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Thinking along the lines of Tom Leinster's fascinating recent question, I'm wondering more generally about how to extend questions about natural numbers to groups, with the cyclic groups representing the natural numbers, and normal subgroups representing divisors. For example, let $\mathbb{G}$ be the set of all isomorphism classes of finite groups. Then an "arithmetic function" could be defined to be simply a function $f:\mathbb{G}\rightarrow\mathbb{C}$. Here are a few analogs I wrote down rather quickly (so perhaps there are better proposals for generalizations than these): $$\text{id}(G)=|G|\quad\quad\quad \epsilon(G)=\begin{cases}1\text{ if $G$ is trivial}\\\ 0\text{ otherwise}\end{cases}\quad\quad\quad z(G)=0$$ $$\sigma_k(G)=\sum_{N\\,\triangleleft \\,G} |N|^k \quad\quad\quad\quad\phi(G)=|G|\prod_{\substack{N\\,\triangleleft \\,G \\\ |N|\text{ prime}}}\left(1-\frac{1}{|N|}\right)$$ Tom Leinster's question is whether there is a solution to $\sigma_1(G)=2|G|$.

Question 1: What is known, if anything, about these functions? What is a good proposal for an analog of the Mobius function (I couldn't think of one offhand)? Can anyone demonstrate that they satisfy formulas that are analogs of their natural-number counterparts, or perhaps instead give examples that would indicate that these functions act weird and aren't nice generalizations to make?

I suspect that if these functions act badly, the most likely fix would be to redefine $\mathbb{G}$ to be isomorphism classes of finite abelian groups. All subgroups are normal, and the structure theorem makes things much more controlled. I would guess that defining (for example) what it means for two groups to be "coprime", and hence what it means for an "arithmetic function" to be "multiplicative", would go over much more smoothly with abelian groups.

Going further: given two "arithmetic functions" $f,g:\mathbb{G}\rightarrow\mathbb{C}$, we can define a "Dirichlet convolution" by $$(f\ast g)(G)=\sum_{N\\,\triangleleft \\,G}f(N)g(G/N)$$ I've got to say, my jaw dropped a bit when I wrote that down. But one immediate difference I can see, somewhat discouraging, is that $\ast$ would not be abelian, since we aren't guaranteed that there are any normal subgroups $M\triangleleft G$ isomorphic to $G/N$ (see this MO question), much less that if $M\cong\\! G/N$ then $G/M\cong\\!\\! N$. However, the function $\epsilon$ is still a left and right identity for $\ast$, and $z$ is still a zero.

Question 2: Can anyone prove or disprove that $\ast$ is associative? If it is, we at least get a non-commutative ring under $\ast$ and pointwise addition (that $\ast$ distributes over pointwise sums is obvious).

Now I suspect I am really getting "greedy" with my generalizing. We could further define "Dirichlet series", such as the zeta function (note that this is not the same thing as the zeta function of a group): $$\zeta_{\mathbb{G}}(s)=\sum_{G\in\mathbb{G}}\frac{1}{|G|^s}=\lim_{n\rightarrow\infty}\sum_{\substack{G\in\mathbb{G}\\\ |G|\leq n}}\frac{1}{|G|^s}$$ I seriously doubt there is any hope for an Euler product-like expression. But perhaps, if we restricted ourselves to finite abelian groups...

Also, I'm familiar with the result that the number of isomorphism classes of groups of order $p^n$ grows as $p^{\frac{2}{27}n^3+O(n^{8/3})}$, and I imagine the growth rate is at least as bad for not-necessarily $p$-groups.

Question 3: Is there any $s>0$ for which $\zeta_{\mathbb{G}}(s)$ converges?

I wrote this question rather fast, and I welcome any feedback on how to improve it, or make it more appropriate for MO. Should I break this up into multiple questions? Is it too open-ended?

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Thanks for the "fascinating", Zev. Re Q2, I think it's associative if you restrict to abelian groups, but not in general, ultimately because a normal subgroup of a normal subgroup need not be normal. (At the moment I'm a bit too lazy to check it properly and type out the details.) Re coprimality, I think the appropriate definition is that two finite groups are coprime if they have no composition factor in common: see section 3 of arxiv.org/abs/math.GR/0104012 –  Tom Leinster Feb 9 '11 at 10:03
...the point about coprimality being that if G and H are coprime in this sense, then every normal subgroup of GxH is (uniquely) of the form XxY where X is normal in G and Y is normal in H. This is like the fact that if m and n are coprime then every divisor of mn is uniquely of the form xy where x|m and y|n. –  Tom Leinster Feb 9 '11 at 10:06
Re Question 2: not associative: check $A_4$. As Tom Leinster remarked, the problems are subnormal subgroups, that aren't normal: It's easy to see that in $((f\ast g)\ast h)(A_4)$ there occurs a summand $f(C_2)g(C_2)h(C_3)$, which doesn't occur in $(f\ast (g\ast h) )(A_4)$. –  Frieder Ladisch Feb 9 '11 at 10:24
The combinatorialists have a well-developed theory of Mobius functions on posets, maybe that could come over to this setting. –  Gerry Myerson Feb 9 '11 at 11:56
Some rough ideas regarding the abelian case of Q3 (may contain some mistakes). If f(n) is the number of abelian groups of order n, then f is multiplicative, and for p prime f(p^n) is the partition function of n. This is closely related to en.wikipedia.org/wiki/Multiplicative_partition (but not equal). Your zeta function thus converges for $s > \sqrt{2}$. Formally (convergence?) you also get the Euler product : $\sum_{n = 1}^{\infty} \frac{f(n)}{n^s} = \prod_{p} \prod_{k=1}^{\infty} (1-p^{-s})^{-1}$ –  Joël Cohen Apr 27 '11 at 20:49

3 Answers 3

Q1. My other answer shows that if we want $\ast$ to be associative, we'd better do something like restrict to abelian groups. So let's do that.

In that context, there is indeed a good analogue of the Möbius function. It's the function $\mu$ given by $$ \mu(A) = \sum_{k = 0}^\infty (-1)^k c_k(A) $$ (for an abelian group $A$), where $c_k(A)$ is the number of chains $$ 1 = A_0 < A_1 < \cdots < A_k = A $$ of proper subgroups.

Why is this a good analogue? Well, write $\zeta$ for the function with constant value 1. The crucial property of the Möbius function is that it's the inverse, with respect to $\ast$, of $\zeta$: $$ \mu = \zeta^{-1}. $$ And this is easily verified by a telescoping sum argument.

Incidentally, I guessed this formula for $\mu$ because something extremely similar is true for Möbius inversion for posets, and more generally categories. I haven't figured out yet whether this is an instance of general results about Möbius inversion for categories.

Edit: This construction appears to be due to Philip Hall: see Mike Spivey's answer.

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Q2: In response to Greg Kuperberg's complaint about answering questions in comments, I'll combine the semi-answer in the comment above with F. Ladisch's specific example and put it here as a proper answer. As it's only half my own work, I'll make it community wiki.

For arbitrary f, g and h, we have $$ ((f \ast g) \ast h)(G) = \sum f(K) g(N/K) h(G/N) $$ where the sum is over all $K \triangleleft N \triangleleft G$, and $$ (f \ast (g \ast h))(G) = \sum f(K) g(N/K) h(G/N) $$ where the sum is over all $K \triangleleft G$ and $N \triangleleft G$ with $K \subseteq N$.

Take f, g and h all to have constant value 1. Then $((f \ast g) \ast h)(G)$ is the number of chains $K \leq N \leq G$ with K normal in N and N normal in G, whereas $(f \ast (g \ast h))(G)$ is the number of chains $K \leq N \leq G$ with K normal in G and N normal in G. The former is greater than or equal to the latter, and in some cases strictly greater: taking F. Ladisch's example, we might have $K = C_2$, $N = C_2 \times C_2$, $G = A_4$. So no, $\ast$ is not associative.

On the other hand, $\ast$ is associative when restricted to abelian groups.

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It looks like some of these ideas appear in Philip Hall's paper, "The Eulerian functions of a group," Quart. J. Math. Oxford Ser. 134-151, 1936. He discusses both an Euler $\phi$ function and a Möbius function defined on groups.

The Möbius function is as Tom Leinster describes in his answer to Q1.

The function $\phi_n(G)$ is the total number of distinct $n$-bases of the group $G$, where an $n$-basis is any ordered set of $n$ elements of $G$ which generate $G$. Thus, if $G$ is cyclic of order $m$, $\phi_1(G)$ is the usual Euler function $\phi(m)$.

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Great. I knew that when Rota defined the Möbius function for posets, he wasn't the first to do so (and made this clear in his paper): at least Hall, Ward and Weisner had done something similar before him. I think this is the paper of Hall's he cites. (Still though, it can be argued that the general attribution to Rota of Möbius inversion for posets is correct; he was the first to really run with it.) –  Tom Leinster Feb 11 '11 at 6:36
@Tom: Rota does cite this paper in his "On the foundations of combinatorial theory I: Theory of Möbius functions," which is how I found it. He also cites an earlier paper of Hall's, "A contribution to the theory of groups of prime power order," which has some of the same ideas in a less developed form and only applied to groups of prime power order (as the title indicates). –  Mike Spivey Feb 11 '11 at 16:59

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