#50 - 1 = 7^2#
Any finite field must have #p^n# elements for some prime #p# and positive integer #n#, where #p# is the characteristic. Otherwise it would have zero divisors - which would be non-invertible elements.
In order to distinguish between ordinary integers and elements of the field, I will use #hat(0)# to be the zero (additive identity) of the field and #hat(1)# to denote the one (multiplicative identity) of the field.
Then for any non-negative integer #n# we can write #hat(n)# to denote #overbrace(hat(1)+hat(1)+...+hat(1))^"n terms"#
The characteristic of the field is the smallest positive integer #n# such that #hat(n) = hat(0)#.
Suppose #hat(n) = hat(0)# for some composite number #n = hk#. Then we find:
#hat(h) hat(k) = hat(hk) = hat(n) = hat(0)#
So at least one of #hat(h) = hat(0)# or #hat(k) = hat(0)#.
Hence the characteristic of any finite field is a prime number.
Give any finite field #F#, there is a subfield of #F# generated by #{ hat(1) }#, which is isomorphic to #GF(p)# for some prime #p#.
If this subfield is the whole of #F#, then #F# has order #p#.
Otherwise, there is some element #a in F# which is not in the subfield. This element satisfies:
#overbrace(a+a+...+a)^"p times" = hat(p) a = hat(0) a = hat(0)#
Hence #{ hat(1), a }# generates a subfield of #F# of order #p^2# with elements of the form #hat(m) + hat(n) a#.
If this subfield is the whole of #F#, then #F# has order #p^2#.
Otherwise, there is some element #b in F# which is not in the subfield. Again we find that #b# is of order #p# and hence the subfield generated by #{ hat(1), a, b }# is of order #p^3#.
Continuing in this way we find that any finite field is of order #p^n# for some prime #p# and positive integer #n# and is of characteristic #p#.
