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For a field $F$ of characteristic $p$, the exact sequence $0\to {\mathbb{F}}_p\to F^{\mathrm{sep}}\stackrel{z\mapsto z^p-z}{\to }{F}^{\mathrm{sep}}\to 0$, together with additive normal-basis acyclicity, gives $H^2(\mathrm{Gal}(F^{\mathrm{sep}}/F),{\mathbb{F}}_p)=0$. Consequently every central finite embedding problem with group homomorphism kernel $C_p$ has a continuous weak solution: its pulled-back two-cocycle is a coboundary, whose correcting one-cochain supplies the lift. Over ${\mathbb{F}}_p((t))$, independent Artin–Schreier classes $t^{-m}$, with $m>0$ prime to $p$, allow the lift to be twisted by characters so that inertia maps onto the group homomorphism kernel as well as onto the original quotient. Induction through central order-$p$ quotients therefore realizes every finite p-group as the Galois group of a totally ramified extension with residue field ${\mathbb{F}}_p$.

A nontrivial normal subgroup $N$ of a finite p-group meets the center of a group nontrivially. The conjugation action of $G$ on $N$ has singleton orbits exactly at $N\cap Z(G)$. The orbit-stabilizer theorem makes every other orbit size divisible by $p$, so $|N\cap Z(G)|\equiv |N|\equiv 0(\mathrm{mod}p)$. Since the intersection contains the identity, its size is at least $p$.

Let $P$ be any Sylow subgroup containing a defect group of a block $D$. Regard the block of a group algebra as a permutation module summand for $G\times G$, acting by left and right multiplication. Its vertex is $\Delta D$. Restricting to $P\times P$ retains a summand with this vertex. Transitive permutation modules for a finite p-group are indecomposable, with their point stabilizers as vertices. Thus some orbit in $G$ has stabilizer $\Delta D$ at a suitably chosen point $c$. This means $c$ centralizes $D$ and $P\cap {}^{c}P=D$. Choosing $P$ to contain a Sylow subgroup of $N_G(D)$ shows that $D=O_p(N_G(D))$; every normal p-subgroup of $G$ is contained in both Sylow factors and hence in $D$.

Realize a finite p-group $P$ as the Galois group of a totally ramified extension $E/{\mathbb{F}}_p((t))$, using Artin-Schreier embedding problems over a characteristic-p field. A uniformizer $u$ of $E$ gives $E\cong {\mathbb{F}}_p((u))$: every element has a unique convergent Laurent expansion with coefficients in the residue field. An automorphism has $u\mapsto a_{1}u+a_2{u}^2+\cdots$, and $a_1$ defines a homomorphism $P\to {\mathbb{F}}_p^{\times }$. Its image is trivial because $|P|$ is a $p$-power and $p\nmid |{\mathbb{F}}_p^{\times }|$. Thus the faithful Galois action lies in the Nottingham group, embedding $P$.

For a finite group, $F(G)$ centralizes its group socle. A nonabelian minimal normal subgroup meets $F(G)$ trivially: otherwise it is nilpotent, and its nontrivial characteristic center makes it abelian. Their group commutators therefore vanish. An abelian minimal normal subgroup $M$ is an elementary abelian p-group; $M\le O_p(G)$ and $M\cap Z(O_p(G))\ne 1$ by the fixed-point counting of a finite p-group on $M$. Minimal normality makes $M\le Z(O_p(G))$, and other p-cores commute with $M$.

Every noncyclic finite finite p-group $P$ has a quotient $C_p\times C_p$. Over any field of characteristic $p$, define a square-zero representation with basis $v_0,\dots ,v_r,w_1,\dots ,w_r$ by $x{v}_i=w_i$ for $1\le i\le r$, $x{v}_0=0$, $y{v}_i=w_{i+1}$ for $0\le i<r$, $y{v}_r=0$, and $xW=yW=0$. Here $x=g-1$, $y=h-1$, and $W$ is the span of the $w_i$. An endomorphism induces the same scalar on the top and on $W$; the remaining maps from the top to $W$ form a square-zero ideal. Thus the endomorphism ring is local and the module is indecomposable. Inflating to $P$ and choosing $p\nmid 2r+1$ gives infinitely many distinct dimensions with vertex $P$: the trace of a relative-trace identity from any proper subgroup would force dimension zero in the field. This construction works over finite fields too.

Every nontrivial finite p-group has a nontrivial center. The conjugation action partitions the group into conjugacy classes whose noncentral sizes are positive powers of $p$; the class equation then implies $|Z(G)|\equiv |G|\equiv 0(\mathrm{mod}p)$.

Every proper subgroup $H$ of a finite p-group is properly contained in its normalizer. Let $H$ act on the left cosets $G/H$ by left multiplication. The fixed cosets are those represented by elements of $N_G(H)$, so there are $[N_G(H):H]$ of them. All other orbits of a group action have sizes divisible by $p$. Since $[G:H]$ is divisible by $p$ and at least one coset is fixed, the fixed-coset count is a positive multiple of $p$. This also proves $[N_G(H):H]\ge p$.

For subgroups $A,B$ write $[A,B]$ for the subgroup generated by their group commutators. A nilpotent group has a central series: a finite chain $1=G_0\le G_1\le \cdots \le G_c=G$ with $[G_i,G]\le G_{i-1}$. Equivalently its upper central series, defined by $Z_0=1$ and $Z_{i+1}/Z_i=Z(G/Z_i)$, reaches $G$. Indeed every central series term $G_i$ lies in $Z_i$ by induction, and the upper central series itself is central. A third equivalent definition is that the lower central series ${\gamma }_1=G$, ${\gamma }_{i+1}=[{\gamma }_i,G]$ terminates at ${\gamma }_{c+1}=1$: reversing this series gives a central series, and induction along any central series gives ${\gamma }_{c+1}=1$.

The least such $c$ is the nilpotency class. Nontrivial abelian groups have class one. Upper-unitriangular $3\times 3$ matrices have class two: their group commutators lie in the central top-right-entry subgroup, and further group commutators vanish. The same example works over a finite field or over the integers, so nilpotence is not a finiteness property. Subgroups and quotients inherit nilpotence by intersecting or mapping a central series; finite direct products of groups have class at most the largest factor class. Every nilpotent group is soluble, since its derived series is contained successively in the lower central series. The converse fails: $S_3$ is soluble but its center is trivial, so it cannot be a nontrivial nilpotent group. Extensions of nilpotent groups need not be nilpotent, as the extension $C_3◃S_3$ with quotient $C_2$ shows.

To prove the finite classification, first note that every nontrivial finite p-group has nontrivial center. In its class equation, every noncentral conjugacy class has size divisible by $p$, so the center's order is divisible by $p$. Induct on the group order: the smaller p-group $P/Z(P)$ is nilpotent, and lifting its central series and adjoining $Z(P)$ proves $P$ nilpotent. A finite direct product of groups of prime-power-order groups is consequently nilpotent.

Conversely suppose $G$ is finite and nilpotent. We need the normalizer condition for nilpotent groups. For any proper subgroup $H<G$, choose the first upper-central term $Z_i$ not contained in $H$. Then $Z_{i-1}\le H$. Every $z\in Z_i$ commutes with $H$ modulo $Z_{i-1}$, so it normalizes $H$. Choosing $z\notin H$ proves $H<N_G(H)$.

Let $P$ be a Sylow subgroup and put $M=N_G(P)$. Since $P$ is normal in $M$, it is the unique Sylow p-subgroup of $M$. Any element normalizing $M$ must therefore normalize $P$, so $N_G(M)=M$. If $M$ were proper, the normalizer condition would contradict this equality. Hence $N_G(P)=G$: every Sylow subgroup is normal. For Sylow subgroups $P,Q$ belonging to different primes, $[P,Q]\le P\cap Q=1$, so they commute. Their product has order the product of their orders, namely $|G|$, and has trivial intersections between each factor and the product of the others. Multiplication therefore givesThis proves finite nilpotent group decomposition in both directions without assuming the classification in the normalizer argument. It also shows that a finite group is nilpotent exactly when all its Sylow subgroups are normal. The finite hypothesis is essential to the prime-power direct-product statement; an infinite cyclic group is nilpotent but is not such a finite product.

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,

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

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

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

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

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

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

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

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

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

Throughout this question $p$ is odd, $A^p$ means the subgroup generated by the $p$-th powers of elements of $A$, and the commutator of group elements convention is $[x,y]=x^{-1}{y}^{-1}xy$. A subgroup $N$ is powerfully embedded in $G$ when $[N,G]\subseteq N^p$. In particular $N$ is normal, since $n^g=n[n,g]\in N$. A finite p-group $G$ is a powerful p-group when $[G,G]\subseteq G^p$.

First put $M=N^p$. It is normal in $G$. By the reduction permitted in the question, it suffices to prove $[M,G]\subseteq M^p$ after factoring out $[M,G,G]$. In that quotient, $[M,G]$ is central. For $n\in N$ and $g\in G$, put $c=[n,g]\in M$. Then $[c,n]$ is central, and the elementary identity $[ab,g]=[a,g{]}^b[b,g]$ gives, by induction,Both $c$ and $[c,n]$ belong to $M$, and $p$ divides $p(p-1)/2$ because $p$ is odd. Hence $[n^p,g]\in M^p$. The subgroup $M^p$ is normal in $G$, so the commutator identity extends this inclusion from the generating elements $n^p$ to every element of $M$. Applying the allowed reduction proves $N^p$ is powerfully embedded in $G$.

For $H=⟨N,x⟩$, normality of $N$ gives $H=N⟨x⟩$. Its quotient by $N$ is cyclic. Modulo $[N,H]$, the image of $N$ is central and the remaining generator is $x$, so that quotient is abelian. ThereforeThus $⟨N,x⟩$ is powerful.

The lower p-series is $G_1=G$, $G_{i+1}=G_i^p[G_i,G]$. For the later calculations write $P_j=G_{j+1}$, so $P_0=G$. Part (i) will show $P_{j+1}=P_j^p$ and $[P_j,G]\subseteq P_{j+1}$. A useful power lifting in a powerful p-group congruence isIndeed, modulo $P_{j+2}$ the commutators $[u,x]$ are central, lie in $P_{j+1}$ and have exponent dividing $p$. Collecting the $p$ copies of $xu$ therefore gives only the correction $[u,x{]}^{\left(\genfrac{}{}{0ex}{}{p}{2}\right)}$, which is trivial. This argument also covers $j=0$. The congruence is the reason roots and coordinates can be corrected one power layer at a time.

For each $j\ge 0$, the elementary abelian group $P_j/P_{j+1}$ is spanned by the images of $a_i^{p^j}$. We construct exponents $e_i$ so that $g$ equals the ordered product $a_1^{e_1}\cdots a_n^{e_n}$ to successively greater precision. Suppose this product agrees with $g$ modulo $P_j$; initially this is automatic for $j=0$. Its error modulo $P_{j+1}$ can be expressed using the images of $a_i^{p^j}$. Add the corresponding digits $p^j{t}_i$ to the exponents $e_i$. Since $[P_j,G]\subseteq P_{j+1}$, moving each new factor $a_i^{p^j{t}_i}$ past the other factors changes nothing modulo $P_{j+1}$. The new ordered product therefore agrees with $g$ modulo $P_{j+1}$. Termination of the lower p-series provesNo uniqueness of this factorization is asserted for a redundant generating tuple.

The printed converse is false without an additional condition. For any odd $p$, take the Heisenberg group $G={\mathrm{UT}}_3({\mathbb{F}}_p)$ withEvery element is uniquely $x^a{y}^b{z}^c$, with $a,b,c\in {\mathbb{F}}_p$, so $G$ is a product of three cyclic subgroups. For $A$ strictly upper triangular, $A^3=0$ andin characteristic $p$, including $p=3$. Therefore $G^p=1$, whereas $[G,G]=⟨z⟩\ne 1$. This finite p-group is not powerful, disproving the literal request in the authoritative PDF.

A precise nearby correction is the minimal cyclic factorization criterion for powerful p-groups: require a product of $d(G)$ cyclic subgroups, where $d(G)$ is the minimal generator number. To prove it, put $d=d(G)$ and $H=G/G^p$. The quotient has exponent $p$ and still has minimal generator number $d$, because $G^p\subseteq \Phi (G)$. Each cyclic factor has image of order at most $p$, so $|H|\le p^d$. But $|H/\Phi (H)|=p^d$ by the Burnside basis theorem. Consequently $\Phi (H)=1$, $H$ is an elementary abelian group, and $[G,G]\subseteq G^p$. A product of exactly $d(G)$ cyclic subgroups is powerful. A direct product of cyclic groups is also powerful, since it is abelian; neither stronger hypothesis appears in the printed converse.

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.

Nagao module theorem states that if $e\in Z(kG)$ is a central idempotent, $eM=M$, and $C_G(Q)\le K\le N_G(Q)$, then ${\mathrm{Br}}_Q(e)$ is central in $kK$ andEvery indecomposable summand of the second term is relatively projective for some subgroup of $K$ not containing $Q$. In particular, if $Q\le K$, the error term has no summand with module vertex containing $Q$. This is the Nagao version of the second main theorem; it identifies the modular block of a module vertex-$Q$ Green correspondent.

Let $e_B$ be the identity of $B$, $N=N_G(D)$, and $b$ its Brauer correspondent of a block in $kN$. Because $D◃N$, the ideal $J(kD)kN$ is nilpotent: normality allows moving $kN$ past $J(kD)$ and the augmentation ideal of the p-group is nilpotent. Hence $b$ has nonzero image in $k(N/D)$. Choose an indecomposable summand $U$ of $bk[N/D]$. It is projective over $k(N/D)$ and belongs to modular block $b$ after inflation.

Since $k[N/D]={\mathrm{Ind}}_D^{N}k$ is a permutation module, its summand $U$ has trivial module source and module vertex contained in $D$. As $D$ acts trivially, all its proper-subgroup traces vanish and its Brauer quotient of a module is $U(D)=U\ne 0$. The module vertex must also contain $D$, so it is exactly $D$. This is inflated quotient projectives have trivial source.

Let $M$ be the Green correspondent of $U$ in $G$, using the theorem proved in the next solution. It has module vertex $D$ and the same trivial module source. For a normal p-subgroup, each central block idempotent of $kN$ equals its Brauer image: the two central idempotents agree modulo the nilpotent ideal $J(kD)kN$, because nonfixed conjugation-orbit sums map to zero in $k(N/D)$; central idempotents congruent modulo a nilpotent ideal coincide. The first main theorem therefore gives ${\mathrm{Br}}_D(e_B)={\mathrm{Br}}_D(b)=b$.

Let $e_C$ be the block identity acting on $M$. Nagao's theorem forces ${\mathrm{Br}}_D(e_C)U=U$, since $U$ has module vertex $D$. But $bU=U$, and for $C\ne B$ orthogonality would give ${\mathrm{Br}}_D(e_C)b={\mathrm{Br}}_D(e_C{e}_B)=0$. Therefore $M$ belongs to $B$, has module vertex $D$ and has trivial module source. For $D=1$, this supplies an indecomposable projective in the defect-zero modular block.

For the representation-type deduction, modules in a block are projective relative to its defect group: the block trace identity acts as a relative trace equal to the module identity, so D. Higman criterion gives split induction from $D$. If $D$ is cyclic of order $d$, then $kD\cong k[t]/(t^d)$, whose indecomposable modules are the nilpotent Jordan chains $k[t]/(t^j)$, $1\le j\le d$. Each indecomposable in $B$ is a summand of the module induction of its module restriction to $D$, and hence occurs among the summands of these finitely many fixed induced modules. Their finite Krull-Schmidt decompositions give finite representation type.

Conversely, the regular defect-group bimodule inside a block givesIndeed this module restriction is a permutation summand. Its Brauer quotient at $\Delta D$ is nonzero, because on the fixed group-basis elements $C_G(D)$ the block projection becomes multiplication by ${\mathrm{Br}}_D(e_B)\ne 0$. Every transitive $D\times D$ orbit in $G$ has stabilizer of order at most $|D|$. A summand with a nonzero quotient at $\Delta D$ must therefore have stabilizer exactly $\Delta D$ up to conjugacy, and is the regular $kD$ bimodule. Tensoring this split inclusion on the right with an arbitrary $kD$-module $V$ yieldsIf $B$ has only finitely many indecomposable modules, their restrictions contain only finitely many indecomposable summands. The displayed inclusion puts every indecomposable $V$ on that finite list, so $kD$ has finite representation type too.

A noncyclic finite p-group has a quotient $C_p\times C_p$, by the Burnside basis theorem for its Frattini quotient. Here is an explicit infinite family over any field, including a finite field. For $s\ge 1$, take basis $v_0,\dots ,v_s,w_1,\dots ,w_s$ and operators $x=g-1$, $y=h-1$ defined byand $x{w}_i=y{w}_i=0$. All products of $x,y$ vanish, so $1+x,1+y$ satisfy the group relations in characteristic $p$. The radical is $W=⟨w_1,\dots ,w_s⟩$. An endomorphism preserves $W$, and commuting with the two shift maps forces its induced maps on $V/W$ and $W$ to be the same scalar. To see this, write the shift matrices as $X=(0I_s)$ and $Y=(I_{s}0)$: the equations $CX=XA$, $CY=YA$ force the off-diagonal entries of $A,C$ to vanish and all diagonal entries to agree, by comparing successive rows, including the zero first column of $X$ and zero last column of $Y$. The remaining linear map sends the top into $W$ and has square zero. Thus the endomorphism ring is a scalar algebra with square-zero radical, hence local; the module is indecomposable. Inflation to $D$ preserves that ring. Its dimensions $2s+1$ are unbounded, contradicting finite representation type. Therefore $B$ has finite representation type if and only if $D$ is cyclic, including trivial defect.

The block statement is Green correspondence for blocks: if $N_G(D)\le H\le G$, blocks of $RG$ and $RH$ with defect group $D$ correspond, and their block bimodules are Green correspondents for the diagonal vertex $\Delta D$. We prove the module correspondence underlying this statement, then specialize it to the block bimodules. This also gives the requested direct proof in the trivial-intersection situation.

Here are the elementary ingredients, including their proofs. Mackey decomposition is obtained by partitioning the tensor basis of $RG{\otimes }_{RL}U$ by the double cosets $H\backslash G/L$; the term for $g$ is induction from $H\cap {}^{g}L$ of the suitably conjugated restriction of $U$. The D. Higman criterion follows from the section constructed in Question 1. Conversely, take a splitting of the induction counit and its coefficient at the identity coset: equivariance reconstructs the section from that $H$-linear coefficient, and the splitting identity is exactly that its relative trace is $1$. Thus the trace criterion is an equivalence.

An indecomposable module here has a local endomorphism ring. Over $k$ this follows from the Fitting lemma: an endomorphism eventually splits its kernel from its image, and indecomposability makes it invertible or nilpotent. Over $\mathcal{O}$, the endomorphism ring is finite over a complete discrete valuation ring; lift orthogonal idempotents from its finite-dimensional residue algebra to see that an indecomposable lattice has a local endomorphism ring too. The Krull–Schmidt theorem follows by the usual exchange argument: composing the inclusion and projection of one indecomposable summand across a second decomposition writes its identity as a sum of endomorphisms; one is a unit in the local endomorphism ring and pairs that summand with a summand of the second decomposition. Remove the pair and continue. Existence follows by successively splitting nontrivial idempotents, which decreases dimension or rank.

A vertex of an indecomposable module is a subgroup minimal for relative projectivity. It can be taken to be a $p$-subgroup because averaging over the index of a Sylow subgroup of any candidate subgroup is invertible. If an indecomposable module is relatively projective for $P$ and $Q$, the Mackey decomposition and its induction splittings make it a summand of a sum of modules induced from $P\cap {}^{g}Q$. In its local endomorphism ring one splitting component must be invertible, so it is relatively projective for one of these intersections. This proves conjugacy of minimal vertices and the useful criterion that a module with vertex $P$ can be relatively projective for $Q$ only when $P$ is conjugate into $Q$. A source of an indecomposable module with vertex $P$ is an indecomposable summand $S$ of its restriction to $P$ from which the original module is an induced summand. Such a summand exists by the induction splitting and Krull–Schmidt theorem; its vertex is $P$, since a proper vertex would make the original module relatively projective for a proper subgroup of $P$.

SetWhen saying a module is $\mathcal{X}$- or $\mathcal{Y}$-projective, allow conjugates in the relevant ambient group and finite sums of modules relatively projective for members of that family. No conjugate of $D$ inside $H$ belongs to $\mathcal{Y}$: if $D\le {}^h(H\cap {}^{g}D)$, equality of orders gives $D={}^{hg}D$, so $hg\in N_G(D)\subseteq H$, contradicting $g\notin H$. Every member of $\mathcal{X}$ is a proper subgroup of $D$.

The two key assertions can now be proved rather than assumed. First, if $W$ is an indecomposable $D$-projective $RH$-module, thenChoose $U$ with $W\mid {\mathrm{Ind}}_D^{H}U$. Mackey decomposition gives ${\mathrm{Res}}_H^G{\mathrm{Ind}}_D^{G}U={\mathrm{Ind}}_D^{H}U\oplus T$, with $T$ $\mathcal{Y}$-projective. Write ${\mathrm{Ind}}_D^{H}U=W\oplus W^{\prime \prime }$ and apply induction and restriction. Both $\mathrm{Res}\mathrm{Ind}W$ and $\mathrm{Res}\mathrm{Ind}{W}^{\prime \prime }$ contain their original modules, by the identity double coset. Comparing multiplicities of every indecomposable which is not $\mathcal{Y}$-projective, their extra multiplicities are nonnegative and have sum zero. Therefore neither has such an extra summand, proving the assertion.

Second, suppose $W$ has vertex $D$, and choose an indecomposable $V\mid {\mathrm{Ind}}_H^{G}W$ whose restriction contains $W$; such a $V$ exists by the first assertion. Write $\mathrm{Ind}W=V\oplus V^{\prime }$. Its restriction has just one summand isomorphic to $W$ and all other summands are $\mathcal{Y}$-projective, so $\mathrm{Res}{V}^{\prime }$ is $\mathcal{Y}$-projective. Any indecomposable $Z\mid V^{\prime }$ has a vertex $P\le D$ after conjugacy. A source $S$ with vertex $P$ is a summand of ${\mathrm{Res}}_{P}Z$. Restricting a $\mathcal{Y}$-projective decomposition of ${\mathrm{Res}}_{H}Z$ to $P$ and using Mackey decomposition forces $P$ to lie in an $H$-conjugate of some $H\cap {}^{g}D$, $g\notin H$. Since $P\le D$, this puts $P$ in $\mathcal{X}$. Thus $V^{\prime }$ is $\mathcal{X}$-projective. Moreover $V$ has vertex $D$, since it is $D$-projective and its restriction contains $W$ with vertex $D$. We have provedIn particular $V$ is unique.

Conversely, let $V$ be an indecomposable $RG$-module with vertex $D$ and source $S$. Decompose ${\mathrm{Ind}}_D^{H}S$ and choose $W$ whose induction contains $V$. Its vertex is contained in $D$ and cannot be proper, since induction would then make $V$ relatively projective for a smaller subgroup. Thus $W$ has vertex $D$, and the previous assertion makes $V$ its unique vertex-$D$ induced summand. Restriction of an $\mathcal{X}$-projective module is $\mathcal{Y}$-projective: in a Mackey term a conjugating element inside $H$ preserves the containment in an $H$-conjugate of $\mathcal{Y}$, and one outside $H$ gives a subgroup of $H\cap {}^{g}D$. Comparing the two decompositions of $\mathrm{Res}\mathrm{Ind}W$ now givesThese assignments are mutually inverse. To check sources, restrict ${\mathrm{Ind}}_D^{H}S$ back to $D$. Its terms of full vertex are conjugates of $S$ by $N_H(D)$; the other terms are induced from proper intersections. Applying the source criterion and Krull–Schmidt theorem shows that the source of $W$ is one of those conjugates. The same argument in $G$, and the uniqueness just proved, matches the source classes on the two sides. Thus Green correspondence preserves sources up to normalizer conjugacy.

To pass to blocks of a group algebra, regard $RG$ as the permutation module $R[(G\times G)/\Delta G]$. A summand of a permutation module has trivial source. Here is why: over a finite p-group, every transitive permutation module is indecomposable, since its socle is the one-dimensional fixed line and every nonzero module has nonzero fixed points. The latter follows by induction on the group order, using a central element of order $p$ and the nilpotence of its difference from $1$. Over $\mathcal{O}$ indecomposability follows by reduction. Restrict a permutation summand to a vertex $P$ and apply Krull–Schmidt theorem; a source must be a transitive permutation module $R[P/Q]$. Its vertex is contained in $Q$, so minimality forces $Q=P$, and its source is the trivial module.

For a permutation module, its Brauer quotient of a module at $P$ has the fixed permutation points as basis: every nontrivial orbit sum is a proper transfer. An indecomposable trivial-source summand with vertex $Q$ has nonzero quotient only at subgroups conjugate into $Q$, by its induced-permutation realization, and its quotient at $Q$ is nonzero because its restriction contains its trivial source. Applied to a block $B=kGe$, the quotient at $\Delta P$ is ${\mathrm{Br}}_P(e)k{C}_G(P)$. Together with the maximal-subgroup characterization of defect from Question 1, this proves that its bimodule vertices are precisely $\Delta D$ for its defect groups $D$. The integral conclusion follows by orbit-sum reduction and idempotent lifting.

FinallyApply the just-proved Green correspondence with ambient group $G\times G$. Its unique vertex-$\Delta D$ summand of $B{\downarrow }_{H\times H}$ lies in $RH$: outside $H$ there is no $\Delta D$-fixed group-basis point, as in Question 2. Since the indecomposable $H\times H$ summands of $RH$ are exactly its block ideals, this summand is a block $b$ with defect group $D$. Conversely induction of $b$ has $B$ as its unique vertex-$\Delta D$ summand. The block bijection is the Brauer correspondence of Question 2, giving the claimed block form over either coefficient ring.

For the final special case put $N=N_G(D)$ with $D$ a trivial-intersection Sylow subgroup. For $g\notin N$, the intersection $N\cap {}^{g}N$ has order prime to $p$. Indeed every $p$-subgroup of it lies both in the normal Sylow subgroup $D$ of $N$ and in ${}^{g}D$, whose intersection is trivial. Mackey decomposition therefore giveswhere every other double-coset term is induced from a group of order prime to $p$. By Maschke's theorem such modules are projective, and induction preserves projectivity, so $P$ is projective.

Every nonprojective $kG$-module has nonprojective restriction to $N$: $[G:N]$ is prime to $p$, so averaging makes the module a summand of the induction of its restriction. A projective restriction would therefore make the original module projective. Decompose ${\mathrm{Ind}}_N^{G}M$ into indecomposables. Each nonprojective summand contributes a nonprojective summand on restriction, whereas the displayed Mackey formula and Krull–Schmidt theorem have exactly one nonprojective summand, $M$, with multiplicity one. There must be one, since $M$ is nonprojective. Hence

A topological group is a group with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous. A profinite group has either of the following equivalent descriptions: a compact Hausdorff totally disconnected topological group, or an inverse limit of finite discrete groups. Totally disconnected means that every connected component is a singleton.

For clarity, an inverse system over a directed set $I$ consists of groups $G_i$ and maps ${\varphi }_{ij}:G_j\to G_i$ for $i\le j$, with ${\varphi }_{ii}=1$ and ${\varphi }_{ik}={\varphi }_{ij}{\varphi }_{jk}$. Its inverse limit iswith coordinatewise multiplication and the subspace topology. Equivalently, it is a group with compatible continuous projections such that every other compatible family of maps into the $G_i$ factors uniquely through it.

The displayed limit is a closed subgroup of the compact Hausdorff product of the finite groups, and is totally disconnected, since two different tuples are separated by a clopen coordinate condition. Conversely a compact Hausdorff totally disconnected group has a neighbourhood basis of open normal subgroups. One way to see the group part of this standard clopen-basis fact is to choose a clopen identity neighbourhood $U$ inside a prescribed neighbourhood. Compactness of $U$ and continuity give a symmetric identity neighbourhood $V$ with $VU\subseteq U$. The stabilizer $H=\{g:gU=U\}$ contains $V$, is open, and is contained in $U$ because $1\in U$. Its index is finite by part (ii); its normal core is the intersection of finitely many conjugates and is open and normal. The canonical mapis injective because these $N$ separate points, and surjective because compatible cosets have the finite intersection property and $G$ is compact. It is a homeomorphism by compactness and the Hausdorff property. This explains the equivalence of the two definitions.

The profinite Frattini subgroup is $\Phi (G)=\bigcap M$, where $M$ ranges over all maximal proper open subgroups; for the trivial group the empty intersection is $G$. For a pro-p group,Here powers and group commutators mean generated subgroups and the bar supplies closure. Every maximal open subgroup in a pro-p group is normal of index $p$, as follows by passing to the finite p-group quotient by its normal core. Such a subgroup therefore contains all powers and group commutators. Conversely the quotient by $D=\overline{G^p[G,G]}$ is an abelian pro-p group of exponent $p$. If an element of this quotient is nontrivial, some finite elementary abelian quotient detects it; a linear functional to ${\mathbb{F}}_p$ then detects it as well. Its group homomorphism kernel pulls back to a maximal open subgroup of $G$. Thus the intersection of these kernels is exactly $D$.

We also use the topological generation criterion: a subset generates $G$ topologically if and only if its image generates $G/\Phi (G)$ topologically. To prove the nontrivial direction, a proper closed generated subgroup $K$ lies in a proper open subgroup $KN$ for a suitably small open normal $N$, and hence in a maximal open subgroup obtained from the finite quotient $G/N$. Since that maximal subgroup contains $\Phi (G)$, the image of $K$ cannot generate the Frattini quotient.

Now suppose $G$ is a nontrivial procyclic pro-p group. If it had distinct maximal open subgroups $M_1,M_2$, the homomorphism to $C_p\times C_p$ induced by the two quotient maps would be onto: its projections are onto, and a proper such subgroup would be a one-dimensional graph, forcing the two kernels to coincide. Hence $G/(M_1\cap M_2)\cong C_p^2$, contradicting procyclicity. Thus there is exactly one maximal open subgroup $M$, and $\Phi (G)=M$. Any $g\notin M$ generates $G/\Phi (G)\cong C_p$, so the criterion givesFor the trivial group use its identity as generator. The additive group ${\mathbb{Z}}_p$ is an infinite example, topologically generated by $1$: its finite continuous quotients are cyclic $\mathbb{Z}/p^n\mathbb{Z}$.

For odd $p$, a subgroup $N$ is powerfully embedded in $G$ if $[N,G]\le N^p$, where $N^p=⟨n^p:n\in N⟩$. The condition implies normality since $n^g=n[n,g]\in N$. The group $G$ is a powerful p-group if $[G,G]\le G^p$, equivalently if $G$ itself is powerfully embedded. For a finite p-group, the Burnside basis theorem gives $d(Q)={\dim }_{{\mathbb{F}}_p}Q/\Phi (Q)$, and $\Phi (Q)=Q^p[Q,Q]$. Thus for powerful $Q$, $d(Q)=\dim Q/Q^p$.

We prove $d(H)\le d(G)$ by induction on $|G|$, with the trivial group immediate. The supplied embedding result applied to $N=G$ makes $G^p$ powerfully embedded, and hence powerful. It is a proper subgroup when $G\ne 1$, because $G^p=\Phi (G)$ is proper in a nontrivial finite p-group. SetAll the displayed quotients are elementary abelian. The supplied onto homomorphism $\theta :V\to W$, $x{G}^p\mapsto x^p{G}^{p^2}$, is therefore linear. Write $v=\dim V=d(G)$, $w=\dim W=d(G^p)$, $r=\dim A$, and $t=\dim \theta (A)$.

Since $H/K$ embeds in $V$, $\Phi (H)\le K$. AlsoHere $(G^p{)}^p=G^{p^2}$ is the iterated-power convention, valid for powerful groups. Hence the natural map $K/\Phi (K)\to W$ is defined. The image under this map of the subspace $\Phi (H)/\Phi (K)$ contains $\theta (A)$, because for every $h\in H$ its power $h^p$ lies in $\Phi (H)\cap K$. It follows thatThe exact sequence $1\to K/\Phi (H)\to H/\Phi (H)\to H/K\to 1$ consists of elementary abelian groups, and $\dim H/K=r$. By induction inside $G^p$, $d(K)\le d(G^p)=w$. ThereforeThis proves $d(H)\le d(G)$, including subgroups $H$ that are not themselves powerful.

The rank of a profinite group is $\mathrm{rk}(P)=\sup \{d(K):K\le P\text{ closed}\}$, where $d(K)$ is its least topological generator number. If $P$ has finite rank then every closed subgroup, including the open subgroup $H$, has no larger rank. Conversely suppose $H$ has rank $r<\infty$. Its normal core $C={\bigcap }_{g\in P}{H}^g$ is an open normal subgroup: there are finitely many conjugates because $H$ has finite index. It has rank at most $r$.

For every closed $K\le P$, the subgroup $K\cap C$ is normal in $K$ and has at most $r$ topological generators. The quotient $K/(K\cap C)$ embeds in the finite group $P/C$. A finite group of order at most $M=[P:C]$ has at most $\lfloor {\log }_{2}M\rfloor$ generators: in a minimal sequential generating list every strict enlargement at least doubles the subgroup order. Lift such quotient generators and adjoin generators of $K\cap C$. Their closed generated subgroup contains the group homomorphism kernel and maps onto the quotient, so equals $K$. ConsequentlyThus an open subgroup has finite rank exactly when the whole profinite group does.

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 $[L_G,L_G]\le p{L}_G$.