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?
Suppose $G$ is a finite group and I know for every $k \leq |G|$ that exactly $n_k$ elements in $G$ have order $k$. Do I know what the group is? Is there a counterexample where two groups $G$ and $H$ have the same number of elements for each order, but $G$ is not isomorphic to $H$? I suspect that there is, but I haven't thought of one.
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