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The integers with addition build a group $\langle \mathbb{Z},+,0\rangle$.

The functions $\operatorname{succ}:\mathbb{Z} \rightarrow \mathbb{Z}$, $\operatorname{pred}:\mathbb{Z} \rightarrow \mathbb{Z}$ with $\operatorname{succ}(x) = x +1$, $\operatorname{pred}(x) = x -1$ are the generators of the (free) group $\langle \{\operatorname{succ},\operatorname{pred}\}^+,\circ,\operatorname{id}\rangle$ with $\{\operatorname{succ},\operatorname{pred}\}^+$ the set of finite concatenations of $\operatorname{succ}$, $\operatorname{pred}$ and $\operatorname{succ}\circ \operatorname{pred} = \operatorname{pred}\circ \operatorname{succ} = \operatorname{id}$. This group is supposed to act on $\mathbb{Z}$.

There is a natural one-to-one mapping between $\langle \mathbb{Z},+,0\rangle$ and $\langle \{\operatorname{succ},\operatorname{pred}\}^+,\circ,\operatorname{id}\rangle$:

  • $+ \leftrightarrow \circ$
  • $0 \leftrightarrow \operatorname{id}$
  • $k > 0 \leftrightarrow \operatorname{succ}^k$
  • $k < 0 \leftrightarrow \operatorname{pred}^{-k}$

What's the official name of the group $\langle \{\operatorname{succ},\operatorname{pred}\}^+,\circ,\operatorname{id}\rangle$?

Compare with the monoid $\langle \mathbb{N},+,0\rangle$.

Consider the functions $\operatorname{succ}':\mathbb{N} \rightarrow \mathbb{N}$, $\operatorname{pred}':\mathbb{N} \rightarrow \mathbb{N}$ with $\operatorname{succ}'(x) = x +1$, $\operatorname{pred}'(0) = 0$ and $\operatorname{pred}'(x) = x -1$ if $x > 0$. The functions $\operatorname{succ}'$ and $\operatorname{pred}'$ are freely composable and $\operatorname{pred}'\circ \operatorname{succ}' = \operatorname{id}$.

But $\operatorname{succ}'\circ \operatorname{pred}' \neq \operatorname{id}$ because $\operatorname{succ}'\circ \operatorname{pred}'(0) = 1$. On the other side $\operatorname{succ}'\circ \operatorname{pred}'$ is somehow "almost" the identity because $\operatorname{succ}'\circ \operatorname{pred}'(n) = n$ for every other $n \neq 0$.

This means: $\operatorname{pred}'$ has a ("strong") right-inverse and $\operatorname{succ}'$ has a ("strong") left-inverse. In turn $\operatorname{pred}'$ has only a "weak" left-inverse and $\operatorname{succ}'$ has only a "weak" right-inverse.

Is this distinction between "strong" and "weak" inverses of any (and in any) use?

What's the name of the kind of algebraic structure $\langle \{\operatorname{succ}',\operatorname{pred}'\}^+,\circ,\operatorname{id}\rangle$ – which is supposed to act on $\mathbb{N}$ – is an example of?

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    $\begingroup$ I may be missing something but isn't that just $\mathbb Z$ with $1$ and $-1$ as generators? (In response to your first question.) The other thing is known to me as a standard example in ring theory or linear algebra of an operator admitting a left inverse but no right inverse. $\endgroup$ – Myself Dec 8 '14 at 12:19
  • $\begingroup$ I'm pretty sure its official name is simply $\mathbb{Z}$. $\endgroup$ – Najib Idrissi Dec 8 '14 at 12:32
  • $\begingroup$ Isn't it the group of the left(or right because is commutative) traslations of $+$ with the composition operation? I guess that we could see it as the ciclyc subgroup $\langle S \rangle$ of the group of bejections from $\Bbb Z$ to itself equipped with the composition. Where $S$ is the successor function. $\endgroup$ – MphLee Dec 8 '14 at 12:33
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    $\begingroup$ As an abstract group it surely is (isomorphic) to $\mathbb{Z}$. $\endgroup$ – Hans-Peter Stricker Dec 8 '14 at 13:21
  • $\begingroup$ Sure @HansStricker because $k \mapsto f_k$ where $f_k(x):=k+x$ is a bijection. And is an isomorphism betwen addition and composition (as you noted). $\endgroup$ – MphLee Dec 8 '14 at 17:39
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If I understand the definition correctly, the structure $\langle \{\operatorname{succ}',\operatorname{pred}'\}^+,\circ,\operatorname{id}\rangle$ is an inverse monoid, i.e. an inverse semigroup with identity, since

  • $\operatorname{succ}' = \operatorname{succ}' \circ\operatorname{pred}'\circ \operatorname{succ}'$

  • $\operatorname{pred}' = \operatorname{pred}' \circ\operatorname{succ}'\circ \operatorname{pred}'$

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