Sums of orders of elements of finite groups

Question

Suppose $G$ is a finite group. Then let’s define $s(G) := \Sigma_{g \in G} ord(g)$ as the sum of orders of all elements of $G$.

My question is:

Suppose $G$ and $H$ are two finite groups, such that $|H| = |G|$ and $s(H) = s(G)$. Does that always mean that $G \cong H$?

This statement seems to be true for groups of order up to $12$

For $1$, $2$, $3$, $5$, $7$ and $11$ this is obvious as there is only one isomorphism class of groups of each of these orders.

For $4$: $$s(C_4)=11$$ $$s(C_2 \times C_2) = 7$$

For $6$: $$s(C_6) = 21$$ $$s(S_3) = 13$$

For $8$: $$s(C_8) = 43$$ $$s(C_4 \times C_2) = 23$$ $$s(C_2\times C_2\times C_2 ) = 15$$ $$s(D_4) = 19$$ $$s(Q_8) = 27$$

For $9$: $$s(C_9) = 61$$ $$s(C_3 \times C_3) = 25$$

For $10$: $$s(C_{10}) = 63$$ $$s(D_5) = 31$$

But is it true in general?

Answer 1

You cannot distinguish between finite groups using just orders of elements. For example, $\mathtt{SmallGroup}(16,5)$ and $\mathtt{SmallGroup}(16,6)$ both have 1,3,4 and 8 elements of orders 1,2,4,8, respectively.

These groups are respectively $C_8 \times C_2$ and $\langle x,y \mid x^8=y^2=1, y^{-1}xy=x^5 \rangle$.