# If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, does the following implication hold?

Let $\sigma(x)$ denote the sum of the divisors of $x$. For example, $\sigma(6)=1+2+3+6=12$. Numbers $N$ satisfying the equation $\sigma(N)=2N$ are called perfect numbers. Euler showed that an odd perfect number must have the form $N = q^k n^2$ where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Here is my question:

If $N = q^k n^2$ is an odd perfect number with Euler prime $q$, does the following implication hold? $$\bigg(q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}\bigg) \implies \left(q^k < n\right)$$

From an answer to this earlier MSE question, I know that the converse $$\left(q^k < n\right) \implies \bigg(q^k < n \iff \sigma(q^k) < \sigma(n) \iff \dfrac{\sigma(q^k)}{n} < \dfrac{\sigma(n)}{q^k}\bigg)$$ holds.