Is a group of order 5 always abelian

Yes. Let G be any group of order 5, and let x be any element of G that isn't the identity. Consider the subgroup H of G generated by the element x (ie, the set {x^n: n is an integer}). By Lagrange's theorem, the order of H has to divide 5, and since 5 is a prime number this tells you the order of H has to be 1 or 5. But H has at least the two elements (the identity) and x, so the order of H cannot be 1. So the order of H is 5 and hence H = G. This shows that G is abelian because G is cyclic. (Any two elements of G are integer powers of x, and integer powers of x commute with one another.)

The same argument shows that more generally any group of prime order is cyclic, for the same reason as above (replace 5 in the paragraph with an arbitrary prime number p). And in fact it shows slightly more than that: it shows that *any* non-identity element of a group of prime order, is a generator of that group.

Every group finite of order prime, is abelian.

This is a theorem.