I am trying to make sense of this proposition. I am fine with part (a), for part (b) however, can you explain what the computation proves? Can you not verify a homomorphism by checking the 3 standard properties? I have had no problem showing H is a field with p elements by using the fundamental homomorphism theorem. However, I am having problems understanding why we necessarily have a bijection?
On
Put another way: $K$ is a vector space over the field $H$. It must be finite-dimensional, since $K$ is finite, and a basis is a subset of $K$.
So say the dimension of $K$ as a vector space over $H$ is $r$. Then $K$ is isomorphic to $H^r$ (which is also a $k$-dimensional vector space over $H$). This may be the part you don't get: that any two finite-dimensional vector spaces over a field $F$ are isomorphic as vector spaces (alternatively, any linear mapping that takes basis to basis is invertible).
As isomorphisms are bijections, this shows that $|K| = |H^r| = |H|^r = p^r$.
To see how this actually plays out, try low-degree extensions of $\Bbb Z_2, \Bbb Z_3$ or $\Bbb Z_5$.
The computation is required by the definition of the homomorphism: basically it is defined first for positive integers, then extended to negative integers. So for formulae involving two integers $m$ and $n$, one has to examine the different combinations. You meet the same problem in mid school on extending the exponent notation from natural numbers to integers.
The bijection comes from the very definition of a basis.