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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.

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.

An idempotent $e$ gives a projective module $Ae$, since $A=Ae\oplus A(1-e)$. It is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. This is equivalent to indecomposability of $Ae$: a projection onto a direct summand is an idempotent of ${\mathrm{End}}_A(Ae)\cong (eAe{)}^{\mathrm{op}}$, and conversely such an idempotent splits $Ae$. Starting with $1$, split any nonprimitive summand repeatedly. The process terminates because each splitting increases the number of nonzero vector-space summands in a finite-dimensional algebra. Thuswith orthogonal primitive idempotents and indecomposable projective modules. Each summand is generated by its idempotent.

Let $D={\mathrm{Hom}}_k(-,k)$. A Frobenius algebra is a finite-dimensional algebra with $A\cong D(A)$ as left modules over $A$, where $(a\varphi )(x)=\varphi (xa)$. Equivalently there is a linear functional $\lambda :A\to k$ for which $(x,y)\mapsto \lambda (xy)$ is nondegenerate. A symmetric algebra satisfies the stronger bimodule isomorphism $A\cong D(A)$, with $(a\varphi b)(x)=\varphi (bxa)$; equivalently the functional can be chosen with $\lambda (xy)=\lambda (yx)$. This is the associative-algebra meaning of symmetric, rather than the symmetric algebra of a vector space.

For a finite-dimensional module,The socle is its largest semisimple submodule, and the radical of a module gives its largest semisimple quotient. With $J=J(A)$ one has $\mathrm{Rad}M=JM$ and $\mathrm{Soc}M=\{m:Jm=0\}$: $A/J$ is semisimple, and $J$ annihilates every simple module.

Now let $P=Ae$ for a primitive idempotent, allowing the field $k$ to be nonsplit. Symmetry identifies $Ae$ with $D(eA)$ through the pairingIt is nondegenerate on these corners. If $v=ve\ne 0$, choose $a$ with $\lambda (av)\ne 0$; then $\lambda (eav)=\lambda (ave)=\lambda (av)\ne 0$. The other variable is treated similarly. The pairing respects the left $A$-action, soThe right-module quotient $eA/eAJ$ is simple. A primitive idempotent remains primitive modulo the nilpotent radical: idempotents lift through that radical, so a decomposition in the quotient would lift to one in $eAe$. In the semisimple algebra $A/J$, the left simple $Ae/JAe$ and the dual of the right simple $eA/eAJ$ belong to the same matrix block. There is exactly one left simple isomorphism class in that block, even over a nonalgebraically closed field. Hence the head-socle identity for a symmetric algebra isThis is an isomorphism of left modules, and need not be canonical.

For the modular examples put $r=(123)$, $s=(12)$, so $r^3=s^2=1$ and $srs=r^{-1}$. The coefficient-of-identity functional on the group algebra is symmetric and nondegenerate, so the identity just proved applies.

In characteristic two, the central idempotents are $b_0=1+r+r^2$ and $b_1=1-b_0$. The first block has basis $b_0,b_{0}s$ and is $k{C}_2\cong k[u]/(u^2)$. For the second, choose a primitive cube root $\omega$ and useThe two $r$-eigenlines are interchanged by $s$, proving irreducibility. Polynomial interpolation in $r$ gives diagonal matrix units, and multiplying by $s$ gives the off-diagonal units. Thus the four-dimensional block maps onto $M_2(k)$ and is isomorphic to it:There are two simple modules, the trivial one-dimensional module $T$ and the two-dimensional module $V$; the sign module equals $T$. Their projective covers are $P_T=k[u]/(u^2)$, with layers $T,T$, and $P_V=V$. The regular module is $P_T\oplus V^{\oplus 2}$, and the Cartan matrix of a group algebra is $\mathrm{diag}(2,1)$. Every module is a sum of $T,P_T,V$: the first block is classified by Jordan chains for a nilpotent linear map of lengths one and two, while the matrix block is semisimple. Thus there are exactly three indecomposable isomorphism classes. This describes the 2-modular blocks of S3.

In characteristic three put $u=r-1$. The ideal $I=uA$ satisfies $I^3=0$, and $A/I\cong k{C}_2\cong k\oplus k$, so $J(A)=I$. The only simples are the trivial and sign modules $S_{+},S_{-}$. Put $e_{\pm }=(1\pm s)/2$. The projectives $P_{\pm }=A{e}_{\pm }$ have dimension three. A convenient radical generator is $v=r-r^{-1}$:The basis $e_{\pm },v{e}_{\pm },v^2{e}_{\pm }$ gives the radical layersThus $A=P_{+}\oplus P_{-}$ and its Cartan matrix of a group algebra isThe elements $e_{-}v{e}_{+}$ and $e_{+}v{e}_{-}$ give one arrow in each direction between the two simples. All paths of length three vanish, and the six paths of lengths zero, one and two are linearly independent. Hence $A$ is the path algebra of that two-vertex oriented cycle modulo the ideal of paths of length three. Its connected quiver also proves there is a single block of S3 in characteristic three.

Any module is a $\mathbb{Z}/2$-graded vector space given by the $s$-eigenspaces, with $v$ interchanging the grades and $v^3=0$. Constructing Jordan chains with homogeneous bases in the kernel and image filtrations and homogeneous lifts decomposes it into graded chains. Each chain has length $1,2$ or $3$ and is determined by its top sign. It is a uniserial module and hence indecomposable. Consequently there are exactly six indecomposables, the length-$\ell$ quotients $P_{\pm }/J^{\ell }{P}_{\pm }$, $\ell =1,2,3$.

The finite representation type of a quiver condition means there are only finitely many isomorphism classes of indecomposable finite-dimensional $kQ$-modules. Equivalently, every finite-dimensional module is a direct sum of members of a finite list of indecomposables.

For the $E_6$ case, we give the reflection argument underlying the Gabriel theorem, rather than using its conclusion as the proof. Since the underlying graph is a tree, every orientation is acyclic. At a sink $i$, writefor the sum of the incoming arrow maps. If an indecomposable $V$ is not the simple $S_i$, then $h$ is surjective: a nonzero complementary subspace to its image at the sink would be a direct summand supported only at $i$. The Bernstein–Gelfand–Ponomarev reflection functor reverses the arrows and replaces $V_i$ by $\ker h$, with reversed maps the component projections. The source reflection replaces that kernel by the cokernel of its inclusion into $⨁V_j$, recovering $V_i$. The inclusion is injective, so the reflected representation has no direct summand $S_i$ at the new source. If it decomposed, applying the inverse would decompose $V$. Thus reflection preserves indecomposability except that it annihilates $S_i$.

On a dimension vector of a quiver representation, it acts by the integral simple reflectionLet $B=2I-\mathrm{Adj}(E_6)$. These reflections preserve the positive-definite quadratic form $d^{\mathsf{T}}Bd$. Positivity can be checked without a classification theorem: label the graph by the chain $1-2-3-4-5$ and attach vertex $6$ to $3$. The leading principal minors are $2,3,4,5,6,3$, all positive.

Choose a topological ordering so arrows point from smaller to larger indices. Reflect successively at sinks $6,5,\dots ,1$. This restores the orientation after one full cycle. The transformation of dimension vectors is the Coxeter element $C=s_1{s}_2\cdots s_6$. If $U$ has diagonal entries one and entry $-1$ above the diagonal at an arrow, direct multiplication of the reflection matrices givesThus $C$ has no fixed vector. It is an integral isometry of a positive definite form, so it has finite order $h$: the image of each lattice basis vector lies in a finite set of vectors of the same norm, giving only finitely many possible isometries. Since $C^h=I$ and $I-C$ is invertible,If an indecomposable survived all $h$ reflection cycles, its dimension vectors $d,Cd,\dots ,C^{h-1}d$ would all be nonzero and componentwise nonnegative, contradicting their zero sum. Therefore it reaches a simple at one of the finitely many reflection steps and is then annihilated. Applying the inverse reflections uniquely reconstructs it from that simple. There are at most $6h$ such stopping positions, provingThis supplies the requested finite-type proof sketch.

Define compactly supported cohomology byFor $K\subseteq L$, the transition map is induced by the identity map of pairs $(X,X\setminus L)\to (X,X\setminus K)$. Equivalently, take the cochain complex of singular cochains that vanish on every chain contained in the complement of some compact set. Directed unions are exact, giving the same definition.

For $\mathbb{R}$, the intervals $[-a,a]$, $a>0$, are cofinal among compact subsets. The complement has two contractible components. The long exact sequence in relative cohomology contains the diagonal map $\mathbb{Z}\to \mathbb{Z}\oplus \mathbb{Z}$, so its cokernel is $H^1(\mathbb{R},\mathbb{R}\setminus [-a,a])\cong \mathbb{Z}$, and all the other relative groups vanish. Enlarging the interval preserves the generator given by the difference of the two ends. Therefore

For the compact-support comparison with a one-point compactification, write $Y=X^{+}$. The assumed Hausdorff one-point compactification is compact; a compact subset $K\subseteq X$ is closed in $Y$. The Excision theorem removes $\{\infty \}$ from the pair $(Y,Y\setminus K)$, because its closure lies inside the open second member. ThusComplements of compact subsets of $X$ are exactly the open neighbourhoods of $\infty$ in $Y$. The hypothesis supplies a cofinal family of contractible such neighbourhoods $U$. For every one, the long exact sequence of the pair identifiesIn degree zero, this is the kernel of evaluation on the component of $\infty$, identified with reduced cohomology by subtracting the constant value there. In degree one the map $H^0(Y)\to H^0(U)\cong \mathbb{Z}$ is surjective; in higher degrees the positive cohomology of $U$ vanishes. These identifications are natural for inclusions of contractible neighbourhoods. Passing to the direct limit proves

For the specified disjoint union of lines, a compact subset meets only finitely many components and is bounded in each. Finite unions $F\times [-a,a]$, with $F\subset \mathbb{Z}$ finite, are cofinal. Applying the preceding relative calculation componentwise givesThe one-point compactification $W$ of this space is the Hawaiian earring: each line becomes a circle by adding the common point $\infty$, and every neighbourhood of $\infty$ contains all but finitely many whole circles. On the remaining finitely many circles it contains neighbourhoods of the common point. This describes exactly the shrinking-circle topology. In particular $W$ is not locally contractible at $\infty$: every such neighbourhood contains a whole circle, whose generator remains nontrivial under the retraction $W\to S^1$ that collapses all the other circles.

For integral singular cohomology, the comparison does not hold. Here is a degree-two obstruction that takes account of the shrinking-circle topology. The standard rational summand in Hawaiian earring homology theorem gives a direct summand $\mathbb{Q}$ in $H_1(W;\mathbb{Z})$. The universal coefficient theorem for cohomology injectsThe summand $\mathbb{Q}$ therefore contributes the nonzero Ext of the rationals with integer coefficients.

For completeness, this last algebraic assertion has an explicit proof. Present $\mathbb{Q}$ using generators $a_n=1/n!$ and relations $a_n-(n+1)a_{n+1}=0$, $n\ge 1$. The corresponding free resolution shows that ${\mathrm{Ext}}_{\mathbb{Z}}^1(\mathbb{Q},\mathbb{Z})$ is the cokernel ofThe constant sequence $(1,1,\dots )$ is not in the image. Otherwise iteration would giveFor large $N$, the factorial sum exceeds $|u_1|$ but is less than $(N+1)!-|u_1|$, making that congruence impossible. Thus the cokernel is nonzero. It follows thatwhich proves the failure of the claimed isomorphism. The ingredient concerning the Hawaiian earring is its singular-homology structure theorem, not the homology of an infinite CW complex wedge of circles; these topologies differ.

The Jacobson radical is $J(A)={\bigcap }_{L}L$, where $L$ runs over the maximal right ideals of $A$; equivalently, it is the intersection of the annihilators of all simple modules. It is a two-sided ideal. A projective module has the lifting property against every surjective R-module homomorphism. A finitely generated module is a projective module precisely when it is a direct summand of a finitely generated free module. An indecomposable module is nonzero and admits no direct sum decomposition into two nonzero submodules.

To calculate the top of an indecomposable projective module, use the right Artinian ring hypothesis, namely the descending chain condition on right ideals. The Hopkins-Levitzki theorem gives finite composition length of the right regular module, hence of its submodule $P$. Also $J=J(A)$ is a nilpotent ideal, and $A/J$ is a semisimple ring. Thus the quotient module $\overline{P}=P/PJ$ is a semisimple module; it is nonzero, since $P=PJ$ would imply $P=P{J}^m=0$ for sufficiently large $m$.

Suppose $\overline{P}$ were not a simple module. A nontrivial direct sum decomposition of this semisimple module would give an idempotent $\alpha \in {\mathrm{End}}_A(\overline{P})$ that is neither zero nor the identity. Writing $\pi :P\to \overline{P}$, the lifting property of the projective module $P$ gives $\beta \in {\mathrm{End}}_A(P)$ with $\pi \beta =\alpha \pi$. The Fitting lemma for an indecomposable module of finite composition length says that $\beta$ is either invertible or a nilpotent element. Its induced map $\alpha$ would then be invertible or a nilpotent element, respectively. Neither is possible for a nontrivial idempotent. Therefore $P/PJ(A)$ is simple. This argument does not assume that an embedded projective module automatically splits off from the ambient module.

A block of an Artinian algebra is a nonzero two-sided direct summand $Ae$ determined by a primitive central idempotent $e$: $e$ cannot be written as a sum of two nonzero orthogonal central idempotents. Its identity is $e$. For a finite-dimensional associative algebra, the blocks of an Artinian algebra give its unique decomposition as a finite product of indecomposable algebras, or equivalently as a direct sum of two-sided ideals.

For the block of S3 in characteristic three, put $A=k{S}_3$, $r=(123)$, $s=(12)$ and $a=r-1$. The group algebra has basis $r^i,r^{i}s$ for $0\le i<3$. In characteristic three,Consequently $I=aA$ is a two-sided nilpotent ideal, $I^3=0$, and $A/I\cong k{C}_2\cong k\times k$. A nilpotent ideal lies in the Jacobson radical, and a quotient that is a semisimple ring forces the reverse inclusion. HenceDefine orthogonal idempotents $e_{+}=(1+s)/2$ and $e_{-}=(1-s)/2$. They sum to one, so the right regular module decomposes asThe two summands are the indecomposable projectives of S3 in characteristic three. Each direct summand is a projective module, with basis $e_{\pm },e_{\pm }r,e_{\pm }{r}^2$. Its quotient module modulo multiplication by $J(A)$ is one-dimensional: the trivial representation for $e_{+}A$, and the sign representation for $e_{-}A$. Each is an indecomposable module, since two nonzero direct summands would each have nonzero quotient module modulo $J(A)$, contradicting its one-dimensional top. For additional detail, the successive factors of the radical series of a module are the trivial representation, sign representation, trivial representation for $e_{+}A$, and the sign representation, trivial representation, sign representation for $e_{-}A$. Indeed, modulo $J^2$ the relation $sa=-as+a^{2}s$ reverses the $s$-sign, whereas $s$ commutes with $a^2$.