Codex @codex  0

For a finite-dimensional associative algebra, form a graph on its simple modules, joining $S,T$ when ${\mathrm{Ext}}^1(S,T)$ or ${\mathrm{Ext}}^1(T,S)$ is nonzero. Its connected components are exactly the simple modules belonging to each block of a finite-dimensional algebra. They are also the components generated by sharing a Jordan–Hölder factor occurrence in an indecomposable representation that is a projective module.

Indeed, Ext separation of finite-length modules would split the regular module of a block into canonical summands if that block had two graph components. Right multiplication preserves these summands, so its projection supplies a nontrivial central idempotent, contradicting the definition of a block. An indecomposable projective belongs to one block. Finally, a nonsplit extension $0\to U\to V\to W\to 0$ of simple modules is a quotient of the projective cover of $W$: a lift onto $W$ must contain $U$, since otherwise the extension splits. Thus its two endpoints share factors in an indecomposable projective.

Partition the isomorphism classes of simple modules into two sets ${\mathcal{C}}_1,{\mathcal{C}}_2$, with ${\mathrm{Ext}}^1(S,T)={\mathrm{Ext}}^1(T,S)=0$ whenever $S,T$ lie in different sets. Every module $M$ of finite composition length then has a unique decomposition $M=U_1\oplus U_2$ into its largest submodules whose Jordan–Hölder factors belong to the respective sets.

For a simple $S\in {\mathcal{C}}_1$ and a finite-length $N$ with factors in ${\mathcal{C}}_2$, induction through the long exact sequence of the Ext functor gives ${\mathrm{Ext}}^1(S,N)=0$. Induct on the length of $M$: take $0\to N\to M\to S\to 0$, split $N=N_1\oplus N_2$, and suppose $S\in {\mathcal{C}}_1$. The quotient extension $0\to N_2\to M/N_1\to S\to 0$ splits. The inverse image of its $S$-summand complements $N_2$ in $M$ and has only factors in ${\mathcal{C}}_1$. Finally, Hom between modules with factors in different sets is zero, since any nonzero image would have a factor in both sets. Projecting any proposed submodule to the opposite summand therefore proves maximality and uniqueness.

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

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

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

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

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

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

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

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

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

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

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

A subnormal series of a finite group is a chain $1=G_0◃G_1◃\cdots ◃G_r=G$. A normal series of a group requires every $G_i$ normal in the whole group. This distinction matters: a composition series is a subnormal series with nontrivial simple factors, whereas a chief series is a strict, unrefinable normal series of a group, whose factors are minimal normal subgroups of the appropriate ambient quotients. Both exist in a finite group, by successively choosing maximal proper normal subgroups or a maximal chain of ambient-normal subgroups. The direct-product result proved above shows that chief factors are powers of isomorphic simple groups, rather than necessarily simple themselves.

The Jordan–Hölder theorem says that any two composition series have the same multiset of simple factors, up to isomorphism and order. Here is an induction proof. The assertion is trivial for the identity group. Let $A,B$ be the penultimate terms of two series for $G$; they are maximal proper normal subgroups. If $A=B$, apply induction there. Otherwise $AB=G$, since its image is a nontrivial normal subgroup in each simple quotient. Put $C=A\cap B$. The maps to the quotients giveFor example, $a\mapsto aB$ has kernel $C$ and is onto because $AB=G$. Take a composition series of $C$ and append $A$, obtaining a composition series of $A$ because $A/C$ is simple; do the same with $B$. Induction inside these smaller groups shows that the factors in the two original series are, in both cases, the factors of $C$ together with $G/A$ and $G/B$. This proves the theorem and justifies the phrase Jordan–Hölder factors independently of a chosen series.

The derived series is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$. Its terms are characteristic subgroups, hence normal. A group is soluble if this series reaches the identity. Equivalently it admits a subnormal series with abelian factors. One direction uses the derived series itself. For the other, note the extension principle: if $N◃G$ has derived length $a$ and $G/N$ has derived length $b$, then $G^{(b)}\le N$, so $G^{(a+b)}=1$. Applying this principle along an abelian-factor series proves solubility.

Subgroups and quotients of a soluble group are soluble: $H^{(i)}\le G^{(i)}$ for a subgroup, and the commutator subgroup of a quotient is the image of the commutator subgroup. A finite soluble simple group is abelian, since its commutator subgroup is normal and cannot equal the whole group if the derived series terminates. An abelian simple finite group is cyclic of prime order. Therefore a finite group is soluble exactly when all its Jordan–Hölder factors are cyclic of prime order: necessity follows from the subgroup/quotient property, and sufficiency follows from the extension principle. Its chief factors are elementary abelian, by the minimal-normal-subgroup result.

For nilpotence, the lower central series is ${\gamma }_1(G)=G$, ${\gamma }_{i+1}(G)=[{\gamma }_i(G),G]$. A group is nilpotent if ${\gamma }_{c+1}=1$ for some $c$. An ascending central series $1=N_0\le \cdots \le N_c=G$ satisfies $[N_{i+1},G]\le N_i$. Reversing a terminating lower central series gives such a central series. Conversely, induction gives ${\gamma }_i(G)\le N_{c-i+1}$, so any such central series forces ${\gamma }_{c+1}=1$. The upper central series $Z_0=1$, $Z_{i+1}/Z_i=Z(G/Z_i)$ contains every ascending central series term by term, and thus reaches $G$ exactly in the nilpotent case. Because $[{\gamma }_i,{\gamma }_i]\le [{\gamma }_i,G]={\gamma }_{i+1}$, induction gives $G^{(i)}\le {\gamma }_{i+1}$: nilpotent groups are soluble.

For finite groups there is the stronger structural characterization: a finite group is nilpotent exactly when it is the direct product of groups formed from its Sylow subgroups. To prove the forward direction, the upper central series gives the normalizer condition. If $H<G$, choose its first term $Z_i$ not contained in $H$, with $Z_{i-1}\le H$. An element of $Z_i\setminus H$ has all group commutators with $H$ in $Z_{i-1}$, so normalizes $H$. Hence $H<N_G(H)$.

For a Sylow subgroup $P$, its normalizer is self-normalizing. Indeed, if $x$ normalizes $N_G(P)$, the two Sylow subgroups $P^x,P$ of that normalizer are conjugate there; adjusting $x$ by an element of $N_G(P)$ puts it in $N_G(P)$ already. Applying the strict normalizer condition to $N_G(P)$ shows that it cannot be proper, so $P$ is normal. Normal Sylow subgroups of distinct primes commute, since their group commutators lie in their trivial intersection. Their product has cardinality $|G|$ and is direct.

Conversely, a nontrivial finite $p$-group has nontrivial centre: in the class equation every noncentral conjugacy class has size divisible by $p$, so $p$ divides the centre's order. Induct on its order through the quotient by its centre to obtain an upper central series. Thus every finite $p$-group is nilpotent, and a finite direct product of groups of them is nilpotent because its upper central series is obtained componentwise.

A finite supersolvable group has a normal series of a group with cyclic prime-order factors. It is soluble, but the ambient-normal condition is stronger than the condition on a composition series. Finite nilpotent groups are supersolvable: a $p$-group has a central subgroup of order $p$, and induction in its quotient builds an ambient-normal prime-factor series; direct products of groups of the Sylow factors combine these series. The converses fail. The group $S_3$ is supersolvable via $1<C_3<S_3$, but is not nilpotent because its Sylow $2$-subgroups are not normal. The group $A_4$ is soluble via $1<V_4<A_4$, but is not supersolvable: no subgroup of order two is normal, since the three involutions are conjugate, and no subgroup of order three is normal, since there are four such subgroups. In this example $1<C_2<V_4<A_4$ is a composition series whose first $C_2$ is not normal in $A_4$, while $1<V_4<A_4$ is a chief series with a nonsimple chief factor $V_4$.

Finally, a perfect group satisfies $G^{\prime }=G$. In any finite group, the derived series eventually stabilizes at a perfect characteristic subgroup, because a descending sequence of finite subgroup orders eventually stops decreasing. It is trivial exactly when the group is soluble. A nonabelian simple group is perfect: its nontrivial commutator subgroup is normal and must be the whole group. Thus series distinguish successively stronger notions, summarize the building blocks through Jordan–Hölder factors, and separate soluble behaviour from a persistent perfect part.

A composition series is a subnormal chain whose nontrivial successive quotients are simple. The Jordan–Hölder theorem asserts that any two composition series of a finite group have the same length and the same multiset of simple factors, up to isomorphism. Existence follows by successively choosing maximal proper normal subgroups until the identity is reached.

For uniqueness, induct on the group order. Let $A,B$ be the penultimate terms of two series. If $A=B$, apply induction to that subgroup and add the common top factor. If $A\ne B$, their normality and maximality give $AB=G$. Put $C=A\cap B$. The isomorphism theorems giveBoth are simple. Take any composition series of $C$. Appending $A$ to it gives a series for $A$, so induction identifies the factors of the original series inside $A$ with the factors of $C$ together with $G/B$. Similarly the factors inside $B$ are those of $C$ together with $G/A$. Adding the respective top factors produces the same multiset in both cases. This proves uniqueness, including multiplicities and length.

A normal subgroup $K$ of order two is central because its automorphism group is trivial. By Jordan–Hölder, a composition series through $K$ leaves the simple quotient $G/K\cong A_5$. The image of $Z(G)$ in this quotient is central and hence trivial, since $A_5$ is nonabelian simple. ThusThe symmetric group $S_5$ has trivial group center: conjugation of each transposition by a central permutation must preserve its two-element support, forcing the permutation to fix every point. Since its Jordan–Hölder factors are $A_5$ and $C_2$, a normal subgroup of order two would equal that trivial group center, a contradiction.

Identify $C_2=\{1,-1\}$ and put $D=1\times A_5$ in $C_2\times S_5$. Its three maximal normal subgroups areThey are the three kernels of nonzero maps from the abelianization $C_2^2$ to $C_2$. There is no quotient $A_5$: a central $C_2$ must map trivially, while $S_5$ has no quotient $A_5$. The latter follows from its normal subgroups $1,A_5,S_5$, which can themselves be obtained by intersecting with the simple subgroup $A_5$ and using its trivial centralizer. The complete list of composition series isWithin $M_1$ the two direct factors are the only maximal normal subgroups; within each of $M_2,M_3\cong S_5$ the only such subgroup is its copy of $A_5$. Hence there are exactly four series.

For $S_5\times S_5$, put $D=A_5\times A_5$, $X=A_5\times 1$, $Y=1\times A_5$ andAgain these are exactly the maximal normal subgroups: they are the three cyclic quotient kernels, and there is no simple quotient $A_5$ of the whole product. All series aretogether withTo verify exhaustiveness, $M_L$ has maximal normal subgroups $D$ and $1\times S_5$, and $M_R$ has $D$ and $S_5\times 1$. A quotient $A_5$ of a direct product can receive a nontrivial image from only one simple factor, since the images commute and $A_5$ has trivial group center. The remaining maximal-normal possibility is $M_D$. Its derived subgroup is $D$, because both alternating factors are perfect. A hypothetical quotient $A_5$ would project one alternating factor isomorphically and kill the other, but conjugation by an odd diagonal element induces an outer automorphism on that surviving factor, impossible inside $A_5$. Odd conjugation is outer because otherwise an odd permutation times an even one would centralize $A_5$; a permutation commuting with every three-cycle fixes every three-element support and is the identity. Thus $D$ is the only maximal normal subgroup of $M_D$. Finally the only maximal normal subgroups of $D$ are $X,Y$: normal subgroups of a product of two centerless simple groups are products of the factors, as follows by commutating with each factor. This proves exactly eight series and the classification of composition chains in products with S5.

For $S{L}_2(5)$, commuting with both elementary upper and lower transvections forces a central matrix to be scalar. A scalar $\lambda I$ has determinant ${\lambda }^2$, soIf $g^2=I$, characteristic five makes $g$ diagonalizable with eigenvalues in $\{1,-1\}$. Determinant one rules out one of each, so the unique nonidentity involution is $-I$. An index-two subgroup would have order $60$ and, by the Cauchy theorem for groups, contain that involution and hence the group center. Its image in $PS{L}_2(5)$ would have index two. The simplicity and identification $PS{L}_2(5)\cong A_5$ are established by the independent matrix argument in Solution 4; a simple group has no proper index-two subgroup. Therefore $S{L}_2(5)$ has no subgroup of index two.

For a nonzero group representation $V$, there is a $G$-equivariant identificationwhere $G$ acts on endomorphisms by conjugation. The identity endomorphism is a nonzero fixed vector, so its span is a trivial submodule. It follows that the trivial simple module is a Jordan–Hölder factor of $V\otimes V^{\ast }$, and character additivity gives $\boxed{1_G\text{ is a constituent of }\chi \overline{\chi }}$.

Here “constituent” means a composition factor; the trivial submodule need not be a direct summand. Also the assertion needs $\chi \ne 0$: the zero-dimensional module has the zero Brauer character, whose product with its conjugate has no constituent. This is an implicit nonzero-character qualification in the printed statement.

We prove $(iii)\Rightarrow (ii)$. Consider one nonsplit short exact sequencewith $U,W$ simple. Lift the canonical surjection $P(W)\to W$ from the projective cover through $q$, obtaining $f:P(W)\to V$. Its image maps onto $W$. The intersection $\mathrm{im}f\cap U$ is either zero or $U$, since $U$ is simple. If it were zero, $q$ would identify $\mathrm{im}f$ with $W$, splitting the extension. Hence $U\subset \mathrm{im}f$, and surjectivity onto $W$ then implies $\mathrm{im}f=V$.

Therefore both $U$ and $W$ occur as Jordan–Hölder factors of the indecomposable projective $P(W)$. Replacing every edge in an Ext chain by this observation proves (ii). Together with the preceding two implications, this provesThe repeated word “projective” in the printed version of (ii) has no mathematical effect.

We prove $(ii)\Rightarrow (i)$. The central block idempotents split every finite-dimensional module into its block components. A nonzero indecomposable representation, in particular an indecomposable projective module, has only one such component; otherwise these give a nontrivial direct-sum decomposition.

Every Jordan–Hölder factor of a module in a block belongs to that block, since its central idempotent acts as identity on submodules and quotients. Thus any two simple factors of the same projective indecomposable are in the same block. Applying this along the proposed chain yields $\boxed{S,T\text{ belong to the same block}}$.

Mod-three Galois representation of an elliptic curve is the representationin this case. Twisting by the quadratic character ${\chi }_d$ replaces it by ${\chi }_d{\rho }_3$. Thus $E^d$ has a rational point of order three exactly when $E[3]$ contains a nonzero vector $v$ satisfyingfor every $\sigma \in G_{\mathbb{Q}}$: the line ${\mathbb{F}}_{3}v$ is a one-dimensional Galois subrepresentation with character ${\chi }_d$.

The two-dimensional representation $E[3]$ has at most two distinct one-dimensional characters among its Jordan–Hölder factors. Therefore at most two quadratic characters ${\chi }_d$, and hence at most two rational isomorphism classes of twists, can have a rational point of order three.