TITLE: Elementary proof that there is no field with 6 elements
QUESTION [13 upvotes]: I am a T.A at an introductory linear algebra course this semester, and before getting into vector spaces, we give them some basic examples and properties of fields. Since this is a first semester course, everything is very elementary and we do not have too many tools.
I'd be happy to give them some elementary proof that there is no field of size $6$ (or some other small, concrete number, like 10... I just do not really care about the general case). They already know that $\mathbb Z _6$ is not a field, but I want to show that there is no other possible field of this size.
By elementary, I mean something which can be taught to math students on the second week to their first semester (so no arguments using groups and such). I can start talking about characteristics of fields and then show that the size has to be some power of the characteristic, but I'd prefer not to get into all of this. I am simply wondering if anyone here knows of an elementary proof for some concrete size (like $6$).
Thank in advance!

REPLY [2 votes]: 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.