# The sum of the orders of all elements of a group G

Let $G$ be a finite group and denote $k(G)$ the sum of the orders of all the elements of the group $G$. I have to determine $\min(k(G))$ and $\max (k(G))$ when $G$ goes through the set of the abelians groups of order $n$, where $n$ is a fixed non-negative integer. Any ideas? Thank you!

• I should think the extreme cases would be the cyclic group on one side, and the direct sum of prime-order groups on the other. – Gerry Myerson Feb 21 '14 at 21:59
• For a cyclic group of order $n$, $k(Z)=\sum_{d|n} d\varphi(d)$. – Nicky Hekster Feb 21 '14 at 22:08
• Are the orders counted with multiplicity or aren't they? – anon Feb 21 '14 at 22:58

Use the classification of finite abelian groups. Write $G=\bigoplus_{p\mid n}G_p$ where $G_p$ is a group of order $p^\nu$ where $p^\nu$ is the maximum power of $p$ dividing $n$. Show that $k(G)=\prod k(G_p)$. So this leaves us with the case that $n=p^\nu$ is a prime power. Show that in this case the minimum $(p^\nu-1)\cdot p+1$ is attained by $(\mathbb Z/p\mathbb Z)^\nu$ and the maximum $$1+(p-1)p+(p^2-p)p^2+\ldots + (p^\nu-p^{\nu-1})p^\nu =1+(p-1)(p+p^3+\ldots+p^{2\nu-1})=\frac{p^{2\nu+1}+1}{p+1}$$ by $\mathbb Z/p^\nu\mathbb Z$
In [1], you can find that $k(\mathbb Z_n)\ge k(G)$ for any group group of order $n$. So $\max k(G)$ is found. For $\min k(G)$, Hagen give you a clue.