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Over a splitting field for finite group representations of characteristic $p$, the irreducible Brauer characters form a complex basis of the class functions on the p-regular elements. Independence is the character form of the Brauer–Nesbitt theorem. To obtain spanning, extend such a class function by zero on the p-singular classes. Ordinary irreducible characters form a basis of all class functions by character orthogonality. Restricting them to p-regular elements yields Brauer characters of reductions of an integral form of a group representation in a compatible splitting p-modular system; if needed, first extend scalars, which does not change the simple-module list under the splitting hypothesis. Each restriction is a nonnegative integral sum of simple Brauer characters by exact-sequence additivity and the Jordan–Hölder theorem. Thus these restrictions span, proving the assertion. Consequently the number of simple modules equals the number of p-regular conjugacy classes, and evaluation identifies the complexified modular representation ring with the product of one copy of $\mathbb{C}$ for each such class.

The composition length of a finite-length module or group is the number of simple factors in a composition series. The Jordan–Hölder theorem makes this independent of the chosen composition series.

The only nontrivial proper normal subgroups of this direct product of groups are its two factors. Indeed a normal subgroup meeting $S\times 1$ nontrivially contains it by the simple group property. Otherwise each of its commutators with $S\times 1$ lies in that trivial intersection. Its first coordinates therefore lie in $Z(S)=1$, so it is contained in $1\times C_p$. Both factors are simple and each quotient by one factor is isomorphic to the other. Hence precisely two composition series occur, one through each factor. This is compatible with the Jordan–Hölder theorem, since the two series have the same simple factors in opposite orders. For $S=A_5,p=2$, the group has order $120$.

A Jordan–Hölder factor is a simple quotient in a composition series. The Jordan–Hölder theorem says that the multiset of such factors is independent of the chosen composition series.

A module has finite length when it admits a composition series of a module, and its length is the number of simple modules among the successive quotients. The Krull–Schmidt theorem concerns decompositions into indecomposable summands; the independence of composition length instead follows from the Jordan–Hölder theorem. A module without a finite composition series is assigned infinite length.

Length is additive in a short exact sequence: $\ell (M)=\ell (N)+\ell (M/N)$ when the outer modules have finite length. For example, $\mathbb{Z}/p^r\mathbb{Z}$ has length $r$, with each successive factor isomorphic to $\mathbb{Z}/p\mathbb{Z}$. The module $\mathbb{Z}$ has infinite length, as its descending chain $2^n\mathbb{Z}$ never stabilizes.

The modular representation ring is the Grothendieck group of finite-dimensional group representations, with relations $[V]=[U]+[W]$ for every short exact sequence $0\to U\to V\to W\to 0$, and multiplication $[V][W]=[V{\otimes }_{k}W]$ using the tensor product of group representations. The Jordan–Hölder theorem makes the classes of simple modules a free integral basis. Tensoring over a field is exact, so this multiplication is well defined. When $k$ is a splitting field for finite group representations, Brauer characters identify $\mathbb{C}{\otimes }_{\mathbb{Z}}{R}_k(G)$ with the algebra of complex class functions on p-regular elements. The splitting hypothesis is essential: ${\mathbb{F}}_2{C}_3\cong {\mathbb{F}}_2\times {\mathbb{F}}_4$ has two simple modules, although $C_3$ has three $2$-regular conjugacy classes.

A composition series is a chain $1=G_0◃G_1◃\cdots ◃G_r=G$ with simple nontrivial factors. Every finite group has one, by recursively choosing a maximal proper normal subgroup. The Jordan–Hölder theorem states that any two such series have the same length and the same multiset of simple factors up to isomorphism.

For a proof induct on $|G|$. Let $M,N$ be the penultimate terms in two series. If $M=N$, apply induction there. Otherwise their maximal normality gives $MN=G$. Set $D=M\cap N$. The second isomorphism theorem givesso these quotients are simple. Choose a composition series of $D$ and extend it once to $M$ and once to $N$. Induction in the smaller groups $M,N$ compares these with the respective original prefixes. Appending $G/M$ or $G/N$ gives in both cases the factors of $D$ together with $G/M,G/N$. This completes the proof, including the case $D=1$.

Exactly one: take $S_5$. Its only nontrivial proper normal subgroup is $A_5$. Indeed a normal $N$ has $N\cap A_5$ either $1$ or $A_5$ by the supplied simplicity. In the first case $N$ injects into $S_5/A_5\cong C_2$, so a nontrivial $N$ would be a normal subgroup of order two and hence central. But $S_5$ has trivial center: conjugating every transposition trivially forces preservation of each unordered pair, and therefore of every point. Thus the unique series is $1◃A_5◃S_5$.

Exactly two: take $A_5\times C_2$. A normal subgroup containing $A_5\times 1$ is that subgroup or the whole product. For a normal subgroup $N$ meeting $A_5\times 1$ trivially, commutators of its elements with $A_5\times 1$ vanish; its first coordinates therefore lie in $Z(A_5)=1$. Hence $N\subseteq 1\times C_2$. The two nontrivial proper normal subgroups are exactly the two factors, each simple, so the two series pass respectively through $A_5\times 1$ and $1\times C_2$.

More than two: take $C_{120}$. A cyclic group has exactly one subgroup of each divisor order, and composition steps have prime indices. The orders of those indices are the distinct permutations of $2,2,2,3,5$, givingAll three example groups have order $120$.

A composition series is a chain $0=V_0⊊V_1⊊\cdots ⊊V_r=V$ of submodules with every factor $V_i/V_{i-1}$ simple; its length is $r$. Suppose $0\subset S\subset V$ is one of length two, and put $T=V/S$.

For any nonzero proper submodule $U$, either $U\cap S=S$ or $U\cap S=0$, by simplicity of $S$. In the first case, simplicity of $V/S$ forces $U=S$. In the second, the quotient map embeds $U$ nontrivially into $T$, so $U\cong T$ and its image is all of $T$. Hence $V=S\oplus U$, and $V/U\cong S$. Thus every nonzero proper submodule is simple and has simple quotient; every composition series has length two.

The Jordan–Hölder theorem states that any two composition series of a finite-length module have equal length and the same multiset of simple factors, up to isomorphism and order. The preceding argument proves this for length two: either $U=S$, giving factors $S,T$, or $U\ne S$, giving factors $T,S$.

The map $A\to S$, $a\mapsto ax$, is a surjective module homomorphism with kernel $I={\mathrm{Ann}}_A(x)$. The first isomorphism theorem for modules gives $A/I\cong S$ as left modules; $I$ need only be a left ideal. Every simple module is cyclic, since any nonzero $x$ generates a nonzero submodule. As a quotient of the finite-dimensional regular module $A$, it is finite-dimensional. Extend $I$ to a composition series of $A$; its simple quotient $A/I$ is one of the factors. Jordan-Hölder shows that its isomorphism type belongs to the fixed finite list of factors of any one composition series of $A$. Hence there are finitely many simple-module isomorphism types.

For the final assertion let $W$ be the image of $\psi$. Its kernel is $\bigcap I_j$. We prove $W=\prod S_j$ by induction on $k$. It is true for $k=1$. Projection onto the first $k-1$ factors is onto by induction. Its kernel on $W$ is a submodule of $S_k$, so is zero or $S_k$. If zero, $W\cong {\prod }_{j<k}{S}_j$. But its projection onto $S_k$ is onto, making $S_k$ a composition factor of that product, contrary to Jordan-Hölder and the pairwise distinct types. Thus the kernel is all of $S_k$. Surjectivity on the first factors together with this kernel proves $W$ is the entire product. The first isomorphism theorem now givesas left $A$-modules, without asserting that the left ideals are two-sided.

The Jordan–Hölder theorem applies to a module of finite length: ifis a composition series of a module, its successive quotients are simple modules. Any two such composition series of a module have the same length $r$, and their composition factors are pairwise isomorphic after a permutation, with multiplicities preserved. Every finite-dimensional module over $A$ has finite length, since strict inclusions increase vector-space dimension. A finite-dimensional algebra over a field does not by itself make every possible module finite-dimensional, so the finite-length hypothesis should be retained in the statement.

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.

The roots-of-unity criterion for a splitting field for finite group representations says that the assumed $m$th roots of unity make $k$ a splitting field for $G$. Fix a multiplicative identification of the group ${\mu }_m(k)$ with the complex $m$th roots of unity. For a p-regular element $g$, its order divides $m$; the operator $\rho (g)$ is diagonalizable because $X^{|g|}-1$ has distinct roots in $k$. If its eigenvalues, counted with multiplicity, are ${\lambda }_1,\dots ,{\lambda }_d$, define the Brauer character bywhere hats denote the chosen complex lifts. This defines a class function on the p-regular elements, not on arbitrary elements of $G$.

A short exact sequence can be represented by block triangular matrices, so the eigenvalue multiset is the union of those on its submodule and quotient. Thus Brauer characters are additive on short exact sequences. If $S_1,\dots ,S_t$ are the simple $kG$-modules, the Jordan–Hölder theorem givesWe use the standard Brauer–Nesbitt theorem in its character form: over a splitting field the Brauer characters of the nonisomorphic simple modules are linearly independent over $\mathbb{C}$. ThereforeBoth the modular splitting-field criterion and this independence theorem are the representation-theoretic results used here.

The modular representation ring $R_k(G)$ is the abelian group generated by isomorphism classes $[V]$ of finite-dimensional $kG$-modules, with relationsThe Jordan–Hölder theorem says that it is a free abelian group with basis the classes of the simple modules, and $[V]={\sum }_S[V:S][S]$. To see independence, composition multiplicity for each fixed simple $S$ is an additive map to $\mathbb{Z}$ and reads off its basis coefficient.

Define $[V][W]=[V{\otimes }_{k}W]$, using the diagonal action in the tensor product of group representations. Tensoring over a field is exact in either argument, so multiplication respects the defining relations. Associativity and the symmetry $v\otimes w\mapsto w\otimes v$ give a commutative ring with identity $[k]$, the trivial representation. Thus $\boxed{R_k(G)\text{ is a commutative ring, freely based on simple modules}}$.

First the product is also split. Write $A_i=k{G}_i$ and $J_i=\mathrm{rad}{A}_i$. The idealis nilpotent: if $J_1^r=J_2^s=0$, every product of $r+s-1$ of its factors vanishes. The quotient isa product of full matrix algebras over $k$, because each $A_i$ is split. A nilpotent ideal lies in the Jacobson radical, and a semisimple quotient forces the reverse inclusion, so $I$ is exactly the radical. Thus $k$ is a splitting field for finite group representations of the product.

For simple modules $S_i$, Schur lemma and splitting give ${\mathrm{End}}_{A_i}(S_i)=k$. The Jacobson density theorem therefore makes the image of $A_i$ on $S_i$ the full ${\mathrm{End}}_k(S_i)$. Tensoring these maps shows that the product algebra acts on $S_1\otimes S_2$ through its full endomorphism algebra. Hence this tensor product of group representations is simple.

On restriction to $G_1$, it is a direct sum of $\dim S_2$ copies of $S_1$. If two external tensor products are isomorphic, their restrictions and the Jordan–Hölder theorem force $S_1\cong S_1^{\prime }$; restricting to $G_2$ similarly forces $S_2\cong S_2^{\prime }$. The converse follows by tensoring the isomorphisms.

Finally, the p-regular conjugacy classes of $G_1\times G_2$ are precisely pairs of such classes in the factors. The corrected result in part (b) counts as many simples for the product as pairs of simples for the two factors. Our pairwise nonisomorphic tensor products already attain that count, so they exhaust all simples. Thus $\boxed{{\mathrm{Irr}}_k(G_1\times G_2)=\{S_1\otimes S_2\}}$, uniquely indexed by pairs of simple isomorphism classes.

For the preliminary definitions, a block is $Re$ for a primitive central idempotent $e$, and a module belongs to it when $eM=M$. For a semisimple algebra, the Artin–Wedderburn theorem identifies its blocks with the factors $M_{n_i}(D_i)$ in its product decomposition. Each factor is a block of a finite-dimensional algebra.

We prove Ext separation of finite-length modules. The extension hypothesis is ${\mathrm{Ext}}^1(S,T)={\mathrm{Ext}}^1(T,S)=0$ for simples in opposite sets. First, for a simple $S\in {\mathcal{C}}_1$ and a module $N$ of finite composition length with factors in ${\mathcal{C}}_2$, we have ${\mathrm{Ext}}^1(S,N)=0$. Induct on the length of $N$: for $0\to N^{\prime }\to N\to T\to 0$ with $T$ simple, the long exact sequence of the Ext functor containswhose outer groups are zero. The same argument works with the two sets interchanged.

Now induct on the length of $M$, with $M=0$ immediate. Take a simple quotient in $0\to N\to M\to S\to 0$. By induction, $N=N_1\oplus N_2$ with factors in the respective sets. Suppose $S\in {\mathcal{C}}_1$; the other case is symmetric. Quotienting by $N_1$ givesThis splits by the preceding Ext vanishing. Let $U_1$ be the inverse image in $M$ of the chosen complementary copy of $S$. Then $U_1\cap N_2=0$, $U_1+N_2=M$, and $0\to N_1\to U_1\to S\to 0$. Thus $U_1$ has only factors in ${\mathcal{C}}_1$, while $U_2=N_2$ has only factors in ${\mathcal{C}}_2$.

If two finite-length modules have factors in disjoint sets, any homomorphism between them is zero: a nonzero image would, by the Jordan–Hölder theorem, have a simple factor belonging to both sets. For any submodule $V$ of $M$ with factors in ${\mathcal{C}}_1$, projection onto $U_2$ is consequently zero, so $V\subset U_1$. The analogous argument applies to $U_2$. HenceIn particular both summands are preserved by every endomorphism of $M$.

In the Klein four-group $V=\{1,a,b,ab\}$, the chainsare distinct composition series. All their composition factors are $C_2$.

The Jordan–Hölder theorem states that any two composition series of a group have the same length and the same composition factors up to permutation and group isomorphism. Thus the subgroup chains may differ, while the multiset of simple quotients is intrinsic.

Supposeare two factorizations into primes. Repeatedly apply the cyclic-subgroup construction of part (c), first removing $p_r$, then $p_{r-1}$, and so on. This constructs a composition series for $C_n$ whose composition factors are $C_{p_1},\dots ,C_{p_r}$. The second factorization constructs another composition series with composition factors $C_{q_1},\dots ,C_{q_s}$.

By the Jordan–Hölder theorem, $r=s$ and these composition factors agree up to group isomorphism and permutation. Since isomorphic finite groups have the same order, the corresponding primes agree. This proves uniqueness of prime factorization up to permutation, using existence of prime factorizations but not their uniqueness.

The cyclic group of order six has two different composition series:Their factor lists are $C_2,C_3$ and $C_3,C_2$ respectively; normality follows from commutativity.

The Jordan–Hölder theorem states that any two composition series of a group have the same length and the same multiset of composition factors up to isomorphism. Equivalently, the factors of one series can be reordered to match those of the other by group isomorphisms. The actual intermediate subgroups need not be the same.

Given a prime factorization $n=p_1\cdots p_r$, repeatedly apply part (c) to the cyclic group of order $n$: it has a chain of subgroups of successive ordersRead in increasing order, this is a composition series with factors isomorphic to the cyclic groups of orders $p_1,\dots ,p_r$. Any second prime factorization of the same integer produces another composition series of the same group $C_n$.

The Jordan–Hölder theorem equates the two multisets of simple factors. Since $C_p\cong C_q$ for primes $p,q$ implies $p=q$ by equality of group orders, the two prime multisets agree. HenceFor $n=1$ both factor lists are empty, corresponding to the trivial group.