Identity of Algebraic Structure is Preserved in Substructure

Theorem
Let $\left({S, \circ}\right)$ be an algebraic structure with identity $e$.

Let $\left({T, \circ}\right)$ be a algebraic substructure of $\left({S, \circ}\right)$.

That is, let $T \subseteq S$.

Let $e \in T$.

Then $e$ is an identity of $\left({T, \circ}\right)$.

Proof
Let $x \in T$.

By the definition of subset, $x \in S$.

Since $e$ is an identity of $\left({S, \circ}\right)$:


 * $e \circ x = x \circ e = x$

Since this holds for all $x \in T$, $e$ is an identity of $\left({T, \circ}\right)$.