Field with 4 Elements has only Order 2 Elements/Proof 1

Proof
By definition of field, both the algebraic structures $\struct {\GF, +}$ and $\struct {\GF^*, \times}$ are (abelian) groups, where $\GF^* := \GF \setminus \set 0$.

By definition:
 * $\struct {\GF, +}$ is of order $4$
 * $\struct {\GF^*, \times}$ is of order $3$.

From Classification of Groups of Order up to 15, there are only two possibilities:
 * $(1): \quad \struct {\GF, +} \cong \Z_4$ and $\struct {\GF^*, \times} \cong \Z_3$
 * $(2): \quad \struct {\GF, +} \cong \Z_2 \times \Z_2$ and $\struct {\GF^*, \times} \cong \Z_3$.

In case $(2)$:
 * $\forall a \in \GF: a + a = 0_\GF$

but in case $(1)$:
 * $\exists a, b \in \GF: a + a = b \ne 0_\GF$ where $b + b = 0_\GF$

as there exists an element of $\struct {\GF, +}$ of order $4$.

$(1)$ describes a field.

Then:

But then:

As $\struct {\GF, +} \cong \Z_4$, both $a + a = b$ and $\paren {-a} + \paren {-a} = b$.

The only way for $\paren {b \times b} + \paren {b \times b} = 0_\GF$ is for $b \times b = b$ or $b \times b = 0_\GF$.

The second case is eliminated as $0_\GF \notin \struct {\GF^*, \times}$.

So it must be the case that $b \times b = b$.

So as $\struct {\GF^*, \times} \cong \Z_3$ it must follow that $a \times b = a, a \times \paren {-a} = b$ and then $a \times a = -a$.

It follows that the Cayley tables of $\struct {\GF, +}$ and $\struct {\GF^*, \times}$ must therefore be as follows:


 * $\begin{array}{c|cccc}

+ & 0_\GF & a & b & -a \\ \hline 0_\GF & 0_\GF & a & b & -a \\ a & a & b & -a & 0_\GF \\ b & b & -a & 0_\GF & a \\ -a & -a & 0_\GF & a & b \\ \end{array} \qquad \begin{array}{c|cccc} \times & 0_\GF & b & a & -a \\ \hline 0_\GF & 0_\GF & 0_\GF & 0_\GF & 0_\GF \\ b & 0_\GF & b & a & -a \\ a & 0_\GF & a & -a & b \\ -a & 0_\GF & -a & b & a \\ \end{array}$

But:

So:
 * $\paren {a + b} \times a \ne \paren {a \times a} + \paren {b \times a}$

demonstrating that $\times$ is not distributive over $+$.

Thus $\GF$ as has been defined:
 * $\struct {\GF, +} \cong \Z_4$ and $\struct {\GF^*, \times} \cong \Z_3$

is not a field.

It follows that our supposition that $\struct {\GF, +} \cong \Z_4$ was false.

Thus, if $\GF$ is a field, then $\struct {\GF, +} \cong \Z_2 \times \Z_2$.

That is:
 * $\forall a \in \GF: a + a = 0_\GF$

as needed to be proved.