Does this "factory of binary sequences" group have a standard name? I was thinking about ways to associate binary sequences with elements of a group. I came up with a group, and I was wondering if it has a standard name, or if it can be described easily in terms of well-studied objects. (The construction seems interesting to me, but I would rather not reinvent the wheel!)
The construction is as follows. Let $T$ be the set of all binary sequences, so any element $t \in T$ is a map $t: \mathbb{Z} \to \{0,1\}$. Define a map $c: T \to T$ (which I think of as "cycle") that changes the zeroth element of a sequence from 0 to 1, or from 1 to 0. That is, for a sequence
$t \in T$, we have:
$$
[c(t)](0)= 
\begin{cases}
0 & \textrm{if } t(0) = 1\\
1 & \textrm{if } t(0) = 0
\end{cases} \\
\textrm{and }[c(t)](n) = t(n) \textrm{ for any } n \neq 0.
$$
Next, define a map $s: T \to T$ (which I think of as "shift") that shifts all the elements of a sequence to the right, so:
$$
[s(t)](n) = t(n-1), \textrm{ for any } n \in \mathbb{Z}.
$$
My reason for considering these maps is that, by using them together, it seems one can construct binary sequences (say, by setting the zero sequence as a starting point and then specifying a list of instructions like "do cycle, then shift, then cycle"). One then can "compose" two binary sequences by concatenating the lists of operations used to generate them. This gives a way of combining binary sequences different from element-wise operations.
Notice that these maps $c$ and $s$ are invertible. We have $c = c^{-1}$, as switching the zeroth element twice of any sequence leaves the sequence unchanged. The inverse of $s$ acts by shifting all elements of an input sequence to the left.
Define the "binary sequence factory group", $F$, as the group generated by these operations $c$ and $s$, with group element composition given by function composition.
This group has an infinite number of elements, corresponding to the fact that there are an infinite number of binary sequences. In addition, this group is not commutative (for example, $ccs$ is different from $csc$). However, it is generated by two pretty simple operations.
Does this group $F$ have a standard name? Or can it be constructed in some standard way from well-known groups? Can this group be simply described in terms of some presentation?
 A: This is indeed a well-studied group known as the "lamplighter group".  You can find plenty of information about it by just looking up that name.  The name comes from imagining an infinite row of lamps and a person walking around turning them on and off.  Your $c$ has the person turn the lamp they are currently at on or off, and your $s$ has the person move to the next lamp in the row.  It has a simple presentation in terms of these two generators: you just need the relations $c^2=1$ and that $c$ commutes with $s^ncs^{-n}$ for any  $n$.
(Note that it is slightly nonobvious that your description of the group as permutations of the set $T$ is isomorphic to the usual definition of the lamplighter group.  The difference is that the usual definition would also keep track of the current location of the person, so it would be acting not on $T$ but on $T\times \mathbb{Z}$ where the second coordinate is the current location (so $c$ ignores it and $s$ shifts it by $1$).  This has the result that the lamplighter group acts freely on $T\times\mathbb{Z}$ (and simply and transitively on $T_0\times\mathbb{Z}$ where $T_0$ is the set of sequences that have only finitely many $1$s), whereas it does not act freely on your $T$, since for instance $s$ fixes any constant sequence.  As a result, a priori the action of the lamplighter group on $T$ could fail to be faithful, so that your $F$ is not the lamplighter group itself but some quotient of it.  However, it's not too hard to see that this is not the case: for instance, if you take any sequence $t\in T$ that is not eventually periodic, then the lamplighter group will act freely on the orbit of $t$.)
