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 $k = \nu_{q}(N) = 1$ to hold. We also prove that $q^k < 2/3{n^2}$, and that $I(q^k) < I(n)$, where $I(x)$ is the abundancy index of $x$.Comment: 10 page