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For any finite group $G$, let $$\theta(G) := \sum_{g \in G} \frac{o(g)}{\phi(o(g))},$$ where $o(g)$ denotes the order of the element $g$ in $G$, and where $\phi$ is the Euler totient function.

It is not too hard to see that if $G$ is nilpotent, then $\theta(G)$ is in fact equal to $\sigma(|G|)$, i.e. the sum of the divisors of $|G|$. However, it seems that $\theta(G)$ is always less than or equal to $\sigma(|G|)$, and that equality holds if and only if $G$ is nilpotent.

My question is twofold: (1) Is this claim true? (2) What kind of "natural" properties of groups (such as nilpotency) are there that can be checked by only looking at the orders of the elements of the group?

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2 Answers 2

Nilpotency seems to be the only easy case. John Thompson asked many years ago whether one could tell if a finite group is solvable by the orders of its elements and as far as I know this is still open (equivalently, embed the two groups H and K of order n into G:=Sym(n) via the regular representation. Having the same collection of orders (as a multiset) is exactly saying that the induced characters from the trivial representation of H or K to G are the same.

I think also that one can prove claim (1) at least for G solvable by taking a minimal normal subgroup N of G that is an elementary abelian p-group and showing that this function over a coset gN is at most the same as for a nilpotent group and if some element in the coset has order prime to p, then this will always be less than the answer in a nilpotent group as long as g does not commute with N. This shows that a minimal counterexample would have trivial Fitting subgroup and perhaps one can reduce to the case of simple groups. At the moment, I do not see how to show this inequality for simple groups.

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Welcome to MathOverflow! –  Lubin Feb 11 '11 at 1:52
    
That's interesting, thanks! Since you wrote that "nilpotency seems to be the only easy case", do you mean that there exists some other known formula (or algorithm) to determine whether a group is nilpotent, only depending on the orders of the elements? –  Tom De Medts Feb 13 '11 at 8:36
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@Tom de Medts: You probably know this, but I mention it nonetheless: Let $e_G(k)$ denote the number of elements of order $k$ in $G$. Since $G$ is nilpotent iff it's the direct product of its Sylow subgroup, $G$ is nilpotent iff $e_G$ is multiplicative on coprime numbers. On the other hand, if you know just the set of occuring orders (without multiplicities), then you can't know: For example, the groups $C_6\times C_2$ and $D_{12}=C_6:C_2$ both have elements of orders 1, 2, 3, 6, only one of them being nilpotent. –  Frieder Ladisch Feb 13 '11 at 17:34
    
@F. Ladish: Thanks -- I wasn't aware of this fact! I even have to confess that I don't see right away why the multiplicativity of $e_G$ implies the nilpotency of $G$. Is there a quick and easy argument? –  Tom De Medts Feb 16 '11 at 9:57
    
@Tom de Metds : yes : if $k,l$ are coprime, the set $P_{kl}$ of elements of order $kl$ is in bijection with the subset of $P_k\times P_l$ consisting of commuting pairs. So the multiplicativity condition says that all such pairs commute, hence the group is the direct product of its Sylows. –  BS. Feb 16 '11 at 16:25

(I post this as an answer, since I don't want to enter the below math into the comment box. It answers Tom de Medt's question on my comment to Robert Guralnick's answer.)
Nilpotency can be checked by looking only at the orders of elements: Write $e_G(k)$ for the number of elements of order $k$ in $G$. As I remarked in that comment, nilpotency of $G$ implies multiplicity of $e_G$ on coprime elements, since $G$ is the direct product of the Sylow subgroups.
Indeed, one can make somewhat more precise statements: Let $p$ be a prime and $|G|_p$ the $p$-part of $|G|$, and note that
$$ \sum_{i\geq 0} e_G(p^i) \geq | G |_p ,$$ with equality if and only if there is only one Sylow $p$-subgroup of $G$. So one can tell from the orders whether $G$ has a normal Sylow $p$-subgroup. $G$ is nilpotent iff all Sylow subgroups are normal.
Looking at the multiplicity of $e_G$ was perhaps more complicated than necessary, but here is a justification of the assertion from my comment: If $e_G$ is multiplicative, then $$ |G| = \prod_{p } |G|_p \leq \prod_{p } \sum_{i\geq 0} e_G(p^i) = \sum_{k\geq 0} e_G(k) = |G|, $$ and thus the "$\leq$" must be a "$=$", which again implies all Sylow $p$-subgroups are normal.

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That's quick and easy indeed, thanks a lot :) –  Tom De Medts Feb 17 '11 at 10:07

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