Zero and Unity of Subfield

Theorem
Let $\left({F, +, \times}\right)$ be a field whose zero is $0$ and whose unity is $1$.

Let $\left({K, +, \times}\right)$ be a subfield of $F$.

Then:
 * the zero of $\left({K, +, \times}\right)$ is $0$

and:
 * the unity of $\left({K, +, \times}\right)$ is $1$.

Proof
By definition, $\left({K, +, \times}\right)$ is a subset of $F$ which is a field.

By definition of field:
 * $\left({K, +}\right)$ and $\left({F, +}\right)$ are groups such that $K \subseteq F$

and:
 * $\left({K^*, \times}\right)$ and $\left({F^*, \times}\right)$ are groups such that $K \subseteq F$.

So:
 * $\left({K, +}\right)$ is a subgroup of $\left({F, +}\right)$

and:
 * $\left({K^*, \times}\right)$ is a subgroup of $\left({F^*, \times}\right)$.

By Identity of Subgroup:
 * the identity of $\left({F, +}\right)$, which is $0$, is also the identity of $\left({K, +}\right)$

and:
 * the identity of $\left({F^*, \times}\right)$, which is $1$, is also the identity of $\left({K^*, \times}\right)$.