 Solution

In a finite group, choose a maximal chain of normal subgroups from $1$ to $G$. Finiteness ensures such a chain exists, and maximality leaves no normal subgroup of $G$ strictly between successive terms. This is a chief series.

The infinite cyclic group has no chief series. In any finite chain starting at zero, the first nonzero subgroup is $n\mathbb{Z}$. But $2n\mathbb{Z}$ is a nonzero proper subgroup of it and is normal in $\mathbb{Z}$. Thus zero is not maximal among the normal subgroups in that first term, contradicting the definition.

For any chief series, $G_{n-1}$ is a maximal proper normal subgroup of $G$. The subgroup correspondence theorem identifies normal subgroups of $G/G_{n-1}$ with normal subgroups of $G$ containing $G_{n-1}$. Consequently the top chief factor is simple.

Other chief factors need only have no proper nonzero subgroup invariant under the action of the whole group. In $A_4$,is a chief series, where $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. Conjugation in $A_4$ permutes the three involutions transitively, so none of its three order-two subgroups is normal in $A_4$. Thus $V_4$ is a minimal nontrivial normal subgroup; the top quotient is $C_3$. But the bottom chief factor $V_4/1\cong C_2\times C_2$ is not simple, since it has proper order-two normal subgroups as an abstract group.