It is trivial to show that if a group $G$ satisfies the maximal condition on subgroups (denoted by $Max$), then $G$ satisfies the maximal condition on its normal subgroups (denoted by $Max-n$). We have a similar statement for the minimal condition ($Min$) and the minimal condition on normal subgroups ($Min-n$).
We claim that the inverse is not true, we have to find an example. In my course we got the following examples:
- Example of group $G$ such that $G \in Max-n$ and $G \notin Max$: Let $p$ prime number and $$A = \left\lbrace \frac{m}{p^n} : m \in \mathbb{Z}, \ n \in \mathbb{N}_0 \right\rbrace.$$ Let $$x: a \in A \mapsto p a \in A ,$$ then $x\in\operatorname{Aut}A$, let $G = \langle x \rangle \ltimes A$.
I have a proof that $G \in Max-n$, but why $G \notin Max$?
- Example of group $G$ such that $G \in Min-n$ and $G \notin Min$: Let $p$ prime number, let $F$ algebraic closure of $\mathbb{Z}/p\mathbb{Z}$. Let $q \neq p$ prime number and $H \leq \bigl( F \setminus \{0\}, \cdot \bigr)$ such that $H \cong Z(q^\infty)$ (there's a proposition that tells me such $q$ exists). Let $A$ the subfield of $F$ generated by $H$, then $$A = \lbrace n_1 h_1 + \cdots + n_t h_t : n_1, \dots, n_t \in \mathbb{Z}, \ h_1, \dots, h_t \in H \rbrace .$$ For all $h \in H$, let $$\theta(h): a \in A \mapsto h a \in A ,$$ we have $\theta(h)\in\operatorname{Aut}\bigl(A, +\bigr)$. Let $$\theta: h \in H \mapsto \theta(h) \in\operatorname{Aut}\bigl(A, +\bigr) ,$$ it is an homomorfism. Let $(G, \cdot) = (H, \cdot) \ltimes_\theta (A, +)$.
I have a proof that $G \in Min-n$, but why $G \notin Min$?
Note: $Z(q^\infty)$ is the Prufer q-group
EXTRA: do you have any other (simpler) example?
$$ 1)\ \langle1\rangle<\langle1/p\rangle<\langle1/p^2\rangle<\ldots $$
$2)$ Pick $t_k\in H$ such that $|t_k|=q^k$. We have $$ \langle t_1,t_2,t_3,\ldots\rangle> \langle t_2,t_3,\ldots\rangle> \langle t_3,\ldots\rangle>\ldots $$