Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Define the GCDs: $$G = \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$ $$H = \gcd\bigg(n^2,\sigma(n^2)\bigg)$$ $$I = \gcd\bigg(n,\sigma(n^2)\bigg).$$
In this preprint, Dris shows that the identity $$G \times H = I^2$$ holds.
The proof of the divisibility constraint $I \mid H$ trivially follows from the GCD property $$\bigg(a \mid b\bigg) \implies \bigg(\gcd(a,c) \mid \gcd(b,c)\bigg).$$
It then follows that $$J = \frac{I}{G} = \frac{H}{I} = \sqrt{\frac{H}{G}}$$ must be an (odd) integer.
In an answer to this question, the anonymous MSE user mathlove shows that the value of $J$ may be given as $$J = \frac{n}{\gcd\bigg(\sigma(q^k)/2,n\bigg)}.$$
Thus, we get $$I = \gcd\bigg(n,\sigma(n^2)\bigg) = \bigg(\frac{n}{\sigma(q^k)/2}\bigg)\cdot\gcd\bigg(\sigma(q^k)/2,n\bigg), \tag{*}$$ since it is known that $$H = \gcd\bigg(n^2,\sigma(n^2)\bigg) = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}.$$
Here is my:
QUESTION: Can you show that Equation (*) unconditionally implies $\sigma(q^k)/2 \mid n$ (since $I$ must be an integer)?