Let $\displaystyle A$ be the set of all strictly increasing sequences of integers. Define a binary operator $\displaystyle +: A\times A \to A$ by $\displaystyle (a_n)_{n\ge 1} + (b_n)_{n\ge 1} = (a_n+b_n-n)_{n\ge 1}$. The pair $\displaystyle (A,+)$ is a monoid. Define a sequence of functions for $\displaystyle k\ge 1$, $\displaystyle \varphi_k:A\to A$ by $\displaystyle \varphi\left( (a_n)_{n\ge 1}\right) = (a_{n+k}-a_k)_{n\ge 1}$. It is easy enough to check that each function in the sequence is a homomorphism, and for each $\displaystyle k$, $\displaystyle \ker(\varphi_k)\subseteq \ker(\varphi_{k+1})$. Let $\displaystyle \displaystyle B=\bigcup_{k\ge 1} \ker(\varphi_k)$. Questions: Is $\displaystyle B$ a sub-monoid of $\displaystyle A$? Is there a way to "mod-out" by $\displaystyle B$?

Edit:

Some additional info:

The identity element is the sequence $\displaystyle (n)_{n\ge 1}$. The sets, $\displaystyle \ker(\varphi_k)=\{(a_n)\in A\mid \forall n>k, a_{n+1}=a_n+1\}$ are all closed under addition. Does the set $\displaystyle B=\{(a_n)\in A\mid \exists N\in \mathbb{N} \forall n>N, a_{n+1}=a_n+1\}$? If so, it should be closed under addition, as well. I forgot to mention that $\displaystyle (A,+)$ is commutative. So, if "modding out" does make sense in this context, cosets would have the form $\displaystyle (a_n)+B$. Without inverses, I have no idea how to tell if two sequences are in the same left coset.