On a finite-dimensional $A$-module $M$, the actions of $X$ and $Y$ are endomorphisms $P,Q$ satisfying $PQ-QP=I_M$. Taking traces and using $\mathrm{tr}(PQ)=\mathrm{tr}(QP)$ givesSince $k$ has characteristic zero, this forces ${\dim }_{k}M=0$. Thus the zero module is the only finite-dimensional module for this Weyl algebra. The obstruction is the Weyl algebra has no nonzero finite-dimensional modules.

Let $M$ be a nonzero simple module for $A=k[X]/(X^3)$ and let $N$ be the action of $X$. Since $X$ is central, $NM$ is a submodule, and $N^3=0$. If $NM=M$, iterating would give $N^{3}M=M$, impossible. Thus $NM=0$.

The action therefore factors through $A/(X)\cong k$. A simple module over a field is one-dimensional: the span of any nonzero vector is already a nonzero submodule. Conversely, $k$ with $X$ acting as zero is simple. Hence there is exactly one simple module up to isomorphism, namelyThis argument covers all simple modules, without assuming their finite dimensionality in advance.

For a two-by-two matrix over a principal ideal domain, its first Smith normal form invariant is the greatest common divisor of its entries, while the product of its two invariants is its determinant, up to a unit. Hereso the entries generate the unit ideal in the Gaussian integers. The first invariant is therefore $1$. AlsoHence a valid Smith normal form isMultiplying the second diagonal entry by a unit $\pm 1,\pm i$ gives an equivalent normal form. The divisibility condition holds because $1$ divides every Gaussian integer.

An Abelian group has the central series $1<G$, since $Z(G)=G$. In any central series, each factor is a subgroup of a center and hence Abelian. A finite normal series with Abelian factors makes the group soluble: inductively, if $N$ is soluble and $G/N$ is Abelian, then $G^{\prime }\subseteq N$, and successive derived groups eventually vanish. Hence every nilpotent group is soluble.

The group $S_3$ has the normal series $1<A_3<S_3$ with cyclic factors of orders three and two, so it is soluble. Its center is trivial, since a permutation commuting with both $(12)$ and $(23)$ must fix all three letters. Every nontrivial nilpotent group has nontrivial center: its first nonidentity central-series term is contained in $Z(G)$. Consequently $S_3$ is soluble but not nilpotent.

For a finite p-group, use induction on its order. The identity group is the base case. A nontrivial $G$ has nontrivial $Z=Z(G)$, so $G/Z$ has smaller order, still a power of $p$. By induction it has a central series. Lift that series to subgroups of $G$ and prepend $1<Z$. The first factor is central by definition of $Z$, and centrality of every lifted factor is exactly centrality of the corresponding quotient factor. This gives a central series for $G$. This proves finite p-groups are nilpotent:

The commutator subgroup is $G^{\prime }=⟨xy{x}^{-1}{y}^{-1}:x,y\in G⟩$. It is characteristic and hence normal, because every automorphism sends a commutator to a commutator. If $G^{\prime }\le H$, then $gh{g}^{-1}{h}^{-1}\in H$ for every $g\in G,h\in H$, so $H$ is normal. Every commutator becomes trivial in $G/H$, making that quotient Abelian. Conversely, the quotient homomorphism to an Abelian group kills each commutator, and hence kills $G^{\prime }$. ThereforeSet $G^{(0)}=G$, $G^{(i+1)}=(G^{(i)}{)}^{\prime }$. The group is soluble exactly when this derived series reaches the identity after finitely many steps.

If a nontrivial simple group is soluble, its normal subgroup $G^{\prime }$ is either $1$ or $G$. The latter would make the derived series constant and nontrivial, so $G^{\prime }=1$ and $G$ is Abelian. Every nonidentity element then generates $G$, since its cyclic subgroup is normal. An infinite cyclic group is not simple, and a finite cyclic group is simple exactly when its order is prime. This proves soluble simple group has prime order.

The multiplication gives $v_1=1_A$ and $v_2^2+3v_2+2v_1=0$. Since $v_1,v_2$ form a basis,by the Chinese remainder theorem. Thus there are exactly two nonisomorphic simple modules, both one-dimensional. With basis vectors $s_{-1},s_{-2}$ their actions areThe two orthogonal primitive idempotents are $e_{-1}=2v_1+v_2$ and $e_{-2}=-v_1-v_2$. They sum to $v_1$ and span the two simple ideals. Consequently the regular module splits as $A=\mathbb{C}{e}_{-1}\oplus \mathbb{C}{e}_{-2}$, so $A$ is semisimple.

Its complete list of composition series isIndeed every proper nonzero submodule has dimension one and is an invariant line for multiplication by $v_2$. The matrix $\left(\begin{array}{cc}0& -2\\ 1& -3\end{array}\right)$ has the two distinct eigenvalues $-1,-2$, so these are exactly its two invariant lines.