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!
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?
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).
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 $|G|\geq 6$
