The Abundancy Index of Divisors of Odd Perfect Numbers

If $N=q^k{n}^2$ is an odd perfect number, where $q$ is the Euler prime, then we show that $n<q$ is sufficient for Sorli’s conjecture that $k = \nu_{q}(N) = 1$tohold\mathrm{.}Wealsoprovethat$q^k < 2/3{n^2}$,andthat$I(q^k) < I(n)$,where$I(x)$istheabundancyindexof$x$.