Ordering finite groups by sum of order of elements

Given two finite groups $G, H$, we are going to say that $G<_oH$ if either

a. $|G|<|H|$

or

b. $|G|=|H|$ and $\displaystyle\sum_{g\in G} o(g)<\sum_{h\in H} o(h)$,

where $o(g)$ denotes the order of the element $g$ (has this ordering a name?).

What is the smallest example (in this ordering) of a pair of nonisomorphic groups such that $G$ and $H$ are incomparable, i.e., such that they have same cardinal and same sum of orders of elements?

Solution 1:

The following Maple function (code here requires the Maple package GroupTheory) takes as argument a group and returns the sum of the orders of its elements:

sumOfOrders := G -> add(u, u in map(PermOrder, convert(Elements(G), list), G));

This function takes as argument a positive integer $n$ and returns the multiset of sums of orders of elements of groups of order $n$:

sumOfOrdersList := n -> sort(map(sumOfOrders, AllSmallGroups(n)));

Checking the lowest orders manually we find that groups incomparable in your sense occur first in order 16: Executing sumOfOrdersList(16) shows there are three groups with sum $47$, three with sum $55$ and two with $87$.

(Cf. https://groupprops.subwiki.org/wiki/Groups_of_order_16 )

Solution 2:

Feed the following code to Magma(http://magma.maths.usyd.edu.au/calc/):

for grouporder in [1..24] do
  printf "Checking sum of order for groups or order %o:", grouporder;
  numgroup := NumberOfSmallGroups(grouporder);

  sset := [];
  for i in [1..numgroup] do
    sumorder := 0;
    G := SmallGroup(grouporder,i);
    for j in G do
      sumorder := sumorder + Order(j);
    end for;
    Append(~sset, sumorder);
  end for;

  sset;
  printf "\n";
end for;

You can see that there are two groups with "signature" $(16,47)$. Much thanks to Travis for pointing it out. Two of them are non-abelian; they have GAP id $(16,3)$ and $(16,13)$. Please refer the groupprop page for the description. The third is $\mathbb Z/4\mathbb Z \times (\mathbb Z/2\mathbb Z)^2$. All have 1 element of order 1, 7 elements of order 2, and 8 elements of order 4.

Also, there are three groups with "signature" $(16,55)$.

The groups are the following:

$G_1 = (\mathbb Z/4\mathbb Z)^2$; $G_2 = \langle a, b \ | \ a^4 = b^4 = 1, ba = ab^3\rangle$ (the semi-direct product of two copies of $\mathbb Z/4\mathbb Z$), $G_3 = Q \times \mathbb Z/2\mathbb Z$, where $Q$ is the quaternion group. All have 1 element of order 1, 3 elements of order 2, and 12 elements of order 4.

Edit: This is copied from Travis' comment below. There are two groups of signature $(16,87)$ as well. One of them is the abelian group $\mathbb Z/8\mathbb Z \times \mathbb Z/2\mathbb Z$, and the other is the non-abelian group with GAP ID $(16, 6)$ or is also called $M_4(2)$. They both have 1, 3, 4, and 8 elements of order 1, 2, 4, and 8, respectively.