If $S$ is all finitary permutations of $\Bbb N$ and $A$ is the alternating subgroup, prove $1\lhd A\lhd S$ is the only composition series of $S$ This is Exercise 3.2.2 of Robinson's, "A Course in the Theory of Groups (Second Edition)". According to this search, it is new to MSE.
Previous exercises of mine from the book include:

*

*Abelian and nonabelian groups with the same composition factors.


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

If $1$ and $G$ are the only normal subgroups of $G$ and $G\neq 1$, then the group $G$ is said to be simple.

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 $S$ is the group of all finitary permutations of the set $\{1,2,3,\dots\}$ and $A$ is the alternating subgroup, prove that $1\lhd A\lhd S$ is the only composition series of $S$.

Thoughts:
That
$$1\lhd A\lhd S\tag{1}$$
is a series for $S$ - not yet necessarily a composition series - is somewhat evident in the fact that $[S:A]=2$ (which I recognise is something I need to prove), meaning $A\lhd S$ (and, perhaps, the index prohibits any $H$ such that $A\lhd H\lhd S$), and $1$ is normal in each group. I don't yet know how to prove that $(1)$ is a composition series, let alone the only such series.
Perhaps the theorem below might help:

Theorem 3.2.1: (Jordan) The alternating group $A_n$ is simple if and only if $n\neq 1,2$, or $4$.

However, strange things happen with infinite groups, so I'm not sure it applies to $A$.
One approach I have in mind to show that $(1)$ is unique is to suppose, for a contradiction, that another, distinct composition series
$$1=G_0\lhd G_1\lhd \dots\lhd G_l=S\tag{2}$$
exists for $S$. I'm not sure where to go from here though. I don't think mentioning this is at all productive but, nonetheless, it's a start . . .
Please help :)
 A: Note that there is a very natural embedding of $S_n$ into $S$. Actually, with this embedding we have $S=\bigcup\limits_{n=1}^\infty S_n$. Similarly, $A=\bigcup\limits_{n=1}^\infty A_n$. Now it is easy to show that $A$ is a simple group. Indeed, let $\{e\}\ne N\trianglelefteq A$. Then for some $n_0\geq 5$ we have $N\cap A_{n_0}\ne\{e\}$. It follows that $N\cap A_{n}\ne\{e\}$ for all $n\geq n_0$. And since $A_n$ is a simple group for all $n\geq n_0$, it follows that $N\cap A_n=A_n$ for all such $n$. Hence:
$$\begin{align}
N\cap A &=N\cap\left(\bigcup\limits_{n=1}^\infty A_n\right)\\
&=N\cap\left(\bigcup\limits_{n=n_0}^\infty A_n\right)\\
&=\bigcup\limits_{n=n_0}^\infty(N\cap A_n)\\
&=\bigcup\limits_{n=n_0}^\infty A_n\\
&=A.
\end{align}$$
And so $A=N$ in that case. $A$ is indeed a simple group.
The next step is showing that the only normal subgroups of $S$ are the $S,A,\{e\}$. So suppose $S\ne N\trianglelefteq S$. Then $N\cap A\trianglelefteq A$, and so $N\cap A=A$ or $N\cap A=\{e\}$. In the first case we get $A=N$, since $A$ is clearly a maximal subgroup of $S$. (it clearly has index $2$). The case $N\cap A=\{e\}$ is a bit more interesting. We have to show that $N=\{e\}$ in that case. Assume otherwise. Note that $N$ can't have exactly two elements, because then it would be contained in the center of $S$, but it is easy to check that $S$ has trivial center. (same proof as for $S_n$). So $|N|\geq 3$, and so it has some nontrivial permutations $\sigma,\tau\in N$. Since $N\cap A=\{e\}$ it follows that $\sigma, \tau$ are both odd. But note that at least one of the elements $\sigma^2, \sigma\tau$ is an even permutation which is not the identity. So $N$ contains a nontrivial even permutation, a contradiction.
Ok, so now your problem is pretty much solved, because you know all the normal subgroups of $S$.
A: To prove it's a composition series, you need to prove that $A$ is  simple. Let $1 \ne N \unlhd A$ and let $1 \ne x \in N$.
We want to prove that $N=A$. So let $y \in A$ and we will  prove that $y \in N$. Let $X$ be a finite subset of ${\mathbb N}$ that contains the supports of both $x$ and $y$, and choose $X$ such that $|X| \ge 5$. Then $x \in N \cap A_X$ (the alternating group on the set $X$).
Since $|X| \ge 5$, $A_X$ is simple, and $N \cap A_X \unlhd A_X$, so $N \cap A_X = A_X$. But $y \in A_X$, so  $y \in N$ and we have proved that $N=A$ and that $A$ is simple.
To prove it is the unique composition series let $1 \lhd N \cdots \lhd S$ be another one, with $N$ simple. If $N=A$ then we have the same series. Otherwise, since $N$ and $A$ are simple, we have $A \cap N = 1$. Since $|S:A|=2$, we must have $|N| = 2$, so $S = A \times N$, and $N \le Z(S)$.
I will leave you to prove that that cannot happen.
