Identity of Submonoid is not necessarily Identity of Monoid

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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 $\struct {T, \times}$.


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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 $\struct {T, \times}$.

$\blacksquare$


Also see


Sources

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