TITLE: Submonoid: Identity
QUESTION [5 upvotes]: There's a common mistake appearing in quite many books in definitions and proofs as well; now my idea is maybe you can safe the day:
Imagine you're having a monoid and a subset being closed under the binary operation inherited from the monoid. Moreover, let it contain an element which looks as an identity element when acting on the subset but being not the identity of the monoid.
For example:
$M:=(\mathcal{M}(2,2; \mathbb{K}),\cdot),\quad N:=(\{\text{all matrices with all but the first entry zero}\},\cdot)$
Then:
$N\leq M\text{(as submonoid)}$
But:
$\mathbb{1}_N\neq \mathbb{1}_M$
My question/idea is now: It should be possible to find an isomorphism in general some submonoid containing the original identity, i.e.:
$N\cong N'\leq M\text{ with }\mathbb{1}_{N'}=\mathbb{1}_M$
...can you proof this?

REPLY [4 votes]: No, it is not always possible.
If you know partial permutations, there is an easy counterexample: Let $M$ be generated by $(1,2,3)(4)(5)$ and $(4,5)$ and let $N$ be generated by $(4,5)$. Then $M$ has 5 elements: $(1)(2)(3)(4)(5)$, $(1,2,3)(4)(5)$, $(1,3,2)(4)(5)$, $(4)(5)$, and $(4,5)$. $M$ contains only two subgroups (subsemigroups that happen to be groups, they can have different identities and inverses): $N=\{ (4)(5), (4,5) \}$ and $N'=\{ (1)(2)(3)(4)(5), (1,2,3)(4)(5), (1,3,2)(4)(5) \}$. Clearly $N \not\cong N'$ since the former has two elements and the latter has three elements. Also $N'$ is the only one that contains $\mathbb{1}_M = (1)(2)(3)(4)(5)$, since $\mathbb{1}_N = (4)(5)$.
Partial perms are fairly easy to understand: they are bijections (rather than self-bijections). The ones in $M$ are even easier to understand, they are self-bijections but of different sets (either $\{1,2,3,4,5\}$ or $\{4,5\}$). Composition is done in the usual way, the product of $f$ and $g$ is defined on the largest subset for which the composition makes sense. $(1,2,3)(4)(5) \cdot (4,5) = (4,5)$ since $(4,5)$ doesn't know what to do with $1$, $2$, or $3$.
They produce nice monoids known as inverse monoids and are fairly group like once you understand idempotents (elements $\mathbb{1}_N$ for subsemigroups of $M$ that happen to be monoids).
Here is the matrix representation:
$$M = \left\langle 
\left[\begin{smallmatrix}0&1&0&0&0\\0&0&1&0&0\\1&0&0&0&0\\0&0&0&1&0\\0&0&0&0&1\end{smallmatrix}\right],
\left[\begin{smallmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&1\\0&0&0&1&0\end{smallmatrix}\right]
\right\rangle, \quad N= \left\langle
\left[\begin{smallmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&1\\0&0&0&1&0\end{smallmatrix}\right]
\right\rangle$$
Then listed explicitly:
$$M=\left\{
\left[\begin{smallmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&1\\0&0&0&1&0\end{smallmatrix}\right],
\left[\begin{smallmatrix}0&0&0&0&0\\0&0&0&0&0\\0&0&0&0&0\\0&0&0&1&0\\0&0&0&0&1\end{smallmatrix}\right],
\left[\begin{smallmatrix}0&0&1&0&0\\1&0&0&0&0\\0&1&0&0&0\\0&0&0&1&0\\0&0&0&0&1\end{smallmatrix}\right],
\left[\begin{smallmatrix}0&1&0&0&0\\0&0&1&0&0\\1&0&0&0&0\\0&0&0&1&0\\0&0&0&0&1\end{smallmatrix}\right],
\left[\begin{smallmatrix}1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\end{smallmatrix}\right]
\right\}$$
$$M=\{(4,5),(4)(5),(1,3,2)(4)(5),(1,2,3)(4)(5),(1)(2)(3)(4)(5)\}$$