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

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Theorem

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


Then:

$\forall a \in F: a + a = 0_F$

where $0_F$ is the zero of $F$.


Proof

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

By definition, $F$ is a Galois field.

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

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

or:

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


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

Then the characteristic of $F$ is $4$.

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

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

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

$\blacksquare$