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Write $C=⟨c⟩$ with $|c|=n$. For each divisor $k$ the subgroup $⟨c^{n/k}⟩$ has order $k$. Conversely, if $H$ has order $k$, Lagrange gives $k\mid n$, and every $c^j\in H$ satisfies $(c^j{)}^k=1$, so $n\mid jk$. Therefore $n/k\mid j$ and $H\subseteq ⟨c^{n/k}⟩$. Equality follows from the common order. Thus there is exactly one subgroup for each divisor of $n$.

For the Abelian assertion, take a surjection ${\mathbb{Z}}^n\to A$ and let $L$ be the inverse image of $B$. Every subgroup of ${\mathbb{Z}}^n$ has at most $n$ generators, by induction: its projection onto the first coordinate is either zero, reducing to ${\mathbb{Z}}^{n-1}$, or $d\mathbb{Z}$ for some $d>0$. In the latter case choose one element with first coordinate $d$; subtracting its multiples reduces every element to the kernel, generated by at most $n-1$ elements by induction. The base case is the elementary description of subgroups of $\mathbb{Z}$ by their least positive element. Images of these generators generate $B$, proving the claim.

In $S_6$, the commuting disjoint transpositions $(12),(34),(56)$ generate $C_2^3$. This is a three-dimensional vector space over ${\mathbb{F}}_2$, so it requires three generators and cannot be generated by two.

For each $n\ge 3$, take $G=S_{2n+2}$, generated by $(12)$ and the cycle $(12\dots 2n+2)$. Its subgroup generated byis $C_2^{n+1}$. It requires $n+1$ generators by the same dimension argument, giving a two-generated group with a subgroup requiring more than $n$ generators.

A simple module is a nonzero module with no proper nonzero submodule. For a finite-dimensional simple complex module $S$, any endomorphism $\phi$ has a complex eigenvalue $\lambda$. The nonzero kernel of $\phi -\lambda I$ is an $A$-submodule, hence all of $S$. Thus $\phi =\lambda I$, proving Schur's lemma.

More generally a nonzero map between two simple modules is an isomorphism, because its kernel is zero and its image is the whole target. The endomorphism algebra of $S_1\oplus S_2$ is the block algebra with entries ${\mathrm{Hom}}_A(S_j,S_i)$. For nonisomorphic summands its off-diagonal entries vanish and its diagonal entries are scalars, giving $\mathbb{C}\times \mathbb{C}$. For isomorphic summands, choose one fixed isomorphism to identify them; every block is scalar and block composition is matrix multiplication, giving $M_2(\mathbb{C})$.

A finite-dimensional algebra is a semisimple algebra when its left regular module is a direct sum of simple modules. The complex Artin–Wedderburn theorem gives $A\cong {\prod }_i{M}_{d_i}(\mathbb{C})$, with the multiset of positive integers $d_i$ uniquely determined. Its dimension is ${\sum }_i{d}_i^2$. For dimension eight only sizes one and two can occur; the number of size-two blocks is zero, one, or two. Thus there are exactly three isomorphism types:They are distinct by the uniqueness clause, or by their numbers of one-dimensional simple modules. If $G$ is nonabelian, $\mathbb{C}G$ is noncommutative because its group elements form a basis, so the first type is excluded. The augmentation $\sum a_{g}g\mapsto \sum a_g$ provides a one-dimensional simple module, excluding the third type. ConsequentlyThere is one simple module for each matrix block, of dimension the block size, so exactly four are one-dimensional. Every nonabelian group of order eight therefore has this same complex group algebra.

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 degree-sum formula for irreducible characters givesThe number of linear characters is the order of the abelianization, so $|G/G^{\prime }|=3$ and $|G^{\prime }|=7$. The intersection of the kernels of all linear characters is $G^{\prime }$. Consequently $g_4,g_5$ are nonidentity elements of this order-seven subgroup and have order seven. The images of $g_2,g_3$ under a displayed linear character have order three, so their orders are divisible by three. Lagrange's theorem leaves three or twenty-one; an element of order twenty-one would make $G$ cyclic, contradicting its degree-three irreducible characters. ThereforeThe center is trivial. A nontrivial proper center would have order three or seven, giving a cyclic quotient and hence an abelian group, again a contradiction. Thus a centralizer of a nonidentity element is proper. It has order three for an order-three element, and seven for an order-seven element. The class sizes are therefore $1,7,7,3,3$.

Column orthogonality says ${\sum }_i|{\chi }_i(g){|}^2=|C_G(g)|$. For $g_2,g_3$, the three linear characters already contribute three, so the remaining two entries must both vanish. Orthogonality with the identity column says ${\sum }_i{\chi }_i(1)\overline{{\chi }_i(g)}=0$ for $g\ne 1$. At $g_4$ this gives $1+\alpha +{\chi }_5(g_4)=0$, so ${\chi }_5(g_4)=\beta$, using $\alpha +\beta =-1$. Similarly ${\chi }_5(g_5)=\alpha$. Thus the completed table isHere $\beta =\overline{\alpha }$, $|\alpha {|}^2=|\beta {|}^2=2$ and $\alpha \beta =2$. These relations, together with $1+\omega +{\omega }^2=0$, verify all weighted row orthogonality relations and the centralizer column norms.

An extension of $A$ by $B$ is a short exact sequence $1\to A\to G\to B\to 1$. It splits if the quotient map has a homomorphic section. Equivalently the embedded $A$ has a subgroup complement isomorphic to $B$, and $G$ is the corresponding semidirect product.

For $t\notin K$ and $a\in K$, both $t$ and $ta$ lie outside $K$, so $t^2=(ta{)}^2=1$. It follows that $tat=a^{-1}$, and $t=t^{-1}$ gives the claimed conjugation identity. Conjugation is an automorphism of $K$, so inversion is an automorphism there. Comparing the image of $ab$ with the product of the images gives $b^{-1}{a}^{-1}=a^{-1}{b}^{-1}$. Therefore $K$ is abelian.

Conversely inversion is an automorphism of any abelian group $A$, of order dividing two. Form the semidirect product $G=A⋊C_2$ with multiplicationFor completeness associativity follows because $b\mapsto b^{(-1{)}^{\epsilon }}$ is an automorphism and these automorphisms compose according to addition modulo two; $(1,0)$ is the identity and inverses follow from this formula. The subgroup $H=A\times \{0\}$ is normal with quotient $C_2$, and every outside element satisfies $(a,1{)}^2=(a{a}^{-1},0)=1$. Thus the generalized dihedral group supplies the required extension, including when $A$ is infinite or has elements of order two.

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 $120$.

For $|G|=p^{r}m$ with $p\nmid m$, Sylow's theorem gives subgroups of order $p^r$, says that every $p$-subgroup lies in a conjugate of one of them, and that all such Sylow subgroups are conjugate. Their number $n_p$ divides $m$ and satisfies $n_p\equiv 1(\mathrm{mod}p)$. A group is soluble when its derived series, obtained by repeatedly taking the commutator subgroup, reaches the identity. Equivalently it has a subnormal series with abelian factors.

For a group of order $p^2$, the class equation shows its center is nontrivial: noncentral conjugacy classes have sizes divisible by $p$. If the center had order $p$, the quotient by the center would be cyclic; a group with cyclic central quotient is abelian, since representatives are powers of one element times central elements. This would make the center the whole group. Thus every group of order $p^2$ is abelian.

Now consider $|G|=p{q}^a$ with $0\le a\le 2$. Cases with a normal Sylow subgroup are soluble: that subgroup has prime or prime-square order and is abelian, and its quotient also has such order and is abelian. This gives a subnormal series with abelian factors.

If $p<q$, the Sylow number $n_q$ divides $p$ and is $1$ modulo $q$, so is one. If $p>q$ and $a\le 1$, $n_p$ divides $q^a<p$, so again is one. For $p>q$ and $a=2$, a nontrivial $n_p$ can only be $q^2$, since $q<p$. The congruence would force $p\mid (q-1)(q+1)$, hence $p\mid q+1$. Since $p>q$, this means $p=q+1$, and primality forces $(p,q)=(3,2)$.

In the remaining order-twelve case, either a Sylow subgroup is already normal, or there are four Sylow subgroups of order three. Those have pairwise trivial intersection and contribute eight distinct elements of order three. Only four elements remain, including the identity. Every Sylow subgroup of order four lies entirely among those four elements, so it is unique and normal. It and its quotient are abelian. Therefore all groups of the required orders are soluble, by elementary Sylow arguments alone.

Each $M_i$ is an ideal because the algebra is commutative. For $m\in M_i$, the product $m{e}_j$ belongs both to $M_i$ and $M_j$, so is zero when $j\ne i$. Multiplying $1={\sum }_j{e}_j$ by $m$ therefore gives $m=m{e}_i$. In particular $e_i^2=e_i$, different $e_i$ have zero product, andFor any idempotent $e$, the set $Ae$ is closed under multiplication and $e$ acts as its identity. With $f=1-e$ the map $a\mapsto (ae,af)$ is an algebra isomorphism $A\to Ae\times Af$: multiplication is componentwise because $ef=0$, and its inverse sends $(u,v)$ to $u+v$.

The $A$-submodules of $Ae$ are exactly its ideals as a unital algebra, since acting by $a$ on $Ae$ is the same as acting by $ae$. A nonzero commutative unital algebra is a simple module over itself exactly when every nonzero principal ideal is the whole algebra, which is equivalent to every nonzero element having an inverse. Thus a nonzero component $Ae$ is simple exactly when it is a field. If $e=0$, both properties fail under the nonzero-field convention.

A semisimple regular module is a finite direct sum of simple ideals. Applying the preceding decomposition gives a finite direct product of fields. Conversely the coordinate ideals of a finite product of fields are simple modules, so its regular module is semisimple. The fields are finite extensions of $K$, because $A$ has finite $K$-dimension.

For a generator $g$ of the cyclic group of order three, evaluation $X\mapsto g$ gives a surjection $\mathbb{Q}[X]\to \mathbb{Q}G$ with kernel $(X^3-1)$, by division and independence of $1,g,g^2$. The factors $X-1$ and $X^2+X+1$ are coprime, and the quadratic is irreducible over $\mathbb{Q}$. The Chinese remainder theorem yieldsThe explicit map sends $g$ to $(1,{\zeta }_3)$, where ${\zeta }_3=(-1+\sqrt{-3})/2$.

A simple module is nonzero and has no submodules other than zero and itself. A semisimple module is a direct sum of simple submodules; the zero module is allowed as the empty sum.

For (1)$\Rightarrow$(2), argue by dimension. Choose a minimal nonzero submodule $S$, which is simple, and split $V=S\oplus C$ using (1). The property passes to $C$: for $U\subseteq C$, write $V=U\oplus D$; then $C=U\oplus (C\cap D)$. Apply induction to $C$ and append $S$. The zero-dimensional case starts the induction. The implication (2)$\Rightarrow$(3) is immediate. For (3)$\Rightarrow$(2), repeatedly append a simple submodule not contained in the sum already chosen. Its intersection with that sum is zero by simplicity. Dimension strictly increases, so after finitely many steps their direct sum is $V$.

For (2)$\Rightarrow$(1), write $V={⨁}_i{S}_i$ and fix $U$. Choose a sum $C$ of a maximal subset of these $S_i$ whose sum is disjoint from $U$. If $U+C\ne V$, some $S_j$ is not contained in $U+C$, so simplicity gives $S_j\cap (U+C)=0$. Then $C+S_j$ is a larger direct sum still disjoint from $U$, a contradiction. Hence $V=U\oplus C$.

Condition (1) implies (1*). Conversely suppose (1*) and let $W$ be the sum of all simple submodules. If $W\ne V$, finite dimension supplies a maximal proper submodule $M$ containing $W$. By (1*), $V=M\oplus C$. Projection to the simple quotient $V/M$ identifies the nonzero complement $C$ with a simple module. Thus $C\subseteq W\subseteq M$, contradicting the direct sum. Consequently $W=V$, which is (3). This proves all four conditions are equivalent.

For the concrete complement, $x^2=x$. Orthogonal idempotents $x$ and $1-x$ give $A=Ax\oplus A(1-x)$: the sum follows from $1=x+(1-x)$, and right multiplication by $x$ shows their intersection is zero. In explicit matrix form,The decomposition of $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ uses second-column remainders $b-\lambda a,d-\lambda c$ in $C$.