Here is a
proof that any two
finite fields of the same order
are isomorphic
^{1}. (The proof in
Haggi's writeup is plainly
bogus
^{2}.)
Let F and F' be two finite fields
of the same order. These fields must have
characteristic p, for a prime p, that is
they contain Z_{p}
as a subfield. I claim that the order of F is p^{n}, for some n.
For, if we choose e_{1},...,e_{n} a basis
for F over Z_{p} then
writing a typical element as a linear combination of these
basis elements there are p possibilities for each of
the n coefficients, making p^{n} elements
in all.
We know that the group of units of F is cyclic (see
a proof that the multiplicative group of a finite field is cyclic).
Let a in F be some generator. Then clearly
F=Z_{p}(a) is the simple field extension generated by
a. Now let f(x) be the minimal polynomial of
a over Z_{p}. By the first isomorphism theorem
F is isomorphic to Z_{p}[x] / (f(x)).
So our aim is to show that F' has the same property.
I claim that the elements of F' are exactly the roots of
the polynomial h(x)=x^{q}x, where q=p^{n}.
To see this, consider an element of the field that is not zero.
It is an element of the group of units so if we raise it to
the power p^{n}1 we must get 1.
The claim is now clear.
Now factorise h(x) into a product of irreducibles. Since
our original a is a root of h(x) it must be that
f(x) is one of the irreducible factors of h(x).
Choose b in F' that is a root of f(x).
Then, again by the first isomorphism theorem, we have that
Z(b) is isomorphic to
Z_{p}[x] / (f(x)). This
subfield of F' therefore has as many elements as F (and
so F') and so this subfield equals F'. We are done.
On the other hand, in order to construct a finite field of
order p^{n} create a splitting field k
of h(x) (as above) over Z_{p}. One can show easily that the zeroes
of h(x) form a subfield of k. The only tricky part is
that if a,b are zeroes then so is ab, but this follows because the appropriate binomial coefficients are divisible by p (cf Frobenius endomorphism). Since this polynomial
is separable (it shares no roots with its derivative) these
zeroes are distinct. So this subfield is a field with p^{n} elements.

A much simpler proof but using deeper technology is to use that a finite
field is a splitting field for h(x)
over Z_{p} (this was shown in the proof above)
and then appeal to the fact that splitting fields are unique.

The problem with Haggis's argument is that he just shows that there
is a vector space isomorphism between the two fields. His "argument" that
you can choose such a vector space isomorphism that preserves multiplication
is unconvincing.