# Identity of Submonoid is not necessarily Identity of Monoid

## Theorem

Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.

Let $\struct {T, \circ}$ be a submonoid of $\struct {S, \circ}$ whose identity is $e_T$.

Then it is not necessarily the case that $e_T = e_S$.

## Proof

Let $\struct {S, \times}$ be the semigroup formed by the set of order $2$ square matrices over the real numbers $R$ under (conventional) matrix multiplication.

Let $T$ be the subset of $S$ consisting of the matrices of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$.

From Matrices of the Form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ we have that $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$.

From (some result somewhere) $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is not in $\struct {T, \times}$.

However, note that:

 $\ds \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ $=$ $\ds \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ $\ds$ $=$ $\ds \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$

demonstrating that $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is the identity of $\struct {T, \times}$.

$\blacksquare$