Field with 4 Elements has only Order 2 Elements

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 1

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 $4$.

Aiming for a contradiction, suppose $(1)$ describes a field.

Then:

 $\ds \paren {b + b} \times b$ $=$ $\ds 0_\GF \times b$ as $b + b = 0_\GF$ from above $\ds$ $=$ $\ds 0_\GF$ as $0_\GF$ is the zero of $\struct {\GF, +, \times}$

But then:

 $\ds \paren {b + b} \times b$ $=$ $\ds \paren {b \times b} + \paren {b \times b}$ as $\times$ is distributive over $+$

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:

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

 $\ds \paren {a + b} \times a$ $=$ $\ds \paren {-a} \times a$ as $a + b = -a$ from above $\ds$ $=$ $\ds b$ as $\paren {-a} \times a = b$ from above

But:

 $\ds \paren {a \times a} + \paren {b \times a}$ $=$ $\ds \paren {-a} + a$ from above $\ds$ $=$ $\ds 0_\GF$ from above

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.

$\blacksquare$

Proof 2

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.

$\blacksquare$

Also see

$\struct {\GF, +} \cong \Z_2 \times \Z_2$ and $\struct {\GF^*, \times} \cong \Z_3$
is actually a field.