That Induction Proof N(n,k)
I stated that the formula for the number of additive subgroups of the additive group of a finite field was given by the formula below.
Where 'n' is the degree of the cartesian product of the prime subfield of F=GF(p,n) isomorphic to F+ and k is the dimension of the kfold cartesian product of subgroups isomorphic to the prime subfield of F (which in a kfold cartesian product form the "group" H in question). The number of such subgroups H of F+ that is given by the formula is exact.
We proceed by induction on k for a given n. I know the case k=1 is correct, as the number of additive groups is simply


(1.1) 
Since each group is disjoint in elements except for the zero.
Now we start assuming that the case k1 is correct and we obtain the desired result after a few calculations.
So, given a group H of order p^{k1} in G (of order p^{n}) every element not in H may extend the group to one of higher degree. These elements partition into their subgroups; as the extensions that may be possible are similar for every element of each porder subgroup not in H. So the number of ways I may extend H to some group in G; (each a subgroup K_{i}) of order p^{k} is in part given by the formula:


(1.2) 
Which simplifies to:


(1.3) 
Yet each one of these possible extensions are not unique: for in the extension of order p^{k} there are some few elements in every K_{i} which generate this group K_{i} not in H, i.e. these elements are in the set K \ H, (otherwise stated as K  H) the number of extensions formed is given by the number of the disjoint pgroups in K \ H which is given by:


(1.4) 
Now, this is the number of p^{k} subgroups generated by each of the p^{k1} subgroups. Yet this is too many as each extension to a p^{k} group is made by more than one p^{k1} group.
So; (1.2) and (1.4) together simplify the count to:


(1.5) 
And this gives us the number of individual groups K_{i} of order p^{k} generated by each of (but not unique to) each p^{k1} subgroup H extended to a K_{i}, one of order p^{k} (a subgroup isomorphic to K). Yet each p^{k} group or K_{i} may very well be generated by several of the p^{k1} groups isomorphic to H from the full collection of N(n, k1). Fortunately, the number of p^{k1} groups in each p^{k} group K_{i} is the same as the number of unique factors of order p in each p^{k} group (see * at bottom). Then, by applying the p^{k} case, the number of unique subgroups is given by


(1.6) 
And this is the multiple upon the number of such H; (our formula N(n, k1)) which is truly the number of p^{k} subgroups of F.
Thus:


(1.7) 
And by counting backward to the case k=1


(1.8) 
Which was what was wanted.
*How could this be otherwise? Given the formula for the number of subgroups N(n, k) it is truly symmetrical in that N(n, k) = N(n, nk). Indeed we simply apply the formula in the lesser case where N(k, 1) = N(k, k1). Our induction hypothesis can be modified to account for this! We may state that for all n > 0, and subsequently all 0 < k < n the induction hypothesis shows that the formula N(n, k) is exact.
Then to prove p(n) we assume only p(k) for 0 < k < n; and the number of subgroups are calculated by simply "counting N(n, k) backwards" from k to 1 as in equation (1.7).
(The case N(n, n) = 1 is trivial, and so the induction hypothesis suffers no trouble.)
Thus, the number of p^{k1} groups that generate the very same p^{k} group is the same as the number of pgroup factors (my Z) in the p^{k} group. The converse, that if two p^{k1} groups are equal and the pgroup factor changes is simply given by the formula (1.4)
Do we have equality from (1.7)? We could state that:


(1.9) 
Yet we do not rely on each p^{k} group in N(n,k) being uniquely generated by one p^{k1} group of N(n, k1). Is it reasonable then to reduce the number by induction down from k=n1 to k=1? I state that yes, it is and this is true by induction, as we are not finding the number of unique groups generated by each p^{k1} group H of the N(n, k1) but are calculating the number of unique subgroups in the extension of the base pgroup as a whole. The multiple is properly a “commensurate” or “ration” and is not a formula for the number of “unique” p^{k} groups so generated from each p^{k1} group.
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