If $G$ has an $\Omega$-composition series, prove that every $\Omega$-subgroup [of $G$ has] a composition series. This is part of Exercise 3.1.4 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to Approach0, it is new to MSE.
The Details:
Since definitions vary, on page 15, ibid., paraphrased, it states that

A subgroup $N$ of $G$ is normal in $G$ if one of the following equivalent statements is satisfied:
(i) $xN=Nx$ for all $x\in G$.
(ii) $x^{-1}Nx=N$ for all $x\in G$.
(iii) $x^{-1}nx\in N$ for all $x\in G, n\in N$.

On page 28, ibid.,

A right operator group is a triple $(G, \Omega, \alpha)$ consisting of a group $G$, a set $\Omega$ called the operator domain and a function $\alpha:G\times \Omega\to G$ such that $g\mapsto (g,\omega)\alpha$ is an endomorphism of $G$ for each $\omega\in\Omega$. We shall write $g^\omega$ for $(g,\omega)\alpha$ and speak of the $\Omega$-group if the function $\alpha$ is understood.
[. . .]
If $G$ is an $\Omega$-group, an $\Omega$-subgroup of $G$ is a subgroup $H$ which is $\Omega$-admissible, that is, such that $h^\omega\in H$ whenever $h\in H$ and $\omega\in\Omega$.

On page 63, ibid.,

Let $G$ be an operator group with operator domain $\Omega$. An $\Omega$-series (of finite length) in $G$ is a finite sequence of $\Omega$-subgroups including $1$ and $G$ such that each member of the sequence is a normal subgroup of its successor: thus a series can be written
$$1=G_0\lhd G_1\lhd\dots\lhd G_l=G.$$
The [. . .] quotient groups $G_{i+1}/G_i$ are the factors of the series.

On page 64, ibid.,

If $\mathbf{S}$ and $\mathbf{T}$  are $\Omega$-series of [an $\Omega$-group] $G$, call $\mathbf{S}$ a refinement of $\mathbf{T}$ if every term of $\mathbf{T}$ is also a term of $\mathbf{S}$. If there is at least one term of $\mathbf{S}$ which is not a term of $\mathbf{T}$, then $\mathbf{S}$ is a proper refinement of $\mathbf{T}$.

On page 65, ibid.,

An $\Omega$-series which has no proper refinements is called an $\Omega$-composition series.
[. . .]
If $\Omega$ is empty, we speak of a composition series.

The Question:

If $G$ has an $\Omega$-composition series, prove that every $\Omega$-subgroup [of $G$ has] a composition series.

Thoughts:
I'm somewhat confused: if $H$ is an $\Omega$-subgroup of $G$, then, in particular, it is a group, so . . .
Yeah, I think that some refinement of
$$1\lhd H$$
would be a composition series of $H$, which would, I suppose, be guaranteed to exist since $G$ has an $\Omega$-composition series, itself a refinement of
$$1\lhd_\Omega G,$$
where $\lhd_\Omega$ means "is a normal $\Omega$-subgroup of".
I don't think I am expressing my thoughts clearly enough. For lack of a better term, perhaps there is some "forgetful functor" to go from $\lhd_\Omega$ to $\lhd$.
I don't know . . .

I first encountered $\Omega$-groups in Roman's "Fundamentals of Group Theory: An Advanced Approach", page 274, a few months ago. They made sense then.
Here is my previous question:

*

*Abelian and nonabelian groups with the same composition factors.

Working through an example might help.
Consider $\Omega=\{x\}$, $G=S_3$, and
$$\begin{align}
\alpha:S_3\times \Omega &\to S_3\\
(\sigma,x)&\mapsto \sigma.
\end{align}$$
The series
$$1\lhd_\Omega \Bbb Z_3\lhd_\Omega S_3$$
is an $\Omega$-composition series by inspection.
The subgroup $H=(\{e, (123), (321)\}, \circ)$ is an $\Omega$-subgroup, since $\tau^x=\tau\in H$ for all $\tau\in H$.
Perhaps this example is not particularly illustrative, because $H$ so clearly has the composition series
$$1\lhd H$$
without recourse to $G$ and its $\Omega$-composition series.

Please help :)
 A: I think your best bet is to use Theorem 3.1.5:

3.1.5 An $\Omega$-group $G$ has an $\Omega$-composition series if and only if it satisfies max-$\Omega$s and min-$\Omega$s.

In other words, if it satisfies ACC and DCC on $\Omega$-subgroups.
If $G$ has an $\Omega$-composition series, then it satisfies ACC and DCC on $\Omega$-subgroups. If $H$ is an $\Omega$-subgroup of $G$, then every $\Omega$-subgroup of $H$ is also an $\Omega$-subgroup of $G$, hence $H$ also satisfies ACC and DCC on its $\Omega$-subgroups, and therefore $H$ has an $\Omega$-composition series.

One is tempted to try to try do this explicitly by taking an $\Omega$-composition series of $G$,
$$1=N_0\triangleleft N_1\triangleleft\cdots\triangleleft N_m=G$$
and trying to use it to obtain an $\Omega$-composition series for $H$, perhaps by intersecting the series with $H$,
$$1 = N_0\cap H \triangleleft N_1\cap H\triangleleft\cdots\triangleleft N_m\cap H = H.$$
The problem is that while this is indeed a series of $\Omega$-subgroups that are normal in $H$, in general we won't get that the successive quotients $(N_{i+1}\cap H)/(N_i\cap H)$ is $\Omega$-simple. For example, taking $\Omega$ empty, so that we are talking about regular composition series, take a proper subgroup of $A_5$. The only composition series for $A_n$ is $1\triangleleft A_n$, since $A_n$ is simple, so that would yield the normal series $1\triangleleft H$; but this will not be a composition series for $H$ unless $H$ is cyclic of prime order or trivial. So you would need to start mucking about with $H$ itself. In the finite group case we obtain the result by induction on the order, but for arbitrary groups, as expected in this exercise and in Robinson's  book in general, I think you really need to go through 3.1.5. Which, by the way, should also deal with the second part of the exercise, dealing with $\Omega$-quotients.
