Field with 4 Elements has only Order 2 Elements/Proof 2

Theorem

Let $\struct {\GF, +, \times}$ be a field which has exactly $4$ elements.

Then:

$\forall a \in \GF: a + a = 0_\GF$

where $0_\GF$ is the zero of $\GF$.

Proof

Let $\struct {\GF, +, \times}$ be a field of order $4$ whose zero is $0_\GF$.

By definition, $\GF$ is a Galois field.

The additive group $\struct {\GF, +}$ of $\GF$ can be one of two:

$(1): \quad$ the cyclic group of order $4$, generated by the identity of $\struct {\GF, +}$ which is $0_\GF$

or:

$(2): \quad$ the Klein $4$-group, whose elements are all of the form $a + a = 0_\GF$.

Aiming for a contradiction, suppose $(1)$ is the additive group of a field.

Then the characteristic of $\GF$ is $4$.

But from Characteristic of Galois Field is Prime it follows that $\GF$ is not a field.

From that contradiction it follows that the additive group $\struct {\GF, +}$ of $F$ cannot be the cyclic group of order $4$.

Hence for $\GF$ to be a field at all, it is necessary for $\struct {\GF, +}$ to be the Klein $4$-group.

$\blacksquare$