True. Every group of order four is Abelian: if it has an element of order four it is cyclic; otherwise every nonidentity element has order two and $(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$, while $(ab{)}^{-1}=ab$, so $ab=ba$.

Over $\mathbb{C}$, the Maschke theorem and the Artin–Wedderburn theorem giveSince $G$ is Abelian, this algebra is commutative, forcing every $d_j=1$. Its dimension is four, so $r=4$. Each factor contributes one simple module; therefore there are exactly four up to isomorphism.

Let $w={\sum }_{g\in G}g$. This is nonzero since its basis coefficients are all one, and $gw=w$ for every $g\in G$. Hence $kw$ is a one-dimensional left ideal of $kG$.

If the regular module is semisimple, $kw$ is a direct summand. Let $\pi :kG\to kw$ be the corresponding $kG$-linear projection, so $\pi (w)=w$. Write $\pi (1)=cw$ for $c\in k$. ThenBecause $w\ne 0$, $c|G|=1$ in $k$. ThusTogether with the Maschke theorem, this gives the exact semisimple algebra criterion for a group algebra.

A semisimple module is a direct sum of simple modules. A finite-dimensional semisimple algebra is one whose left regular module is semisimple; equivalently, all its finite-dimensional modules are semisimple, or every submodule of a finite-dimensional module has a complementary submodule.

The Maschke theorem states that, for a finite group $G$ and a field $k$ with $|G|\ne 0$ in $k$, every finite-dimensional $kG$-module is semisimple, so $kG$ is a semisimple algebra. The assumption is equivalent to saying that the characteristic of a field does not divide $|G|$. The averaging construction makes a projection onto a submodule equivariant by replacing $P$ with $|G{|}^{-1}{\sum }_{g\in G}gP{g}^{-1}$.

Given a group representation $\rho :G\to {\mathrm{GL}}_n(k)$, define an action on $k^n$ byIt is bilinear. The identity acts as $I_n$, andbecause $\rho$ is a group homomorphism. Hence the product action agrees with successive action, proving the module axioms. Thus every group representation extends uniquely by linearity to a $kG$-module.

The group algebra $kG$ is the vector space with basis $G$, whose elements are formal sums ${\sum }_{g\in G}{a}_{g}g$ with $a_g\in k$. Extend the group multiplication bilinearly:This is an associative unital $k$-algebra, with multiplicative identity the group identity, and ${\dim }_{k}kG=|G|$.

The general Schur's lemma says that a homomorphism $f:M\to N$ between simple modules over any unital ring is either zero or an isomorphism. Indeed, $\ker f$ is a submodule of $M$ and $\mathrm{im}f$ is a submodule of $N$. If $f\ne 0$, simplicity gives $\ker f=0$ and $\mathrm{im}f=N$. In particular, ${\mathrm{End}}_R(M)$ is a division ring.

For a complex algebra and a finite-dimensional simple module $M$, the stronger conclusion isFor $f\in {\mathrm{End}}_R(M)$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. The endomorphism $f-\lambda I$ has nonzero kernel, so the general Schur's lemma forces it to be zero. Consequently a homomorphism between two such simple modules is zero if they are nonisomorphic, and their homomorphism space is one-dimensional if they are isomorphic.

The finite-dimensional qualification is essential for this scalar conclusion. For example, the field $\mathbb{C}(t)$ is a complex algebra, simple as a module over itself, and its endomorphisms include multiplication by $t$, which is not a complex scalar.

A composition series of a module is a finite chainof submodules whose factors $M_j/M_{j-1}$ are simple modules. A finite-dimensional module has such a series by induction on its dimension, choosing a maximal proper submodule at each step.

The Jordan–Hölder theorem for modules states that any two composition series of a module have the same length and the same simple factors up to isomorphism and reordering. Thus the multiset of composition factors is an invariant of the module, even when the series itself is not unique.

A simple module is a nonzero module with no submodules other than zero and itself. Equivalently, every nonzero element generates the entire module: $Am=M$ whenever $m\ne 0$.

The minimal polynomial factors over $\mathbb{C}$ as $(t-3)(t-3i)(t+3i)$ with distinct roots. The diagonalizability criterion using the minimal polynomial therefore makes $\alpha$ diagonalizable. Its characteristic polynomial gives multiplicities two, one and one, so its Jordan normal form isup to reordering the one-dimensional Jordan blocks.

The first block is the companion matrix of $t-3$. The last block is the companion matrix of $f(t)=t^3-3t^2+9t-27=(t-3)(t^2+9)$. Since $t-3$ divides $f$, these are the ordered invariant factors of a linear operator. ThusThe minimal polynomial is the largest invariant factor, and the characteristic polynomial is their product.

Give $V$ a $k[t]$-module structure by $t\cdot v=\alpha v$. A finite vector space basis generates it as a module. It is a torsion module: the Cayley-Hamilton theorem gives a nonzero polynomial annihilating all of $V$. Since $k[t]$ is a Euclidean domain, cyclic decomposition giveswith monic nonconstant invariant factors of a linear operator $f_j$ and no free summand. On $k[t]/(f_j)$, use the basis $1,t,\dots ,t^{\mathrm{deg}{f}_j-1}$. Multiplication by $t$ then has the companion matrix $C(f_j)$, because the relation $f_j(t)=0$ expresses the last power in terms of earlier ones. Concatenating these bases yieldsthe rational canonical form of $\alpha$. In particular, its minimal polynomial is $f_s$ and its characteristic polynomial is ${\prod }_j{f}_j$.

The structure theorem for finitely generated modules over a principal ideal domain applies because every Euclidean domain is a principal ideal domain. In its cyclic decomposition form, every finitely generated module has a decompositionwhere the $d_i$ are nonzero nonunits. The nonnegative integer $r$ and the invariant factors of a finitely generated module $d_i$ are unique, with each $d_i$ determined up to multiplication by a unit. The zero module corresponds to no summands. Equivalently, factor the $d_i$ into powers of prime elements and use the Chinese remainder theorem to obtain a direct sum of cyclic prime-power torsion modules and a free summand.

An $R$-module $M$ is a finitely generated module if there are $m_1,\dots ,m_s\in M$ such that every element of $M$ has the form $r_1{m}_1+\cdots +r_s{m}_s$ with $r_i\in R$. Equivalently, $M=R{m}_1+\cdots +R{m}_s$, or $M$ is a quotient module of $R^s$.

By Cauchy theorem for groups, choose $a$ of order $p$ and $b$ of order two. The subgroup $P=⟨a⟩$ has index two, hence is normal, and $b\notin P$. Thus $a,b$ generate $G$. Conjugation by $b$ has $ba{b}^{-1}=a^r$ with $r\not\equiv 0(\mathrm{mod}p)$ and $r^2\equiv 1(\mathrm{mod}p)$. Since $p$ is prime, $r\equiv 1$ or $-1$.

If $r=1$, the two generators commute and $ab$ has order $2p$, so $G$ is cyclic. If $r=-1$, then $(ab{)}^2=e$ and $ab\ne b$. The two involutions $b,ab$ generate $G$, because $(ab)b=a$. Hence $G$ is dihedral in the stated convention. Thuswhere $D_{2p}$ denotes the dihedral group of order $2p$. In both cases $P$ is a proper nontrivial normal subgroup, so $G$ is not simple.

By Lagrange's theorem, a nonidentity element in a group of order four has order two or four. In the latter case the group is cyclic. Otherwise every nonidentity element has order two, and $(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$ while $(ab{)}^{-1}=ab$, so the group is Abelian. Choose distinct nonidentity elements $a,b$. Their cyclic subgroups meet trivially and $ab$ is the fourth element, so $(i,j)\mapsto a^i{b}^j$ gives an isomorphism $C_2\times C_2\to G$. HenceNo classification of finite Abelian groups is used.

The map $g\mapsto g^{-1}$ partitions $S$ into disjoint two-element sets $\{g,g^{-1}\}$, so $|S|$ is even. Since $|G|$ is even, $|G\setminus S|$ is even as well. This complement contains the identity and consists of elements with $g^2=e$. It therefore contains at least one further element $h\ne e$, and that element has order two. Thus every finite group of even order contains an involution, without using Cauchy theorem for groups.

For the Abelian case of Cauchy theorem for groups, use induction on $|G|$. Choose $a\ne e$. If $p$ divides $m=|a|$, then $a^{m/p}$ has order $p$. Otherwise $⟨a⟩$ is a nontrivial normal subgroup of order prime to $p$, and $G/⟨a⟩$ has smaller order still divisible by $p$. Induction gives a coset of order $p$; any lift $b$ has order divisible by $p$, because the order of its image divides its own order. A suitable power of $b$ then has order $p$. This uses no classification of finite Abelian groups.

A semisimple module is a direct sum of simple modules. An algebra is semisimple when all its modules are semisimple. Suppose the left regular module $A$ is semisimple. Every module $M$ is a quotient of a free module ${⨁}_{j\in J}A$, by sending its basis elements to a generating set of $M$. This free module is a direct sum of simple modules.

A quotient of a direct sum of simple modules is semisimple: the images of its simple summands are simple or zero and span the quotient. Choose a maximal direct sum among these simple images, allowing all simple submodules in the choice. If it does not span the whole quotient, some simple image is not contained in it. Its intersection with that direct sum is zero by simplicity, so it could be added, contradicting maximality. For infinite index sets, Zorn's lemma supplies the maximal direct sum; in finite dimension the construction terminates directly.

Hence every $M$ is semisimple, and $A$ is a semisimple algebra. This proves semisimplicity from the regular module.

Choose $s\ne 0$ in the simple module $S$. The map $\varphi :A\to S$, $a\mapsto as$, is an $A$-module homomorphism whose image is the nonzero submodule $As$, hence all of $S$. Thus the first isomorphism theorem for modules givesThe kernel is a maximal left ideal because the quotient is simple. Whenever the regular module has finite length, as it does for a finite-dimensional algebra, take a composition series of the kernel and append $\ker \varphi <A$. Its last factor is $S$, so $S$ occurs as a composition factor of the regular module; the Jordan–Hölder theorem for modules makes that independent of the chosen series.

The quotient conclusion holds for every unital algebra. If an arbitrary infinite-dimensional regular module has no finite composition series, the universally valid statement is that $S$ is a simple subquotient; a claim about the factors of a finite series requires finite length.

A simple module is nonzero and has no proper nonzero submodule. A composition series of a module is a finite strict chain $0=V_0<V_1<\cdots <V_n=V$ of submodules with simple quotients $V_i/V_{i-1}$. These quotients are the composition factors of the series.

The Jordan–Hölder theorem for modules says that if a module has a composition series, then any two such series have the same length and the same simple factors up to isomorphism and reordering. In particular, every finite-dimensional module has a well-defined multiset of composition factors.