Proper isometry group/Finite subgroup/Numerical properties/Possibilities/Fact/Proof

Proof

For ${}m=1$, we muss have ${}n_{1}={\frac {n}{-n+2}}$ but this does not have a solution for ${}n\geq 2$.  For ${}m=3$, the condition becomes

${}1+{\frac {2}{n}}={\frac {1}{n_{1}}}+{\frac {1}{n_{2}}}+{\frac {1}{n_{3}}}\,$

with ${}n_{1}\leq n_{2}\leq n_{3}$. The left-hand side is ${}>1$. Because of ${}n_{i}\geq 2$, at least one them is ${}n_{i}=2$. So suppose ${}n_{1}=2$. For ${}n_{2}=4$, the right-hand side is again ${}\leq 1$, so that ${}n_{2}=3$ holds. The value ${}n_{3}=3$ leads to the solution ${}n=12$, the value ${}n_{3}=4$ leads to the solution ${}n=24$, and the value ${}n_{3}=5$ leads to the solution ${}n=60$. For ${}n_{3}\geq 6$, the right-hand side is again ${}\leq 1$, so that there is no further solution.
 For ${}m\geq 4$, the condition has the form

${}m-2+{\frac {2}{n}}={\frac {1}{n_{1}}}+{\frac {1}{n_{2}}}+{\frac {1}{n_{3}}}+{\frac {1}{n_{4}}}+\cdots +{\frac {1}{n_{m}}}\,.$

This has no solution, because the right-hand side is ${}\leq m-2$, since the first four summands yield at most ${}2$, and the further summands can be bounded by ${}m-4$.

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