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Partition a basis of a module $E$ into disjoint subsets $E^{\prime },E^{\prime \prime }$. Splitting the finite expression of each element gives $M=⟨E^{\prime }⟩+⟨E^{\prime \prime }⟩$. An element in both generated submodules would give, by subtraction, a relation among the disjoint parts of $E$; linear independence in a module forces every coefficient to vanish. The intersection is zero, proving the direct sum of modules decomposition. Each part is itself a basis of a module for the submodule it generates.

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.

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.

An $A$-module is completely reducible if it is a direct sum of modules each of which is a simple module:Here a simple module is nonzero and has no submodules other than zero and itself. This is precisely the definition of a semisimple module. The indexing set need not be finite unless $V$ has finite dimension; the zero module is included as the empty direct sum of modules.

Use the corrected Lie algebra representation $E=-e$, $H=h$, $F=f$ from part (i). A vector subspace is invariant under $E,H,F$ exactly when it is invariant under the printed $e,h,f$, since changing the sign of a linear map does not change its invariant subspaces. DefineThis is a nonzero proper submodule: $H$ preserves each $k{v}_i$, $F$ increases the index, $E(v_1)=0$, and $E(v_i)=-(i-1)v_{i-1}\in W$ for $i\ge 2$.

We must rule out submodules consisting only of mixtures of weight vectors. Let $U\ne 0$ be any submodule, and choose $u={\sum }_{j\in S}{a}_j{v}_j\in U$ with $S$ finite and every $a_j\ne 0$. The eigenvalues $-2j$ of $H$ are distinct. For $i\in S$, the Lagrange interpolation polynomialsatisfies $p_i(H)u=a_i{v}_i$. Since $U$ is invariant under $H$, it is invariant under every polynomial in $H$, so $v_i\in U$.

If $i=0$, repeated applications of $F$ give $F^m{v}_0=m!v_m$, and $U=V$. If $i\ge 1$, repeated applications of $E$ give a nonzero multiple of $v_1$, because the successive coefficients $-(i-1),\dots ,-1$ are nonzero in characteristic zero. Repeated applications of $F$ to $v_1$ then give every $v_j$ with $j\ge 1$, so $W\subseteq U$.

Finally, if such a $U$ contains any vector outside $W$, that vector has a nonzero coefficient of $v_0$; the same polynomial interpolation extracts $v_0$, forcing $U=V$. Hence the complete list of submodules is $0,W,V$. In particular, $W$ is a simple module. The quotient module $V/W$ has basis ${\overline{v}}_0$; $E{\overline{v}}_0=H{\overline{v}}_0=0$ and $F{\overline{v}}_0={\overline{v}}_1=0$. It is the one-dimensional trivial Lie algebra representation.

The module extension does not split. If $V=W\oplus U$ for a submodule $U$, then $U$ would be a one-dimensional trivial Lie algebra representation, since it maps isomorphically to $V/W$. But $\ker H=k{v}_0$ and $F{v}_0=v_1\ne 0$, so no such $U$ exists. ThusHere semisimple module means a direct sum of modules that are simple modules, equivalently a module in which every submodule has a complementary submodule. This is the nonsplit zero-weight Verma extension for sl2: rescaling to $w_i=i!v_i$ gives $F{w}_i=w_{i+1}$ and $E{w}_i=i(1-i)w_{i-1}$. Its infinite-dimensional vector space is outside the hypotheses of the Weyl complete reducibility theorem.

The exact orders printed in the question are inconsistent. The source PDF says that the orders of group elements $a,b,c$ are $4,12,24$, respectively, and also imposes $3a+2b=0$. Multiplying this relation by $6$ and using $12b=0$ gives $18a=0$. Since $4a=0$ also holds, subtracting $4(4a)=0$ gives $2a=0$, contradicting the stated order of a group element $4$. Thus

The natural abelian group presentation interpretation is to impose the relations $4a=12b=24c=0$ and $3a+2b=0$, with no further relations, without requiring the orders of the images of the generators to stay $4,12,24$. Under this interpretation the calculation is as follows. Since $3a=-a$ follows from $4a=0$, the mixed relation gives $a=2b$. Consequently $8b=0$, and together with $12b=0$ this gives $4b=0$. Conversely, $a=2b$, $4b=0$, and $24c=0$ imply every original relation. Eliminating $a$ leaves two independent cyclic groups:To verify that these relations force no further identifications, map $a,b,c$ to $(2,0),(1,0),(0,1)$ in ${\mathbb{Z}}_4\oplus {\mathbb{Z}}_{24}$. All defining relations hold, and the inverse group homomorphism sends the two standard generators to $b,c$. These group homomorphisms are mutually inverse. The divisibility condition $4\mid 24$ gives the requested invariant factors of a finitely generated module over $\mathbb{Z}$. In this presented group, $a,b,c$ actually have orders $2,4,24$.

Equivalently, the relation matrixhas nonzero Smith normal form entries $1,4,24$, with the trivial ${\mathbb{Z}}_1$ factor omitted. The displayed direct sum of modules is the answer to the presentation interpretation; the inconsistency of the literal exact-order formulation remains.

Define the group homomorphismBy part (b)(i), its kernel of a group homomorphism isIt is surjective. Indeed, choose $r,s\in \mathbb{Z}$ with $rd+sh=1$ by Bézout's identity. For any prescribed classes $u+d\mathbb{Z}$ and $v+h\mathbb{Z}$, the integersatisfies $t\equiv u(\mathrm{mod}d)$ and $t\equiv v(\mathrm{mod}h)$. The first isomorphism theorem for groups therefore givesThis proves the required isomorphism and is the Chinese remainder theorem for coprime integers. It also shows that $(1+d\mathbb{Z},1+h\mathbb{Z})$ generates the direct sum of modules when these additive groups are viewed as $\mathbb{Z}$-modules.

Let $N=|G|$ be a square-free integer. If $N=1$, then $G$ is the trivial cyclic group. Otherwise, finiteness means that a direct sum of modules decomposition into cyclic groups has only finitely many nontrivial factors, each of finite group order. Omit trivial factors and writeThe integers $n_i,n_j$ are coprime integers whenever $i\ne j$. If a prime number $p$ divided both, then $p^2\mid N$, contradicting the square-free integer hypothesis.

Moreover, $n_1\cdots n_{j-1}$ and $n_j$ are coprime integers: any common prime divisor would divide one of the earlier factors as well as $n_j$. Thus part (b)(ii) can be applied repeatedly to combine factors, givingIn particular $G$ is a cyclic group. Under the displayed identifications, the element whose coordinate in every factor is the class of $1$ is a generator of a group.

In the direct sum of modules $M=N\oplus P$, every $m\in M$ has a unique expression $m=n+p$, with $n\in N$ and $p\in P$. DefineThe uniqueness makes this well-defined. For $m=n+p$, $m^{\prime }=n^{\prime }+p^{\prime }$ and $r\in R$,so $\pi$ is a module homomorphism. Finally, $\pi (n+p)=0$ exactly when $p=0$, equivalently $m\in N$. Therefore $\boxed{\ker \pi =N}$. Its image is $P$, and ${\pi }^2=\pi$, so it is the projection onto that direct summand.