It is well-known that any rational number can be represented using a sum of distinct Egyptian fractions (that is, rational fractions of the form $1/n$ with $n\in\mathbb{N}$). This may be proven by establishing a greedy algorithm that constructs a sequence of such decompositions. For instance, \begin{align} 1 &= \frac12+\frac13+\frac16 \\ &= \frac12+\frac13+\frac17+\frac1{42}\\ &= \frac12+\frac13+\frac17+\frac1{43}+\frac1{1806}\\ &= \cdots\\&=\sum_{k=1}^\infty \frac{1}{x_k} \tag{1} \end{align} where $x_1=2$ and $x_k = 1+\prod_{j=1}^{k-1}x_j$ for $k>1$ (i.e. Sylvester's sequence A000058).

As noted by Bergner & Walker [1], such Egyptian fractions are relevant in the context of groupoid cardinality. To briefly review: Given a groupoid $G$, the groupoid cardinality is defined [2] by $|G|=\sum_{[\bullet]\in G}\frac{1}{\#\text{Aut}(\bullet)}$ where $[\bullet]$ is a component of $G$ and $\#\text{Aut}(\bullet)$ is the order of the automorphism group of $\bullet$. If $G$ is a group, then this reduces to $|G| = \frac{1}{\# G}$, and if we form the groupoid $G\coprod H$ as a direct union of groups $G,H$ we have $\textstyle |G\coprod H| =\frac{1}{\#G}+\frac{1}{\#H}$.

Here the link to Egyptian fractions arises: Since every rational number can be represented in terms of Egyptian fractions, so too can every rational fraction be obtained as the cardinality of some groupoid. As a key example, since $|\mathbb{Z}/n|=1/n$ the representations in $(1)$ imply $$1 = |\mathbb{Z}/2\coprod \mathbb{Z}/2|=|\mathbb{Z}/2\coprod \mathbb{Z}/3\coprod \mathbb{Z}/6|=\cdots.$$

But this construction, while valid, is quite artificial. In particular, it is far from obvious to me why each successive groupoid has unit cardinality. This suggests the following question: Is there a natural sequence of groupoids $\{G_k\}$, each of unit cardinality, which generate $(1)$?


[1] John Baez, James Dolan, "From Finite Sets to Feynman Diagrams" (arXiv)

[2] Julia Bergner, Christopher Walker, "Groupoid Cardinality and Egyptian Fractions" (JSTOR)

  • $\begingroup$ Your definition of Egyptian fractions omits the usual requirement that the cardinalities be distinct, without which the representation theorem would be trivial (any $m/n$ is the sum of $m$ fractions $1/n$). Still in your context you might need to repeat some fractions. $\endgroup$ Commented May 31, 2016 at 22:59
  • $\begingroup$ @NoamD.Elkies Edited to include the distinctness condition. $\endgroup$ Commented May 31, 2016 at 23:08
  • 2
    $\begingroup$ I think it's going to be difficult to find a satisfactory answer, because it would depend on it looking natural to you, and that has no precise mathematical meaning. $\endgroup$ Commented May 31, 2016 at 23:59
  • $\begingroup$ Well, my main reason for using the word 'natural' is because the construction I gave above is so artificial: it's engineered to give the right arithmetic. Hence I was holding out hope that some higher-minded approach would exist---say, a groupoidification of Sylvester's sequence that would serve as a non-arithmetic proof of the Egyptian fraction reps. But that was always an ambitious speculation, and it would hardly shock me if it was wrong. @FernandoMuro $\endgroup$ Commented Jun 1, 2016 at 13:22

1 Answer 1


First a notational issue: you shouldn't write $G$ for the one-object groupoid corresponding to $G$. A much better name for this groupoid is $BG$, or $\text{pt} / G$.

A natural way to write down a groupoid whose groupoid cardinality is $1$ is to write down a groupoid / homotopy quotient $X/G$ where $|X| = |G|$. In turn, a natural class of such quotients are the adjoint quotients $G/G$ where $G$ acts on itself by conjugation. This is equivalent to the groupoid

$$\coprod_{[g]} BC_G([g])$$

where $C_G(-)$ denotes the centralizer and the coproduct runs over the conjugacy classes of $G$. We get a potential Egyptian fraction representation

$$1 = \sum_{[g]} \frac{1}{|C_G([g])|}$$

if the sizes of the centralizers are distinct. The largest $n$ such that $\frac{1}{n}$ appears in this sum will always come from the conjugacy class of the identity, where $C_G(e) = G$, so $n = |G|$, and everything else will divide $|G|$. This identity is just the usual class equation divided by $|G|$.

To get the first sum this way we want a group of order $6$ with $3$ conjugacy classes having centralizers of orders $2, 3, 6$, or equivalently $3$ conjugacy classes of sizes $3, 2, 1$. There is a unique such group, namely $D_3$.

To get the second sum this way we want a group of order $42$ with $4$ conjugacy classes having centralizers of orders $2, 3, 7, 42$, or equivalently $4$ conjugacy classes of sizes $21, 14, 6, 1$. Such a group necessarily has elements of orders $2, 3, 7$, so these must be the non-identity conjugacy classes (in the same order as before, since everything centralizes itself). There are $6$ elements of order $7$, so the Sylow $7$-subgroup $C_7$ is normal, and so by Schur-Zassenhaus our group $G$ must be a semidirect product $C_7 \rtimes H$ where $|H| = 6$. We must have $H = D_3$ or else $G$ has an element of order $6$, so $G$ must be the semidirect product $C_7 \rtimes D_3$. There is a unique nontrivial such semidirect product because the only nontrivial action of $S_3$ on $C_7$ is the one where $\sigma \in S_3$ acts by $\text{sgn}(\sigma)$; unfortunately, this group has an element of order $21$. So in fact no group with the desired condition on conjugacy classes exists.

Nevertheless, the second sum is still related to some geometry, specifically to the geometry of the (2,3,7) triangle group, or rather to the orbifold quotient of the upper half plane by this group. I don't know about the third sum, though.

  • 1
    $\begingroup$ Thanks for the answer! I should note that, in principle, I'm interested in not just the first 'Egyptian sums' that I listed but the entire family generated by Sylvester's sequence (including the infinite series). But the construction you offered doesn't inspire much hope that this ambition was a sensible one---which, then, so much the worse for that ambition. $\endgroup$ Commented Jun 1, 2016 at 13:13
  • $\begingroup$ After looking up some things, I think no further examples of the kind I desired can be constructed in the above manner; the number of conjugacy classes required is simply too small. For example, in the five-term case one needs a group of order 1806, but the largest finite group with five conjugacy classes is $A_5$ with 60 elements. (Presumably one can give a general result of this kind, but I don't know enough to do so.) $\endgroup$ Commented Jun 2, 2016 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.