Element structure of groups of prime-fourth order

Particular cases
The prime $$p = 2$$ and $$p = 3$$ behave somewhat differently from the other primes.

Grouping by conjugacy class sizes
Because of the small order, it turns out that the nilpotency class completely determines the number of conjugacy classes of each size in terms of the underlying prime.

In particular, we have:

Grouping by cumulative conjugacy class sizes (number of elements)
Note that it is true in this case that the number of elements in conjugacy classes of size dividing any number itself divides the order of the group (in particular, all these numbers are powers of $$p$$). However, this is not true for all groups and in fact an analogous statement fails for groups of prime-sixth order (see element structure of groups of prime-sixth order). For more, see:


 * All cumulative conjugacy class size statistics values divide the order of the group for groups up to prime-fifth order
 * There exist groups of prime-sixth order in which the cumulative conjugacy class size statistics values do not divide the order of the group

Correspondence between conjugacy class sizes and degrees of irreducible representations
For groups of order $$p^4$$, it is also true that the nilpotency class determines the degrees of irreducible representations in terms of the underlying prime.

In fact, the nilpotency class, conjugacy class size statistics, and degrees of irreducible representations all determine each other. Knowing any one of these determines the other two once we fix the underlying prime. Below is the correspondence:

Note that the phenomenon of the conjugacy class size statistics and degrees of irreducible representations determining one another is not true for all orders:


 * Degrees of irreducible representations need not determine conjugacy class size statistics
 * Conjugacy class size statistics need not determine degrees of irreducible representations

1-isomorphism
We say that two groups are 1-isomorphic groups if there is a bijection between them that restricts to an isomorphism on every cyclic subgroup of either side.

Case $$p$$ at least 5
In this case, all groups of order $$p^4$$ are 1-isomorphic to abelian groups via the Lazard correspondence. This is because all the groups have nilpotency class at most three, which is less than the underlying prime, and hence the Lazard correspondence applies.

Further, the nature of the Lazard correspondence is similar for all choices of $$p$$. More detailed information is presented in the table below:

Case $$p = 3$$
Of the 15 groups of order 81, 5 are abelian, 6 have nilpotency class two, and 4 have nilpotency class three. Via the Baer correspondence, each of the groups of class two has a Baer Lie ring, and in particular is 1-isomorphic to the additive group of that Lie ring. Of the 4 groups of nilpotency class three, only one (SmallGroup(81,8)) is 1-isomorphic to an abelian group. There are no 1-isomorphisms between pairs where both members are non-abelian.

Case $$p = 2$$
Of the 14 groups of order 16, 10 are not 1-isomorphic to any other group. The remaining four come in pairs. These two pairs are described below.

Case $$p$$ at least 5
For groups of order $$p^4$$ with $$p \ge 5$$, the equivalence classes of groups based on order statistics coincide precisely with the equivalence classes based on 1-isomorphism, and every equivalence class contains exactly one abelian group (see the section ). The convention for order statistics is as follows: the comma-separated values give the number of elements of order 1, $$p$$, $$p^2$$, $$p^3$$, and $$p^4$$ respectively. For cumulative order statistics, these are the number of elements of order dividing 1, $$p$$, $$p^2$$, $$p^3$$, and $$p^4$$ respectively::

Case $$p = 3$$
Here, we discuss the equivalence classes of groups of order 81 up to being order statistics-equivalent finite groups and up to the stronger notion of being 1-isomorphic groups (which means there is a bijection that restricts to isomorphisms on cyclic subgroups). See also order statistics-equivalent not implies 1-isomorphic.

Case $$p = 2$$
Note that for groups of order $$2^4 = 16$$, not all groups are order statistics-equivalent to an abelian group, and even for those that are, the group need not be 1-isomorphic to an abelian group. Here, we discuss the equivalence classes of groups of order 16 up to being order statistics-equivalent finite groups and up to the stronger notion of being 1-isomorphic groups (which means there is a bijection that restricts to isomorphisms on cyclic subgroups). See also order statistics-equivalent not implies 1-isomorphic.

Here are the GAP commands to produce a list sorted by order statistics:

gap> F := List(AllSmallGroups(16),G -> List(Set(G),Order));; gap> K := List(F,L->[Length(Filtered(L,x -> x = 1)), > Length(Filtered(L,x -> x = 2)),Length(Filtered(L,x -> x = 4)), > Length(Filtered(L,x -> x = 8)),Length(Filtered(L,x -> x = 16))]);; gap> M := List([1..14], i ->[K[i],i]);; gap> S := SortedList(M);

Here is GAP's output (note that this does not distinguish based on 1-isomorphism):

[ [ [ 1, 1, 2, 4, 8 ], 1 ], [ [ 1, 1, 10, 4, 0 ], 9 ], [ [ 1, 3, 4, 8, 0 ], 5 ], [ [ 1, 3, 4, 8, 0 ], 6 ], [ [ 1, 3, 12, 0, 0 ], 2 ], [ [ 1, 3, 12, 0, 0 ], 4 ], [ [ 1, 3, 12, 0, 0 ], 12 ], [ [ 1, 5, 6, 4, 0 ], 8 ], [ [ 1, 7, 8, 0, 0 ], 3 ], [ [ 1, 7, 8, 0, 0 ], 10 ],  [ [ 1, 7, 8, 0, 0 ], 13 ], [ [ 1, 9, 2, 4, 0 ], 7 ], [ [ 1, 11, 4, 0, 0 ], 11 ], [ [ 1, 15, 0, 0, 0 ], 14 ] ]