Do there exist simple groups of order $n(n+1)$ for some integer $n>1$?

Solution 1:

The alternating groups $A_m$ are ruled out for sufficiently large $m$ by the abc conjecture. If $n(n+1) = \frac{m!}{2}$, then by the abc conjecture we have (starting from the identity $n + 1 = (n+1)$) for any $\varepsilon > 0$ a constant $K_{\varepsilon}$ such that

$$n < K_{\varepsilon} \text{rad}(n(n+1))^{1 + \varepsilon}$$

for all $n$, but $\text{rad}(n(n+1)) = \text{rad} \left( \frac{m!}{2} \right)$ is the product of all the primes less than or equal to $m$. This is asymptotically about $e^m$, whereas $n$ is asymptotically about $\left( \frac{m}{e} \right)^{\frac{m}{2}}$ by Stirling's approximation.