None:
Polyps:
Strongs:

 Metamath Divergence So far, we have considered the number of proper subgroups of the additive group of the Galois field of eight elements GF(8). Conveniently enough, these are all seven. (Seven of order two, seven of order four.) there is but one of order one, and one of order eight. For GF(4), there are three order two. So, are the number of proper subgroups of GF(16), GF(32) and those fields with prime subfields of other primes, 3, 5, 7 etc of orders powers of these primes also equal to the order of their multiplicative groups? Well, for example, whilst the number of order two subgroups in GF(16) is fifteen, (and also order eight subgroups, since each corresponds to a factor group of  the image of a morphism with each two-group as the kernel) The number of four groups in GF(16)  is 35. This is not one less than a power of any prime number.   Without giving a proof for the following identity, the number of subgroups N of a given prime power ordered field's additive group GF(p,n)+  of a lesser power k, the number of additive subgroups is given by: (In Product notation, much like sigma summation notation, but using Pi for product, not sum.) The top power of 'p' counts down from 'n' and the bottom power counts up from 1, both by the same degree of 'k'. The minimum number of proper subgroups is given by k=1 and k=n-1, (when all powers except for r=1 cancel) This identity is readily constructed, and therefore this construction as applied to the number of subgroups, is a proof that this identity always returns for N an integer value. The top is always greater than or equal to the denominator. By reducing for a smaller parent group, evaluating a substitution with m=n-1, reveals by induction from the case m=k that N for the larger parent group is greater than (or minimally equal to if m=n ) the N for the smaller. Thus, as n increases, the result N is divergent in a near-exponential fashion For other primes, the number of subgroups may not equal the order of the multiplicative group. for N to equal the nth power of p, we require a denominator of one, with k=1. yet (p-1 = 1) only when p = 2. Thus other primes diverge from the 'closed' set.Continue To Next Page Return To Section Start