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Let $b$ be a block of a group algebra of $kH$ with defect group of a block $D$, where $D{C}_G(D)\le H\le G$. Apply Brauer first main theorem inside $H$ to obtain its correspondent in $N_H(D)$. Composing that correspondent's central character of a block with the Brauer morphism from $Z(kG)$ selects a unique block of a group algebra $b^G$. It is the unique block of a group algebra whose restriction to $H\times H$ contains the block bimodule $b$ as a direct summand. Its defect group of a block can be larger than $D$ when $H$ does not contain $N_G(D)$.

A block of an associative algebra with a finite decomposition of its identity is a two-sided direct summand $Ae$ corresponding to a primitive central idempotent $e$. Its identity is $e$, and it cannot be split into two nonzero algebras by central idempotents. For a group algebra of a finite group, this agrees with a block of a group algebra. Orthogonal primitive central idempotents give $A={⨁}_{i}A{e}_i$.

A submodule $N$ of a free module with basis of a module $E$, generated by a countable set, is contained in the submodule generated by some countable subset $E^{\prime }\subseteq E$. Take $E^{\prime }$ to be the union of the finite coordinate supports in a free module of a countable generating family for $N$. This union is countable, and every generator belongs to $⟨E^{\prime }⟩$, which proves the inclusion. Combining this with basis partition gives a module direct sum places $N$ inside a countably generated direct summand.

A finite sum of simple modules inside one module is a semisimple module. Inductively, if $M$ is the sum already retained, then $M\cap S_j$ is a submodule of the simple module $S_j$, hence zero or $S_j$. In the first case append $S_j$ as a direct summand; in the second it adds nothing. This gives a direct sum of modules selected from the original family.

With $s=(12)$, the idempotents $e_{\pm }=(1\pm s)/2$ split the right regular module into two three-dimensional projective modules. Their tops are respectively the trivial representation and sign representation. The nilpotent ideal $J(k{S}_3)$ ensures that any nonzero direct summand has nonzero top, so the one-dimensional tops make these projective modules indecomposable. Their successive radical series of a module factors are trivial, sign, trivial and sign, trivial, sign. This decomposition concerns right modules; the block of S3 in characteristic three remains a single algebra.

Let $n\in {\mathbb{R}}^3$ be a unit vector, let $P=n{n}^T$ be the orthogonal projection onto its axis, put $Q=I-P$, and let the cross-product matrix $J$ satisfy $Jx=n\times x$. A real axis-angle decomposition has the form $A=aP+bQ+cJ$. It multiplies the axial component by $a$ and acts on the perpendicular plane as the complex number $b+ic$. It is an invertible matrix exactly when $a\ne 0$ and $b^2+c^2>0$, and thenTo check the formula, use $P^2=P$, $Q^2=Q$, $PQ=PJ=JP=0$ and $J^2=-Q$, the last identity following from the vector triple product. Thus $(bQ+cJ)(bQ-cJ)=(b^2+c^2)Q$. The axial and perpendicular direct summands can be inverted independently. For a pure rotation through an angle $\theta$, $a=1$, $b=\cos \theta$, $c=\sin \theta$, and the inverse reverses that angle.

A locally free module is one whose associated sheaf is locally a free module. In the finite constant-rank case, it is a finite projective module whose localization at a prime ideal has rank $r$ everywhere. Equivalently, it becomes $A_{f_i}^r$ on a finite cover by principal open subsets. Its associated sheaf is a locally free sheaf and defines an algebraic vector bundle.

A surjection onto such a module splits. If its source is also a finite projective module, the kernel is finite projective as a direct summand. A source hypothesis is essential: the projection $\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}\to \mathbb{Z}$ splits and has a free target, but its kernel $\mathbb{Z}/2\mathbb{Z}$ is not projective.

An integral Seifert form on a free abelian group $L$ of rank $2g$ is metabolic if there is a rank-$g$ direct summand $M\subset L$ on which the entire bilinear form vanishes:Over a field, replace the direct summand by a $g$-dimensional vector subspace. Such a half-rank vector subspace is a metabolizer. In a basis starting with a basis of the metabolizer, the Seifert matrix has a zero upper-left $g$-by-$g$ block.

The knot-theoretic nonsingularity condition concerns the matrix $V-V^T$, which defines a unimodular intersection pairing; $V$ itself need not be an invertible matrix. For example,has a metabolizer spanned by the first basis vector and has $\det (V-V^T)=1$, despite $\det V=0$. Requiring $V$ itself to be nonsingular would exclude this legitimate metabolic example. Algebraic concordance of knots identifies two forms when their difference is metabolic.

A metabolizer for an integral Seifert form $\theta$ on a free abelian group $L$ of rank $2g$ is a rank-$g$ direct summand $M\subseteq L$ on which the form vanishes:This is the data witnessing a metabolic Seifert form. Over a field, the corresponding definition uses a half-dimensional vector subspace. The same term also denotes a half-dimensional isotropic subspace for a metabolic bilinear form.

For $L={\mathbb{Z}}^2$ with Seifert matrix $V=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, the subgroup spanned by the first basis vector is a metabolizer. Its self-pairing is zero and it has the complementary summand spanned by the second basis vector. The subgroup generated by twice the first basis vector is also isotropic and has rank one, but is not a direct summand: its quotient in $L$ has torsion.

Thus half rank alone does not suffice over the integers. The direct-summand condition ensures that a basis of the metabolizer extends to a basis of the whole lattice, exposing a zero $g$-by-$g$ block in the Seifert matrix.

The module of integer sequences consists of all functions ${\mathbb{N}}_0\to \mathbb{Z}$, with pointwise addition and integer scalar multiplication. Its submodule of finitely supported sequences is the direct sum of modules ${⨁}_{j\ge 0}\mathbb{Z}$, a free module with coordinate basis of a module $(e_j)$. The full product is not a free module.

To prove this, suppose it had a basis of a module $E$. The countable basis support lemma places every finitely supported sequence in a submodule $C$ generated by a countable subset of $E$. The basis partition gives a module direct sum lemma supplies a projection $\pi$ onto its complementary direct summand, with kernel $C$. By countability of countably generated abelian groups, $C$ is countable. The Cantor diagonal argument makes the family of all sequences $w(j)=\pm 2^j$ uncountable, so choose one with $\pi (w)\ne 0$. Truncating $w$ before coordinate $i$ writes $w=f_i+2^i{g}_i$, where $f_i$ is finitely supported and $g_i$ is an integer sequence. Thus $\pi (w)=2^i\pi (g_i)$ for every $i$. The vanishing of infinitely divisible integer sequences forces $\pi (w)=0$, a contradiction.

Construct the sum successively, retaining a simple module only when it adds a new direct summand. Suppose $S=V_1+\cdots +V_{j-1}$ is already expressed as a direct sum of modules selected from these simple modules. The intersection $S\cap V_j$ is a submodule of the simple module $V_j$, so it is either zero or all of $V_j$. In the latter case $S+V_j=S$. In the former caseStarting with $S=0$ and proceeding by mathematical induction gives a direct sum of modules equal to $V_1+\cdots +V_n$. Thus the sum is a semisimple module. The argument proves the useful finite sum of simple submodules lemma without claiming that the originally listed submodules are themselves independent.

The matrix with rows $u,v,e_3$ has determinantIts inverse has integer entries by the adjugate formula, so these rows form an integer basis of $L$. The unimodular basis test for an integer direct summand now givesThus $N$ is a direct summand, with the displayed complement. This proves integral splitting, not just real or rational linear independence.

A proper finite-index subgroup of a torsion-free group cannot be a direct summand when the complement is a subgroup of that same free abelian group. Indeed, if $L=M\oplus K$, every $k\in K$ has $18k\in M$ because the quotient has order eighteen. But also $18k\in K$, so $18k=0$ by the trivial intersection. The ambient group ${\mathbb{Z}}^3$ is torsion-free, forcing $k=0$ and hence $K=0$. This would give $M=L$, contrary to index eighteen. Thus $M$ is not a direct summand.

The ring form of the Artin–Wedderburn theorem states that the following conditions on a unital ring $R$ are equivalent: $R$ is a semisimple ring; $R$ satisfies the descending chain condition on left ideals and $J(R)=0$; andwhere $s$ is finite, each $n_i\ge 1$, and each $D_i$ is a division ring. The factors are unique up to permutation and isomorphism. The zero ring may be included as the empty product. This theorem also implies the corresponding right-handed statements.

First prove the implication from the chain condition and zero Jacobson radical. Among finite intersections of maximal left ideals, choose a minimal one, say $L=L_1\cap \cdots \cap L_t$. Such a minimal intersection exists by the descending chain condition. Intersecting with any other maximal left ideal cannot make $L$ smaller, so $L$ is contained in every maximal left ideal. Thus $L\subseteq J(R)=0$. It follows that the map of left modulesis injective. Each quotient is a simple module.

For completeness, a submodule $N$ of a finite direct sum $V={⨁}_j{T}_j$ of simple modules is itself a finite direct sum of simple modules and is a direct summand of $V$. Choose a sum $W$ of coordinate summands, maximal among those with $N\cap W=0$. For any omitted coordinate summand $T_j$, simplicity gives $T_j\cap (N+W)=0$ or $T_j\subseteq N+W$. In the first case $N\cap (W\oplus T_j)=0$, contradicting maximality; hence every coordinate summand lies in $N+W$. Therefore $V=N\oplus W$. Projection onto $N$ shows $N$ is a finite sum of images of the $T_j$, each image zero or simple. Add these images successively, discarding any image already in the sum: simplicity makes each new intersection zero. This yields a direct-sum decomposition. Applying this argument to the displayed injection proves that the left regular module is a finite sum of simple modules.

Conversely, if the regular module is a direct sum of simple modules, that sum is finite because $1$ lies in a finite subsum and generates the whole regular module. Every left ideal, being a submodule, has the preceding decomposition and splitting property. Its chain lengths are bounded by the number of simple summands, so the descending chain condition holds. The Jacobson radical annihilates every simple summand of the regular module, hence annihilates $1$ and is zero. This identifies the first two conditions.

Write the finite decomposition using pairwise nonisomorphic simple left modules:Every simple module is a quotient of $R$, by sending $1$ to any nonzero vector, so this list includes every simple isomorphism class. The Schur lemma gives ${\mathrm{Hom}}_R(S_i,S_j)=0$ for $i\ne j$, and $E_i={\mathrm{End}}_R(S_i)$ is a division ring: any nonzero endomorphism has zero kernel and full image by simplicity. ConsequentlyEvery endomorphism of the left regular module is right multiplication by its value on $1$. Composition reverses multiplication of those values, so the left side is $R^{\mathrm{op}}$. Taking the opposite ring, and using transpose to identify $M_n(E{)}^{\mathrm{op}}$ with $M_n(E^{\mathrm{op}})$, gives the theorem's decomposition with $D_i=E_i^{\mathrm{op}}$. This keeps the left/right conventions explicit.

Conversely, the standard column module $D^n$ over $M_n(D)$ is simple: a nonzero coordinate of a vector can be inverted to construct a matrix sending it to any prescribed vector. The left regular module decomposes as the sum of its $n$ column ideals, each isomorphic to $D^n$. A finite product of these matrix rings is therefore a semisimple ring. This proves all implications. For uniqueness, the isomorphism classes of simple modules and their multiplicities in the regular module determine the numbers $n_i$ and the division rings $E_i^{\mathrm{op}}$. The multiplicity can also be recovered as the dimension of ${\mathrm{Hom}}_R(S_i,R)$ as a right $E_i$-module, so it does not depend on the chosen decomposition.

Now let $A=\mathbb{C}[G]$ for the finite group $G$. It has dimension $|G|$, with the elements of $G$ as a basis, so its left ideals satisfy the descending chain condition by their dimensions. The permitted assumption $J(A)=0$ makes the Artin–Wedderburn theorem applicable. Its decomposition is a complex-algebra decompositionwith each $D_i$ finite-dimensional over the central copy of $\mathbb{C}$. Each such division algebra is just $\mathbb{C}$. Indeed, for $d\in D_i$, finite dimensionality gives a nonzero polynomial relation $p(d)=0$ with complex coefficients. Factor $p$ into linear factors over $\mathbb{C}$. Their product at $d$ is zero, so the absence of zero divisors forces $d-\lambda =0$ for some scalar root $\lambda$. Thus every element is scalar.

The pairwise nonisomorphic simple modules of ${\prod }_i{M}_{n_i}(\mathbb{C})$ are the column modules ${\mathbb{C}}^{n_i}$, with all other factors acting as zero. Hence ${\dim }_{\mathbb{C}}{S}_i=n_i$. Taking complex dimensions gives

Since $M$ is a finitely generated module, choose a surjection from a finite free module, $F_0\twoheadrightarrow M$. A finite free module over a left Noetherian ring is a Noetherian module, so the kernel $K_1$ is finitely generated. Choose a finite free module $F_1$ surjecting onto $K_1$, then compose its map with $K_1\hookrightarrow F_0$. Its kernel is again finitely generated. Repeating constructs an exact sequencewith all $F_i$ finite free modules, hence finitely generated projective modules. Thus the requested projective resolution exists, without asserting that it has finite length.

Assume now that $M\ne 0$ and ${\mathrm{pd}}_{R}M=n<\infty$. We may arrange a length-$n$ projective resolution with all terms finite projective. Here is why the preceding construction gives one. Set $K_0=M$, with short exact sequences $0\to K_{j+1}\to F_j\to K_j\to 0$. Applying the Ext functor, whose positive degrees vanish on a projective first argument, gives by repeated dimension shiftingfor every left module $N$. The final equality follows from the existence of any length-$n$ projective resolution. Vanishing of these degree-one groups means that every short exact sequence ending in $K_n$ splits: its extension class is zero. In particular a free surjection onto $K_n$ splits, so $K_n$ is a projective module. It is finitely generated by the Noetherian kernel construction. Use $P_n=K_n$ and $P_j=F_j$ for $j<n$ to obtainFor a left finite projective module $P$, put $P^{\ast }={\mathrm{Hom}}_R(P,R)$, a right module under $(fr)(p)=f(p)r$. It is a right finite projective module, and its natural evaluation mapis an isomorphism. These claims hold for a finite free module by coordinates, and pass to direct summands, proving them for every finite projective module without any right-Noetherian hypothesis.

Suppose $n\ge 1$ and ${\mathrm{Ext}}_R^n(M,R)=0$. Computing the Ext functor from this resolution shows that its last dual mapis surjective. Since $P_n^{\ast }$ is projective as a right module, this map has a right-module section $h:P_n^{\ast }\to P_{n-1}^{\ast }$, with $d_n^{\ast }h=1$. Dualizing gives $h^{\ast }{d}_n^{\ast \ast }=1$. Under the natural double-dual identifications, $d_n^{\ast \ast }$ is $d_n$, so this gives a left-module retraction of $d_n$.

Therefore $P_{n-1}=d_n(P_n)\oplus Q$ for a finite projective module $Q$. The next differential has kernel $d_n(P_n)$, so restricting it to $Q$ gives an exact resolution with $Q$ replacing $P_{n-1}$ and the initial $P_n$ removed. This has length $n-1$, contradicting the definition of projective dimension. For $n=0$, the nonzero $M$ is a finite projective module and a summand of some $R^d$. A nonzero vector in that summand has a nonzero coordinate, so one coordinate projection restricts to a nonzero map $M\to R$. Thus the degree-zero Hom functor is nonzero as well. We have provedThis is top Ext detects finite projective dimension over a left Noetherian ring. The nonzero-module convention matters when $n=0$: if one assigns projective dimension zero to the zero module, then ${\mathrm{Hom}}_R(0,R)=0$ is an exception. With the convention ${\mathrm{pd}}_{R}0=-\infty$, it is outside the asserted finite nonnegative case.

For an example of infinite projective dimension, take the dual numbers $R=k[\epsilon ]/({\epsilon }^2)$ over a field $k$, and $M=k=R/(\epsilon )$. The periodic resolution over dual numbers isIt is exact because both the kernel and image of multiplication by $\epsilon$ are the ideal $(\epsilon )$, and that is also the kernel of $R\to k$. Applying ${\mathrm{Hom}}_R(-,k)$ gives a copy of $k$ in every degree and zero differentials, since $\epsilon$ acts as zero on $k$. ConsequentlyA finite projective resolution would make these groups zero above its length, so

An abelian category is an additive category with kernels in a category and cokernels in a category in which the canonical comparison from coimage to image of every morphism is an isomorphism. This is the standard image-coimage form of the definition, equivalently expressed by normality of every monomorphism and epimorphism. Additivity means abelian-group hom-sets with bilinear composition, a zero object and finite biproducts. A sequence is exact at an object when the image of the incoming morphism is the kernel in a category of the outgoing one.

For $f:X\to Y$, construct $q:X\to \mathrm{coim}f=\mathrm{coker}(\ker f)$ and $j:\mathrm{im}f=\ker (\mathrm{coker}f)\to Y$. The universal properties give $f=j\bar{f}q$: first $f$ kills its kernel in a category and factors through $q$; then the resulting map is killed by $\mathrm{coker}f$ because $q$ is epic, so it factors through $j$. The comparison $\bar{f}$ is invertible by the abelian axiom. Cokernels in a category are epic and kernels in a category monic, directly by uniqueness in their universal properties, so $e=\bar{f}q$ is epic and $j$ monic. Thus $\boxed{f=je\text{ is an epimorphism-monomorphism factorization}}$.

In a short exact sequence $0\to X\stackrel{i}{\to }Y\stackrel{p}{\to }Z\to 0$, we have $i=\ker p$ and $p=\mathrm{coker}i$, up to the canonical isomorphisms. If $r:Y\to X$ retracts $i$, the map $1_Y-ir$ kills $i$ and factors as $sp$ for some $s:Z\to Y$. Composing with $p$ gives $psp=p$, so epicity gives $ps=1_Z$. Conversely if $ps=1_Z$, then $1_Y-sp$ is killed by $p$ and factors as $ir$. Composing with $i$ gives $iri=i$, so monicity gives $ri=1_X$. ConsequentlyIn that case $Y\cong X\oplus Z$.

For the derived-functor requests, use additive right exact functors, as customary in abelian homological algebra. An abelian category with enough projectives permits an augmented projective resolution $\cdots \to P_2\to P_1\to P_0\to X\to 0$, obtained by repeatedly taking an epimorphism from a projective onto the previous kernel in a category. Apply $F$ and defineProjective lifts construct maps of resolutions; the comparison and chain-homotopy results make the construction independent of choices and functorial. For a short exact sequence, the horseshoe construction produces a degreewise split short exact sequence of resolutions, hence of their images under additive $F$. Its homology objects give the long exact sequenceThis supplies the requested sketch; detailed comparison and functoriality proofs are not needed here.

A projective object $X$ has a resolution concentrated in degree zero, so $L_{1}F(X)=0$ for every such $F$. Conversely choose $0\to K\stackrel{i}{\to }P\to X\to 0$ with $P$ projective, and take $F=\mathcal{A}(-,K)$ with target ${\mathbf{Ab}}^{\mathrm{op}}$. Precomposition makes this covariant to the opposite category. The usual left exactness of contravariant Hom, which follows directly from the cokernel in a category universal property, says exactly that this $F$ is right exact in that opposite target. If $L_{1}F(X)=0$, the long exact sequence and $L_{1}F(P)=0$ make $F(K)\to F(P)$ monic in ${\mathbf{Ab}}^{\mathrm{op}}$. Equivalently,is epic, hence surjective, in abelian groups. The identity of $K$ has a preimage, giving a retraction of $i$. The sequence splits, and $X$ is a direct summand of $P$, hence projective: a lifting problem for $X$ extends along the projection $P\to X$, is solved for $P$, and restricts along the summand inclusion. ThereforeThis is projectivity detected by all first left derived functors. The following parts prove the three requested equivalent formulations of a hereditary abelian category with enough projectives.

Use the usual Krull–Schmidt theorem setting, in which indecomposable summands have local endomorphism rings; the paper's rings $\mathcal{O},k$ satisfy it for the subgroups involved. The relative projective module condition for $H$ means that $M$ is a direct summand of a module induced from $H$, equivalently, $M$ is a summand of ${\mathrm{Ind}}_H^G{\mathrm{Res}}_H^{G}M$. A vertex of an indecomposable module is an inclusion-minimal such subgroup. A source of an indecomposable module at its module vertex $D$ is an indecomposable $RD$-summand $S$ of $M{\downarrow }_D$ such that $M\mid S{\uparrow }^G$.

Existence of a module source follows by decomposing $M{\downarrow }_D$ and using Krull–Schmidt theorem on the induced decomposition: the indecomposable summand $M$ must occur in the induction of one indecomposable summand $S$. Such $S$ has module vertex $D$ as a $D$-module, since relative projectivity of $S$ for a proper subgroup would make $M$ relatively projective for that subgroup.

In the paper's modular coefficient rings, an indecomposable module is a trivial source module when its module source is the rank-one trivial module $R$ of its module vertex. It is then a summand of $R[G/D]$, a permutation module. Conversely suppose $M$ is an indecomposable summand of a permutation module. Restriction to a module vertex $D$ is a sum of modules $R[D/E]$, one for each $D$-orbit of the permutation basis. For a $p$-group $D$, each transitive permutation module is indecomposable: over $k$ it has simple head of a module over the local ring $kD$, and over $\mathcal{O}$ a nontrivial splitting would give a nontrivial splitting after reduction. A module source $S$ must be one of these summands by Krull–Schmidt theorem. Its module vertex is $D$, whereas $R[D/E]$ is relatively $E$-projective. Thus $E=D$, and $S\cong R$. Therefore trivial module source is equivalent to being an indecomposable permutation-module summand.

The rank-one formulation uses the paper's coefficient rings. If the earlier general coefficient-ring wording were applied to a disconnected ring, it would need qualification: for $R=k\times k$ and $G=1$, $(k,0)$ is an indecomposable summand of the permutation module $R$, but is not the rank-one trivial module $R$. One can instead allow indecomposable summands of the trivial coefficient module as module sources in that generality.

For the homomorphism request, take $M_i=\mathcal{O}[G/H_i]$. Frobenius reciprocity and Mackey decomposition giveThe summand for $g$ is generated by the sum of the cosets in the $H_1$-orbit of $g{H}_2$. The identical orbit-sum construction over $k$ gives one copy of $k$ per double coset. These are free bases in both characteristics, since invariance means coefficients are constant on orbits, with no division by orbit size. HenceThe same argument applies to direct sums of coset permutation modules.

If $M$ is a trivial module source $kG$-module, choose a permutation lattice $X$ with $M\mid \bar{X}$. Its projection is an idempotent $f\in {\mathrm{End}}_{kG}(\bar{X})$. The endomorphism reduction is surjective by the preceding calculation. Its kernel is $\mathfrak{p}{\mathrm{End}}_{\mathcal{O}G}(X)$, and the endomorphism algebra is complete over $\mathcal{O}$. By idempotent refinement theorem, $f$ lifts to an idempotent $\hat{f}$. The lattice $\hat{f}X$ is a permutation-module summand, hence a trivial module source lattice, and reduces to $M$.

To prove uniqueness within the class of trivial module source lattices, let $L,L^{\prime }$ be two such lifts. They are summands of permutation lattices. Lift a modular isomorphism and its inverse between them by the same homomorphism-surjectivity argument, inserting the summand projections on either side. The lifted composites are identity modulo $\mathfrak{p}$ and therefore invertible over $\mathcal{O}$, by Nakayama lemma or the determinant criterion. The lifted maps are consequently inverse up to an invertible composite and give $L\cong L^{\prime }$. Thus each trivial module source module has a unique trivial module source lift up to isomorphism. This is permutation homomorphisms lift through modular reduction; uniqueness is not claimed among arbitrary lattice lifts.

Let $B=kGe$ have defect group $D$. Use two vertex-and-source facts, with their relevant content made explicit. Every indecomposable module in $B$ has a module vertex conjugate into $D$. Also $B$, as a $k[G\times G]$-module with $(g,h)a=ga{h}^{-1}$, is an indecomposable permutation-module summand with module vertex $\Delta D=\{(d,d):d\in D\}$ and trivial module source. The second fact is the bimodule characterization of a defect group of a block; the first follows by applying its relative-$\Delta D$ induction splitting to $B{\otimes }_{kG}M=M$.

If $D$ is cyclic, every module source for every module vertex $Q\le D$ occurs among the finitely many indecomposable modules for a cyclic p-group. Each module in the block is a summand of an induction of one of these finitely many module sources. There are finitely many subgroups $Q$ and each induced representation has finitely many indecomposable summands by Krull–Schmidt theorem. Thus cyclic defect implies finite representation type of the block.

For the converse, the restriction $B{\downarrow }_{D\times D}$ has the regular $kD$-bimodule as a direct summand. To see this from module vertices and module sources, its Brauer quotient of a module at $\Delta D$ is nonzero. Its summands are permutation modules for the $p$-group $D\times D$ and have module vertices of order at most $|D|$. A summand detected at $\Delta D$ therefore has module vertex $\Delta D$ and trivial module source, hence isHere a transitive permutation module for a $p$-group is indecomposable: it has one-dimensional head of a module over the local group algebra and cannot split into two nonzero modules. This also explains the Brauer-quotient detection used in the argument.

For any $kD$-module $U$, tensor that split bimodule inclusion with $U$. It giveswhere $\mid$ denotes being a direct summand. If $B$ has only finitely many indecomposable modules $M_1,\dots ,M_s$, each $B{\otimes }_{kD}U$ is a sum of these. Therefore every indecomposable $U$ must occur among the summands of the finite list $M_i{\downarrow }_D$. Thus $kD$ has finite representation type.

A noncyclic $p$-group has a quotient $C_p\times C_p$: its Frattini quotient has rank at least two, since a one-generator Frattini quotient forces the group itself to be cyclic. Inflation along this quotient embeds all the indecomposable modules of that elementary abelian group, preserving isomorphisms and indecomposability. Part (b) rules out finite type when $k$ is infinite.

The conclusion remains valid if the splitting field $k$ is finite. For completeness, replace the scalar parameter in part (b) by an $m\times m$ companion matrix $T_f$ of an irreducible polynomial $f$. On $k^m\oplus k^m$ useAn endomorphism commuting with these has form $\left(\begin{array}{cc}A& 0\\ C& A\end{array}\right)$ with $A{T}_f=T_{f}A$. Its endomorphism ring has a square-zero ideal of the $C$-matrices and quotient $k[T_f]\cong k[t]/(f)$, a field, so it is local and the module is indecomposable. Finite fields have irreducible polynomials of arbitrarily large degrees, giving infinitely many dimensions. This also excludes noncyclic $D$ in that case. Altogether,

First show that free modules are flat. If $F={⨁}_{j\in J}A{e}_j$, distributivity of the tensor product of modules givesExplicitly, $u\otimes {\sum }_j{a}_j{e}_j$ maps to $(a_{j}u{)}_j$, and the inverse sends a finitely supported tuple $(u_j{)}_j$ to ${\sum }_j{u}_j\otimes e_j$. For an injection $u:U\hookrightarrow V$, the tensor map is therefore the coordinatewise injection ${⨁}_{J}U\hookrightarrow {⨁}_{J}V$. This proves flatness for any free rank.

By part (c), a projective module $P$ is a direct summand of a free module $F$. The latter is flat by the preceding argument, and part (b) proves that its summand $P$ is flat. Therefore $\boxed{\text{every projective }A\text{-module is flat}}$. This proves projective modules are flat directly from the exactness and splitting definitions.

Write $A=kG$ and $J=J(A)$, its Jacobson radical. First suppose the characteristic is $p>0$. We prove the counting statement through the cocenter of a finite-dimensional algebrawhere $[A,A]$ is the linear span of commutators, not an ideal. In a group algebra, this span is exactly the span of differences of conjugate group elements: $gh-hg$ is such a difference, and $xg{x}^{-1}-g=[x,g{x}^{-1}]$. Consequently $C(A)$ has one basis vector for each conjugacy class.

There is a well-defined Frobenius map on an algebra cocenterHere $F$ is additive and $F(c[a])=c^{p}F([a])$. To prove additivity, expand $(a+b{)}^p$ into words of length $p$. Cyclic rotations of a word are equal modulo commutators. Every nonconstant word has a rotation orbit of length $p$, so its orbit contributes zero; only $a^p,b^p$ remain. Alsobecause the two products are cyclic rotations. Additivity then shows that the whole commutator subspace maps to zero, proving well-definedness on the quotient.

Let $s$ be the number of simple modules up to isomorphism. The semisimple algebra $A/J$ is ${\prod }_{\nu =1}^s{M}_{d_{\nu }}(k)$: over an algebraically closed field, a finite-dimensional division algebra is the field itself, since the minimal polynomial of any element splits into linear factors and a division algebra has no zero divisors. In a matrix algebra, the commutator subspace is precisely the traceless matrices. Off-diagonal matrix units are commutators, and $E_{ii}-E_{jj}=[E_{ij},E_{ji}]$; these span that subspace in every characteristic. ThusThe map induced by $F$ on each coordinate is $z\mapsto z^p$, since $\mathrm{tr}(B^p)=(\mathrm{tr}B{)}^p$. This is bijective because $k$ is a perfect field.

The natural surjection $C(A)\twoheadrightarrow C(A/J)$ has kernel $(J+[A,A])/[A,A]$. Every kernel class has a representative in $J$. Since $J$ is nilpotent, $F^N$ kills that kernel for sufficiently large $N$. On the quotient it remains bijective, so its image has dimension exactly $s$. On the conjugacy-class basis, however,For large $N$, every $g^{p^N}$ is a p-regular element; moreover the $p^N$-power map permutes the p-regular conjugacy classes, because an inverse exponent can be chosen modulo the prime-to-$p$ part of the exponent of $G$. The stable image therefore has precisely those conjugacy classes as its basis. Comparing dimensions provesThis argument uses dimensions of semilinear images: a bijective scalar Frobenius map preserves dimension, so the usual rank count applies.

In characteristic zero every group element is relevant. Averaging a linear projection onto a submodule over $G$ makes that projection $G$-equivariant, proving Maschke's theorem and semisimplicity of $kG$. The same matrix-algebra cocenter calculation now gives $s=\dim C(kG)$, the total number of conjugacy classes.

For the second request, $M$ is a relative projective module for $H$ when every $G$-equivariant epimorphism onto $M$ which splits after restriction to $H$ already splits over $G$. Equivalently $M$ is a direct summand of $kG{\otimes }_{kH}M$, the induced representation of the restriction of a representation to $H$. We can verify the defining splitting property directly: if $q:N\twoheadrightarrow M$ has an $H$-linear section $u$, choose representatives $T$ of the left cosets $G/H$ and setThe sum is independent of representatives by $H$-linearity. Reindexing $gt=t^{\prime }h$ proves $G$-linearity, while $qv(m)=[G:H{]}^{-1}{\sum }_{t}m=m$. Hence every $M$ is relatively $H$-projective when the index is invertible in $k$.

For completeness the canonical induced-module epimorphism $\epsilon (g\otimes m)=gm$ has the explicit $G$-linear sectionIt exhibits the promised direct summand and is the projectivity detected on a subgroup of invertible index construction.