinite abelian group $G$, its Iwasawa algebra is the completed group algebrawhere $U$ runs over open subgroups. The transition map sums the coefficients over the fibers of $G/V\to G/U$ when $V\subset U$. Consequently an element gives compatible values $\mu (gU)\in {\mathbb{Z}}_p$ on cosets, equivalently a finitely additive ${\mathbb{Z}}_p$-valued p-adic measure on the clopen sets of $G$. Conversely these values specify every finite-quotient coefficient. Integration of a continuous function with values in ${\mathbb{Z}}_p$ is obtained by uniform approximation by locally constant functions; the integrals converge because all measure values have $p$-adic absolute value at most $1$. Multiplication in the Iwasawa algebra becomes convolution of p-adic measures. This is its measure realization of an Iwasawa algebra.

The Mahler theorem states that each continuous function $g:{\mathbb{Z}}_p\to {\mathbb{Z}}_p$ has the unique expansion converging in the sense of uniform convergenceIts Mahler coefficients areand $\Vert g{\Vert }_{\infty }={\sup }_n|b_n{|}_p$. The same assertion holds for continuous functions with values in ${\mathbb{Q}}_p$ with coefficients in ${\mathbb{Q}}_p$.

Given $f(T)={\sum }_{n\ge 0}{a}_n{T}^n\in R$, define the bounded linear functionalThe sum converges because $a_n\in {\mathbb{Z}}_p$ and $b_n\to 0$. Its values on indicator functions of clopen sets give a p-adic measure, denoted $\lambda (f)$, withConversely, a p-adic measure determines these coefficients $a_n\in {\mathbb{Z}}_p$, and continuity allows termwise integration of every Mahler expansion. Thus the inverse iswhich is the Amice transform. These two constructions are inverse, proving the canonical bijection $\lambda :R\simeq \Lambda ({\mathbb{Z}}_p)$. In fact it is an isomorphism of rings: the identity $\left(\genfrac{}{}{0ex}{}{x+y}{n}\right)={\sum }_{i+j=n}\left(\genfrac{}{}{0ex}{}{x}{i}\right)\left(\genfrac{}{}{0ex}{}{y}{j}\right)$ identifies convolution of p-adic measures with multiplication of formal power series. For example, $\lambda (1)={\delta }_0$ and $\lambda (1+T)={\delta }_1$, where ${\delta }_a$ is the Dirac measure at $a$.

For the moment identity, let $\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$ denote a Stirling number of the second kind. The finite polynomial identitygives $\int x^{k}d\lambda (f)={\sum }_{n=0}^k{a}_{n}n!\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$. On the other hand, $D=(1+T)d/dT$ satisfies $D^k(1+T{)}^x=x^k(1+T{)}^x$ for every nonnegative integer $x$. Expanding $(1+T{)}^x$ and evaluating at $T=0$ yieldsComparison in the basis of binomial polynomials gives $(D^k{T}^n)(0)=n!\left\{\genfrac{}{}{0ex}{}{k}{n}\right\}$. Terms with $n>k$ have zero constant term after at most $k$ differentiations, so no infinite-series interchange is needed. ThereforeThe case $k=0$ says that the total mass is $f(0)$.

Choose $m\ge \ell (\lambda )$, pad the partition of an integer with zeros, and define its beta numbers byThey form a strictly decreasing set of nonnegative integers. Conversely, from its decreasing enumeration recover ${\lambda }_i={\beta }_i-m+i$. With $m=\ell (\lambda )$ these are exactly the first-column hook lengths, since $h_{i1}={\lambda }_i+{\lambda }_1^{\prime }-i={\lambda }_i+m-i$. Padding changes the numbers: increasing $m$ by one replaces $B$ by $\{0\}\cup (B+1)$, and leaves the represented partition of an integer unchanged.

Arrange positions $0,1,2,\dots$ on $p$ runners, putting position $r+pk$ on runner $r$ at height $k$. A bead occupies each position in the beta set. This is the abacus of a partition. The hook criterion in a beta set says that a bead $b$ and a gap $a<b$ correspond to a hook of length $b-a$; replacing $b$ by $a$ removes the corresponding rim hook. In particular removing a rim $p$-hook moves a bead from $b$ to the vacant position $b-p$ immediately above it on the same runner. Its leg length counts the beads in the intervening positions of the full linear numbering.

For an explicit example, take $\lambda =(4,2,1)$, $p=3$, and $m=3$. Its beta set is $\{6,3,1\}$. Runner zero has beads at $3,6$, runner one has a bead at $1$, and runner two is empty. Packing upwards gives $\{3,1,0\}$, which represents $(1)$; thus the three-core is $(1)$ and the weight is two.

Every bead slide reduces the sum of occupied positions, so repeated slides terminate. The number $n_r$ of beads on each runner is unchanged. At termination they must occupy exactlyon runner $r$, since any gap above a bead would allow a further slide. Hence the final beta set is uniquely determined by the initial runner counts, independently of the order of hook removals. Increasing the number of beads by $p$ replaces $B$ by $\{0,\dots ,p-1\}\cup (B+p)$, adds one bead to each runner, and gives exactly the same represented core after packing. General changes in padding follow by applying $B\mapsto \{0\}\cup (B+1)$; packing commutes with this operation up to the cyclic relabelling of runners. Therefore the result is independent of both removal order and padding:The weight of a partition is $w_p(\lambda )=(n-|C_p(\lambda )|)/p$.

A block of a group algebra is an indecomposable two-sided summand $F{S}_{n}e$, where $e$ is a primitive central idempotent. The identity decomposes into pairwise orthogonal such idempotents; a module belongs to a block when its idempotent acts as identity and the other block idempotents act as zero. For ordinary characters, use a splitting p-modular system $(K,\mathcal{O},F)$, or extend the residue field first: the block idempotents lift to $\mathcal{O}{S}_n$, and determine which ordinary irreducible $K{S}_n$-modules lie in each $p$-block. The symmetric group is split over its prime field, so extending $F$ introduces no change to the resulting partition of an integer labels.

The Nakayama block theorem, traditionally called the Nakayama Conjecture for symmetric groups, isThe simple modules in that block are exactly the $D^{\nu }$ with regular labels having that core. At fixed $n$, the common core also fixes the weight. Here is an essay proof route through central characters, specifying the main algebraic and combinatorial steps and why each is needed.

First introduce the commuting Young–Jucys–Murphy elementsTheir commutators vanish because contributions supported on each triple of letters cancel. Symmetric polynomials in the $J_i$ are central: for adjacent transpositions $s_i$, the relations $s_i{J}_i=J_{i+1}{s}_i-1$, $s_i{J}_{i+1}=J_i{s}_i+1$ show that $s_i$ commutes with symmetric expressions in $J_i,J_{i+1}$, and it commutes with the other $J_j$. Since these transpositions generate $S_n$, the symmetric expressions commute with the whole group algebra.

The substantial generation ingredient is the integral Jucys–Murphy description of the center of a symmetric-group algebra: every conjugacy-class sum is an integral symmetric polynomial in the $J_i$. Its proof expands symmetric monomials in these sums of transpositions and organizes the resulting class sums by the number of moved letters and cycle filtration; triangular elimination supplies integral spanning, rather than merely rational spanning. A useful first identity in this construction isIndeed, when adjoining the letter $n$, either choose $t$ and leave it as a singleton cycle, or choose a transposition $(jn)$ and insert it into the cycle containing $j$; this gives every permutation uniquely. The full generation step distinguishes the individual class sums which this first identity groups together. Its integrality is essential: a description only over $\mathbb{Q}$ could use denominators divisible by $p$ and would not prove generation of the modular center. Since class sums form the integral basis of the center, this integral theorem reduces to the same generation statement over $F$.

Next determine their scalar action. In the characteristic-zero tableau basis of $S^{\lambda }$, $J_i$ is diagonal with entry the Content of a Young-diagram cell occupied by $i$, namely column minus row. This follows recursively along the multiplicity-free branching chain: the commuting operators preserve each path line, and $J_{i+1}=s_i{J}_i{s}_i+s_i$ determines the successive eigenvalues. Consequently a symmetric polynomial $f(J_1,\dots ,J_n)$ acts on $S^{\lambda }$ byThis is independent of the ordering of the cells. The scalar identity holds on the integral Specht lattice, so after modular reduction it depends only on the multiset of contents modulo $p$. Every composition factor has this same central character of a block.

Two simple modules are in the same block exactly when their central characters agree. The center of a block is a commutative local finite-dimensional algebra; its semisimple quotient gives one scalar character, while the orthogonal block idempotents separate different blocks. By the generation theorem, equality on all symmetric polynomials in the $J_i$ is enough. Conversely, different residue multisets are detected by elementary symmetric polynomials: the monic polynomial ${\prod }_{(i,j)\in \lambda }(z-(j-i))$ over $F$ recovers the multiplicity of each residue by unique factorization, including multiplicities larger than $p$. Thus block equivalence is exactly equality of the integersThis gives the residue-content criterion for symmetric-group blocks without assuming that the sum of contents alone distinguishes blocks.

Finally identify residue content with the core. A removable rim $p$-hook has consecutive contents modulo $p$, hence removes one cell of each residue. Therefore shapes with the same core and the same size have the same residue counts. For the converse choose $m$ divisible by $p$ and at least $n$, and write $q_r=n_r-m/p$ for the runner charges. Starting from the empty partition of an integer gives $q_r=0$. Adding a cell of residue $r$ moves a beta-number bead from residue $r-1$ to residue $r$, increasing $q_r$ and decreasing $q_{r-1}$. The same changes occur in $c_s-c_{s+1}$ when $c_r$ increases by one. Induction on the cells therefore givesEqual residue counts give equal charges and hence equal packed runner configurations, so they give the same core. This completes both directions of the block criterion. The argument exposes the three essential bridges: integral center generation, its scalar content evaluation, and the abacus identification of residue data with a unique core.

For a cell $(i,j)$ in the Young diagram of $\lambda$, its hook of a Young diagram consists of that cell, the cells to its right in row $i$ and the cells below it in column $j$. Its length isA removable rim hook is a connected border strip $\lambda /\mu$ containing no $2\times 2$ square, whose removal leaves a Young diagram. Its leg length is the number of rows it meets minus one. The hook graph is the Young diagram labeled at each cell by $h_{ij}$; here “graph” means this labeled array, not a graph of possible hook-walk moves.

Choose $l\ge \ell (\lambda )$, pad with zeros and form the beta numbersThey are distinct nonnegative integers in decreasing order and recover ${\lambda }_i={\beta }_i-l+i$. For $l=\ell (\lambda )$ they equal the first-column hook lengths, since $h_{i1}={\lambda }_i+l-i$. If $l$ is increased, old beta numbers shift upward by one and a new bead at zero is inserted.

For a prime-modulus partition abacus arrange the nonnegative positions in $p$ runners: position $pq+r$ lies at level $q$ on runner $r$, $0\le r<p$. Put beads at the beta numbers and leave the other positions empty. Replacing a bead at $b$ by a bead at a vacant $b-k\ge 0$ removes a rim hook of length $k$; its leg length is the number of beads strictly between these positions. This follows by tracing the boundary of the diagram: bead positions record vertical steps and gaps record horizontal steps, and exchanging one bead and gap removes the intervening border strip. In particular a removable $p$-hook is exactly a bead that can slide up one place on its own runner.

The partition core at modulus $p$ is the result of sliding beads up until there are no gaps above any bead on a runner. If runner $r$ initially has $b_r$ beads, its final occupied levels must be $0,1,\dots ,b_r-1$, independently of slide order. Thus its packed beta set, and hence its core partition of an integer, is unique. Increasing the number of beta numbers merely re-expresses the same boundary by shifting positions and adding the new initial bead; the packed configuration represents the same partition of an integer. This proves well-definedness both with respect to removal order and to the chosen padding. The number of slides isThe partition quotient components record how far each runner's beads lie above their packed positions, giving the hook-removal interpretation used in Question 5.

A block of a group algebra in characteristic $p$ for $F{S}_n$ is a two-sided ideal $B=F{S}_{n}e$ for a primitive central idempotent $e$, meaning a nonzero central idempotent that is not a sum of two nonzero orthogonal central idempotents. The identity is the sum of these block idempotents and the group algebra is their direct product. An indecomposable module $V$ belongs to $B$ when $eV=V$, equivalently every other block idempotent acts as zero. Indeed $V={⨁}_{e}eV$; indecomposability makes exactly one summand nonzero.

The Nakayama block theorem, historically called the Nakayama Conjecture, statesEquivalently the ordinary characters assigned to one modular block have a common partition core at modulus $p$. Its simple modules are the $D^{\lambda }$ with regular $\lambda$ having that core. Only cores $\kappa$ for which $n-|\kappa |$ is a nonnegative multiple of $p$ occur. The block weight is $(n-|\kappa |)/p$; a core alone determines the block at this fixed degree.

One route to the proof separates the combinatorial core invariant from the algebraic center invariant. Let $c_r(\lambda )$ be the number of cells of residue $r=j-i(\mathrm{mod}p)$. A length-$p$ rim hook has one cell of every residue: along its border strip the contents are consecutive integers. Therefore removal subtracts one from every $c_r$. More precisely, choose the number $l$ of beta numbers divisible by $p$ and write the runner charges as $q_r=b_r-l/p$. Reading the boundary, or adding a cell and checking its two changed beta positions, givesThe charges determine the packed runners and hence the core. For fixed $n$, two equal cores have the same weight, so their full residue multisets are equal. Conversely equal residue multisets give equal charges and the same core. This establishesfor partitions of an integer of the same integer. The fixed-degree condition matters.

For the center invariant, use the commuting Jucys–Murphy elements $J_k={\sum }_{i<k}(ik)$, with $J_1=0$. In the Young seminormal form basis, $J_k$ has eigenvalue equal to the content of the cell containing $k$. A symmetric polynomial in them therefore acts on $S^{\lambda }$ as the same scalar for every Young tableau: its value on the multiset of all contents of $\lambda$. The integral Specht lattice makes this a scalar identity before modular reduction, so modulo $p$ these central scalars depend only on the residue multiset, and every composition factor inherits them.

The substantive center theorem used in this proof is that the center of the symmetric-group algebra, integrally and over a field, consists of symmetric polynomials in the Jucys–Murphy elements. Centrality of elementary symmetric polynomials follows fromthe coefficients are central class-sum combinations. The reverse inclusion requires the integral triangular class-sum argument: expand symmetric monomials in the $J_k$ as products of transpositions, group the leading terms by reduced cycle type, and eliminate these leading class sums in the resulting unitriangular order. The integral coefficients, with no division by multiples of $p$, are crucial; knowing the center only over $\mathbb{Q}$ would not prove the modular assertion. This is the central algebraic ingredient of this proof route.