Herzog's conjecture on the number of involutions

About 9 years old · traced to

Let GG and HH be simple groups, and let I2(G)I_2(G) and I2(H)I_2(H) denote their numbers of involutions.

Herzog's conjecture. If I2(G)=I2(H)I_2(G)=I_2(H), then ∣G∣=∣H∣|G|=|H|.

The conjecture was proposed by M. Herzog in 1979. The paper gives a counterexample, so the assertion is false.

References

Primary source

Mohammad Zarrin, “A counterexample to Herzog's Conjecture on the number of involutions”, arXiv:1802.08162 (2018).

Additional references

2 papers in this index state this conjecture (2017–2018). The statement above is taken from the most recent of them; the others are arXiv:1705.00575.

Progress summary

Refreshed
Claimed solved

A 2018 paper claims the conjecture is false by exhibiting two simple groups with equally many involutions but different sizes.

M. Herzog proposed the conjecture in 1979: equal numbers of involutions should force equal group orders. The assertion is now contradicted by a claimed explicit pair of finite simple groups.

May 29, 2018 counterexample

M. Zarrin's paper A Counterexample to Herzog's Conjecture on the Number of Involutions claims

I2(PSp⁡(4,3))=I2(PSL⁡(3,4))=315,I_2(\operatorname{PSp}(4,3))=I_2(\operatorname{PSL}(3,4))=315,

while

∣PSp⁡(4,3)∣=25920≠20160=∣PSL⁡(3,4)∣.|\operatorname{PSp}(4,3)|=25920\ne 20160=|\operatorname{PSL}(3,4)|.

Thus the conjecture is false if the computation is accepted. The paper also records the earlier motivating example involving A8A_8 and PSL⁡(3,4)\operatorname{PSL}(3,4), where both the group orders and involution counts agree.

Current status (as of September 2026): Zarrin's published preprint claims a counterexample, so the conjecture is regarded here as refuted but remains unverified by this scan; no later challenge, withdrawal, or independent verification was found.

Sources

Solutions 0

No solutions have been posted yet.