The Prüfer $p$-group is Artinian but not Noetherian

Let $C_n=\{a\mathrm{/}{p}^n+\mathbb{Z}:a\in \mathbb{Z}\}\le \mathbb{Q}\mathrm{/}\mathbb{Z}$. The class $1\mathrm{/}{p}^n+\mathbb{Z}$ generates $C_n$ and has order $p^n$, while $C_n<C_{n+1}$. If an element of ${\bigcup }_k{C}_k$ has order dividing $p^n$, cancelling its denominator shows that it lies in $C_n$. Hence every cyclic subgroup of order $p^n$ is $C_n$, and $\mathbb{Z}(p^{\mathrm{\infty }}):={\bigcup }_{n\ge 0}{C}_n$ is the Prüfer $p$-group.

The group is not finitely generated, which is the failure of [L1] directly: a finite subset of ${\bigcup }_k{C}_k$ lies in a single $C_N$, because each of its finitely many members lies in some $C_k$ and the $C_k$ are nested, so the submodule it generates is contained in $C_N$ and is proper. Equivalently, the strict ascending chain $C_0<C_1<C_2<\cdots$ of step 1.1 does not stabilize.

If a subgroup $H$ contains elements of unbounded order, then for every $n$ it contains an element whose cyclic subgroup contains the unique $C_n$, so $C_n\le H$ and $H=\mathbb{Z}(p^{\mathrm{\infty }})$. Otherwise the element orders in $H$ are bounded by some $p^n$, and step 1.1 gives $H\le C_n$. The subgroups of the cyclic group $C_n$ are the unique $C_m$ for $0\le m\le n$, so every proper subgroup of the Prüfer group is one of these finite cyclic groups.

A descending chain either remains at the whole group or enters some finite $C_n$, after which it stabilizes because $C_n$ has only the chain $0=C_0<C_1<\cdots <C_n$ of subgroups. Thus DCC holds. The argument includes $C_0=0$ and is unchanged for $p=2$. This proves the stated claim.