Infinitude of Leinster Groups

Infinitely Many Leinster Groups

Definitions and mathematical statement

Shared notation.

Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

For a finite group $G$, let $\sigma_N(G)=\sum_{H\trianglelefteq G}|H|$, including both the identity subgroup and $G$. A Leinster group satisfies $\sigma_N(G)=2|G|$. Infinitude means infinitely many isomorphism classes of finite groups with this property, not infinitely many differently labelled copies of one group. [S1] [S2]

Statement recorded in the supplied source.

Are there infinitely many Leinster groups? [S1]

Residual task reported by the archive.

The embedded review $2026-08-17$ records: Construct an infinite family or prove only finitely many exist. [S1]

Short English statement

Are there infinitely many finite groups for which the orders of all normal subgroups add up to twice the group’s order?

Research attempt: coprime products and arithmetic obstructions

Status. No infinite family is obtained. The inspected abstract of Leinster's original paper [S2] connects cyclic examples with perfect numbers. We prove that connection and a product criterion, then identify precisely why they do not yield unconditional infinitude.

1. Cyclic groups and a conditional infinite supply

A cyclic group $C_n$ has one subgroup of each divisor order $d\mid n$, all normal. Hence $\sigma_N(C_n)=\sigma(n)=\sum_{d\mid n}d$, and $C_n$ is Leinster exactly when $n$ is a perfect number. If $q=2^p-1$ is prime, then $p>1$, $q$ is odd, and coprime divisor factorization gives \[\sigma(2^{p-1}q)=(1+2+\cdots+2^{p-1})(1+q) =(2^p-1)2^p=2(2^{p-1}q).\] Thus infinitely many Mersenne primes would imply infinitely many pairwise nonisomorphic Leinster groups. The implication is proved; its antecedent is not proved here. The first orders from $p=2,3,5,7$ are $6,28,496,8128$, which the diagnostic verifies by direct divisor sums.

2. Normal subgroup sums multiply under a precise hypothesis

Let $G,H$ have coprime orders. Every subgroup $L\leq G\times H$ is a product. To see this, for $(g,h)\in L$ choose an integer $e$ congruent to 1 modulo $|G|$ and 0 modulo $|H|$. Then $(g,h)^e=(g,1)\in L$; a complementary exponent gives $(1,h)\in L$. Thus $L=L_G\times L_H$. Such a product is normal exactly when each factor is normal, as follows by conjugating separately in each coordinate. Consequently \[\sigma_N(G\times H)=\sigma_N(G)\sigma_N(H),\qquad \rho(G\times H)=\rho(G)\rho(H),\quad \rho(G)=\frac{\sigma_N(G)}{|G|}.\tag{1}\] This supplies an exact construction test: coprime factors must have ratios multiplying to 2. It does not allow repeated multiplication of existing examples; the product of two coprime Leinster groups has ratio 4.

3. Excluding tempting factors

No nontrivial finite $p$-group is Leinster. Its identity subgroup contributes 1 to $\sigma_N$, whereas every other subgroup order is divisible by $p$. Thus $\sigma_N\equiv1\pmod p$, while $2|G|\equiv0\pmod p$. For a nonabelian finite simple group $S$, only $1$ and $S$ are normal, so $\rho(S)=1+1/|S|<2$; it is not itself a solution either. Appending a coprime cyclic prime-order factor multiplies a ratio by $1+1/p>1$, and therefore cannot preserve ratio 2.

The coprimality hypothesis in $1$ is indispensable. For example, $C_p\times C_p$ has $p+1$ subgroups of order $p$, so \[\sigma_N(C_p\times C_p)=1+(p+1)p+p^2=1+p+2p^2, \] whereas $\sigma_N(C_p)^2=(1+p)^2$. Their difference is $p(p-1)>0$. The extra diagonal subgroups invalidate naive multiplicativity when orders overlap.

4. Verification and what remains

The diagnostic checks cyclic perfect orders below 10000, the displayed elementary-abelian calculation for $p=2,3,5,7$, and coprime cyclic multiplicativity for small orders. These tests confirm the arithmetic formulas rather than the existence of infinitely many groups. An unconditional infinite sequence of coprime ratio factorizations, or a construction controlling the additional normal subgroups when factors share primes, remains absent. Reducing infinitude to an unproved prime-producing assertion is not a solution of the original target.

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 1193 (ALG-029), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath.

• S2 T. Leinster, Perfect numbers and groups, arXiv:math/0104012 $2001$. https://arxiv.org/abs/math/0104012. Bibliographic information from the supplied record.

• S3 T. Demedts et al., Perfect numbers and finite groups, preprint (). https://cage.ugent.be/~tdemedts/preprints/leinster.pdf. Bibliographic information from the supplied record.