The set of non-zero elements of a finite field of order #q# form a cyclic group of order #q-1# under multiplication. If #q > 2# then this cyclic group has at least one non-identity element #alpha# that generates it.
Choose #alpha in F_16# that generates all the #15# non-zero elements.
Then #alpha^5# generates the cyclic subgroup of order #3# corresponding to the subfield isomorphic to #F_4#.
#alpha^3# generates the subgroup of order #5#, but #5+1 = 6# is not a power of #2#, so there is no corresponding subfield.
#alpha^15 = 1# generates the subgroup of order #1#, corresponding to the subfield isomorphic to #F_2#.
In general, #F_(p^n)# has subfields isomorphic to #F_(p^k)# where #k# is a divisor of #n#.
