Remark

A group with operators does not always possess a Jordan-Hölder series; an example is given by the additive group $\mathbf{Z}$ of rational integers: the sequence $(2^n \cdot \mathbf{Z})_{n \geq 0}$ is a strictly decreasing infinite sequence of (normal) subgroups of $\mathbf{Z}$; for all $p$, the first $p$ terms of this sequence form with the group $\{0\}$ a strictly decreasing composition series; if there existed a Jordan-Hölder series for $\mathbf{Z}$, it would have at least $p+1$ terms, by the Corollary to Theorem 6; absurd, since $p$ is arbitrary.

On the other hand, there exists a Jordan-Hölder series in every finite group with operators G: if $G \ne \{e\}$, among the normal stable subgroups of G distinct from G, let $H_1$ be a maximal subgroup; similarly define $H_{n+1}$ by induction as a maximal element in the set of normal subgroups of $H_n$, distinct from $H_n$, when $H_n \ne \{e\}$; the sequence of orders of the $H_n$ is strictly decreasing, hence there exists $n$ such that $H_n = \{e\}$ and the sequence consisting of G and the $H_i$ ($1 \leq i \leq n$) is by its formation a Jordan-Hölder series.