Let $N = q^k n^2$ be an odd perfect number given in the so-called Eulerian form, where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.
Consider the GCDs $$G = \gcd\left(\sigma(q^k),\sigma(n^2)\right)$$ $$H = \gcd\left(n^2,\sigma(n^2)\right)$$ $$I = \gcd\left(n,\sigma(n^2)\right).$$
It is known that $$H = \frac{\sigma(n^2)}{q^k};$$ therefore, since $\gcd(q,n)=\gcd(q^k,\sigma(q^k))=1$ and $\sigma(n^2)$ is odd, then the above GCDs for $G$ and $I$ can be rewritten as $$G = \gcd\left(\sigma(q^k)/2,H\right)$$ $$I = \gcd\left(n,H\right).$$
It is known that the identity $$G \times H = I^2$$ is true, and that the divisibility constraint $$G \mid I$$ holds.
Now, consider $$\gcd(\sigma(q^k)/2,H)=G=\gcd(G,I)=\gcd\left(\gcd(\sigma(q^k)/2,H),\gcd(n,H)\right)=\gcd\left(\gcd(\sigma(q^k)/2,n),\gcd(H,H)\right),$$ where the last equality holds by GCD Associative Property. It follows that $$\gcd(\sigma(q^k)/2,H)=G=\gcd\left(\gcd(\sigma(q^k)/2,n),H\right). \tag{1}$$ Equating GCD function arguments, we obtain $$\sigma(q^k)/2=\gcd(\sigma(q^k)/2,n), \tag{2}$$ which is equivalent to $\sigma(q^k)/2 \mid n$. But since $N = q^k n^2$ is (odd) perfect, then $$\sigma(q^k)\sigma(n^2)=\sigma(q^k n^2)=\sigma(N)=2N=2q^k n^2$$ whence we obtain $$\frac{\sigma(n^2)}{n}=\frac{q^k n}{\sigma(q^k)/2};$$ hence, $n \mid \sigma(n^2)$ is also equivalent to $\sigma(q^k)/2 \mid n$.
To conclude, we have derived $$G = \gcd\left(\sigma(q^k)/2,H\right) = \sigma(q^k)/2$$ and $$I = \gcd\left(n,H\right) = n.$$
Here is my:
QUESTION: Is the derivation of the step marked with a $(2)$ from the step marked with a $(1)$ above logically valid? If not, how can it mended so as produce a correct proof?
This answer is copied verbatim from Bill Dubuque's comment, so that this question does not remain in the unanswered queue.
Per Bill Dubuque: