On odd perfect numbers and a GCD (Note:  This question is closely related to this other one in MSE.)
Let $N = q^k n^2$ be an odd perfect number.
From this paper in NNTDM, we have the equation
$$i(q) := \frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd\left(n^2,\sigma(n^2)\right).$$
In particular, we know that the index $i(q)$ is an integer greater than $5$ by a result of Dris and Luca.
We now attempt to compute an expression for $\gcd\left(\sigma(q^k),\sigma(n^2)\right)$ in terms of $i(q)$.
First, since we have 
$$\sigma(q^k)\sigma(n^2) = \sigma({q^k}{n^2}) = \sigma(N) = 2N = 2{q^k}{n^2}$$
we obtain
$$\sigma(q^k) = \frac{2 q^k n^2}{\sigma(n^2)} = \frac{2n^2}{\sigma(n^2)/q^k} = \frac{2n^2}{i(q)}$$
and
$$\sigma(n^2) = \frac{2 q^k n^2}{\sigma(q^k)} = {q^k}\cdot\bigg(\frac{2n^2}{\sigma(q^k)}\bigg) = {q^k}{i(q)},$$
so that we get
$$\gcd\left(\sigma(q^k),\sigma(n^2)\right) = \gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg).$$
Now, since $\gcd(q, n) = \gcd(q^k, 2n^2) =  1$ and $i(q)$ is odd, we get
$$\gcd\bigg(\frac{2n^2}{i(q)}, {q^k}{i(q)}\bigg) = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$
Hence, we conclude that $G:=\gcd(\sigma(q^k),\sigma(n^2))=\gcd\bigg({n^2}/{i(q)}, i(q)\bigg)$.
(Edited Sept 12 2017) Here are my questions:
Original Question

I seem to recall that somebody (was it Pomerance [?] et. al) proved that
  $$G \neq 1.$$  Does anybody here happen to know a reference?  Additionally, does
  $G \neq 1$ imply that $G = i(q)$?

Additional Question (Added Sept 12 2017)

Some authors have already considered the possibility that $i(q)$ may be a square.  This would imply that $G$ is also a square.  Would $G$ a square mean that $G = i(q)$ holds?

(End Edit)
Thank you.
 A: It turns out that
$$G \text{ is a square } \iff i(q) \text{ is a square.}$$
The proof is essentially contained in this answer to a closely related MSE question.
Thus, we have the implication
$$G \text{ is a square } \implies k=1$$
by a result of Broughan, Delbourgo, and Zhou (Improving the Chen and Chen Result for Odd Perfect Numbers).
A: 
Here is a conditional proof that 
  $$G = \gcd(\sigma(q^k),\sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2)).$$

As derived in the OP, we have
$$G = \gcd\bigg(\frac{n^2}{i(q)}, i(q)\bigg).$$
This is equivalent to
$$G = \frac{1}{i(q)}\cdot\gcd\bigg(n^2, (i(q))^2\bigg) = \frac{1}{i(q)}\cdot\bigg(\gcd(n, i(q))\bigg)^2.$$
But we also have
$$\gcd(n, i(q)) = \gcd\bigg(n, \gcd(n^2, \sigma(n^2))\bigg) = \gcd\bigg(\sigma(n^2), \gcd(n, n^2)\bigg) = \gcd(\sigma(n^2), n).$$
Consequently, we obtain
$$G = \frac{\bigg(\gcd(n, \sigma(n^2))\bigg)^2}{\gcd(n^2, \sigma(n^2))}.$$
In particular, we get
$$\gcd(\sigma(q^k), \sigma(n^2)) = i(q) = \gcd(n^2, \sigma(n^2))$$
if and only if
$$\gcd(n, \sigma(n^2)) = \gcd(n^2, \sigma(n^2)) = i(q).$$
