Zero of Subfield is Zero of Field/Proof 2

Theorem

Let $\struct {F, +, \times}$ be a field whose zero is $0$.

Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.

The zero of $\struct {K, +, \times}$ is also $0$.

Proof

By definition, $\struct {K, +, \times}$ is a subset of $F$ which is a field.

By definition of field, $\struct {K, +}$ and $\struct {F, +}$ are groups such that $K \subseteq F$.

So, by definition, $\struct {K, +}$ is a subgroup of $\struct {F, +}$.

By Identity of Subgroup, the identity of $\struct {F, +}$, which is $0$, is also the identity of $\struct {K, +}$.

$\blacksquare$