My goal is to show that groups of order < 6 are abelian in a more elegant way than just listing all possible Latin squares. To do this, my first problem where I got stuck, is how to prove that all groups of order p prime are abelian. Can someone help me?
Thanks & have a nice evening!
On
The proof of this is literally anywhere with just a Google search. It follows from Lagrange's theorem: any non-identity element $x$ generates a subgroup, which has order either $1$ or $p$; but it cannot be $1$ since $x$ is not the identity element. Can you finish it from here?
On
You do not even need to go there.
Another way to prove it is that a finite ordered non-abelian group must have order 6 or greater which can be done like this
if $|G|=1$ then it's trivial, So assume $|G|>1$ then there exist $a\in G$, but all groups have inverses so $a^{-1}\in G$, which means $|G|\geq 3$, but this case would be abelian so there must be another element $b\in G$ and equally so it's inverse, so we have that $|G|\geq 5$, But we also have that $ab\in G$, which would give one of the following $ab=a$, $ab=b$, $ab=e$, $ab=a^{-1}$ or $ab=b^{-1}$, the first ones can be eliminated as they imply that either $a$ or $b$ are identity or inverse, the latter two can be eliminated for the reason that they would make it abelian as we get $aba=a^2b=e$, with left and right a multiplication on $ab=a^{-1}$. This gives us that $ab=ba$ for the elements which clearly doesn't work as it is a non-abelian group and the remainder would be the same (with the inverses etc), so there must exist some other element then. So $|G|\geq 6$
For small groups follow the strategy that is laid out here and complies with the very basic axioms of a group: A group with five elements is Abelian. (Scroll a bit down to see my solution, that also works for groups of order 2, 3 and 4.) You will not need the fact that groups of prime order are cyclic (hence abelian).