Elementary proof that there is no field with 6 elements

Solution 1:

Let $F$ be a field with six elements. Since $|F|=6$, the sequence $0$, $1$, $1+1$, $1+1+1$, ..., $6\times 1$ must have a repetition. If $n\times 1=m\times 1$ for $n>m$, then by subtraction we obtain $(n-m)\times 1=0$. Thus the first repetition must actually be $n\times 1=0$ in the list above for some $2\leq n\leq 6$.

If $n\times 1=0$, then $n\times a=0$ for all $a\in F$, and similarly if $m\times a=0$ for some $a\neq 0$ then $m\times 1=0$ by dividing by $a$. This shows that for any $a$, $a$, $a+a$, ..., $(n-1)\times a$ are all distinct.

(This idea is due to Eric Wofsey in the comments below). If $n>2$ then $a+a\neq 0$ for any $a\neq 0$, as we have seen above. This shows that $a\neq -a$ for all non-zero $a$. Thus, pairing $a$ with $-a$ for all $a\in F\setminus\{0\}$, we see that $|F\setminus\{0\}|$ is even. This contradiction means that $n=2$.

Thus $n=2$. Let $a\in F\setminus\{0,1\}$, and notice that $0,1,a,a+1$ are all distinct. (If $a+1=0$ then $a=1$, if $a+1=a$ then $1=0$, if $a+1=1$ then $a=0$.) Notice that adding $1$ moves the elements $\{0,1,a,a+1\}$ around. If $b\in F\setminus\{0,1,a,a+1\}$ then $F$ must be $$F=\{0,1,a,a+1,b,b+1\}.$$

Where is $a+b$? It cannot equal $0$ ($a=b$), it cannot equal $1$ ($a=b+1$), or $a$ ($b=0$) or $b$ ($a=0$), or $a+1$ ($b=1$) or $b+1$ ($a=1$). (All of these are by cancellation.) So $a+b\not\in F$, a contradiction.

Solution 2:

A modification of David Craven's fine answer.

Assume that such a field $F$ exists. Its multiplicative group has order five. Therefore (Lagrange) $F$ cannot have an element of multiplicative order two. In other words, we must have $-1=1$ implying that every element of $F$ is its own additive inverse.

Let $a\in F$ be distinct from $0$ and $1$. We know that $a+a=0$ and that $a+1$ is distinct from all of $0,1,a$. It follows that $\{0,1,a,a+1\}$ is an additive group. The remaining detail is that $a+(a+1)=(a+a)+1=0+1=1$. But, again by Lagrange, the group $(F,+)$ cannot have a subgroup of order four.

Solution 3:

Let F = {0, 1, -1, a, b, c}. All a, b and c must have unique multiplicative inverses, that is, they must be paired with eachother. There's only 3 elements therefore at least one does not have an inverse. You can fairly easily generalise this to even numbers perhaps? Sorry for lack of formalism, this is intended as a comment but my score isn't high enough

Edit: Thank you in comments for pointing out my grave error and thank you for the insights I think I actually have a proof now though I could be wrong again. As said below in comments, the above proves this for the case $1 \neq -1$ so lets consider the case $1=-1$ and take $F = \{0, 1, a_1, a_2, a_3, a_4\}$ to be a field for such a case.

Because of the existence of the multiplicative inverse, we know that the inverse of $a_1$ must be one of $a_2, a_3, a_4$. Let's pick one arbitrarily, say $a_3$, and consequently $a_2$ must be the inverse of $a_4$. Thus we have:

$a_1^{-1} = a_3$,

$a_2^{-1} = a_4$.

Now let's look at $a_1 a_4$. It is certain that $a_1 a_4 \neq a_1$ or $a_1 a_4 \neq a_4$ since that implies $a_4 = 1$ or $a_1 = 1$ which violates uniqueness of identity. Similarly, $a_1 a_4 \neq 0$ or $a_1 a_4 \neq 1$ since that would contradict uniqueness of zero element and inverse respectively. This means that $a_1 a_4 = a_2 \space or \space a_3$. It doesn't matter which one we pick, the proof is pretty much the same for both from here on out, so lets pick $a_1 a_4 = a_3$. From this equality and using a tad of manipulation:

$a_2 = a_1^2$

Because of the existence of the additive inverse, the inverse of $a_1$ must be one of $-a_2, -a_3, -a_4$. These 3 cases are considered briefly as follows (using $1 = -1$):

1.) $a_1 = -a_2 = a_2 = a_1^2$ which implies $a_1 = 1$

2.) $a_1 = -a_3 = a_3 = a_1^{-1}$ which implies $a_2 = a_1^2 = 1$

3.) $a_1 = -a_4 = a_2 = a_1^2$ which implies $a_1 = 1$

This contradicts uniqueness of multiplicative identity. Furthermore, since choices above were arbitrary and/or near identical, we are forced to conclude that F is not in fact a field.

I don't know if this is the right kinda proof for you as its a tad long but should be understandable enough right?