(Preamble: This inquiry is an offshoot of this MSE question.)
Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ as $I(x)=\sigma(x)/x$.
Recall that an odd perfect number $N = q^k n^2$ is said to be given in Eulerian form if $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Also, elsewhere we showed that $$\gcd(n^2,\sigma(n^2)) = \frac{\sigma(n^2)}{q^k} = \frac{n^2}{\sigma(q^k)/2}.$$
In the hyperlinked post, we showed that we have the equation $$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q},$$ from which it follows that $$\frac{N}{\sigma(n^2)/q} = \frac{1}{I(n^2) - \frac{2(q - 1)}{q}} = \frac{q\left(q^{k+1} - 1 \right)}{2(q - 1)} = \frac{q\sigma(q^k)}{2},$$ since $$I(n^2) = \frac{2}{I(q^k)} = \frac{2q^k (q - 1)}{q^{k+1} - 1}.$$
This shows that the divisibility condition $\sigma(n^2)/q \mid N = q^k n^2$ holds.
Under the assumption $k=1$, since $$\gcd\left(q,\frac{\sigma(n^2)}{q}\right) = \gcd\left(q,\frac{\sigma(n^2)}{q^k}\right) = \gcd\left(q,\gcd(n^2,\sigma(n^2))\right) = \gcd(\gcd(q,n^2),\sigma(n^2)) = \gcd(1,\sigma(n^2)) = 1,$$ then it follows from $\sigma(n^2)/q \mid q^k n^2$ and $\gcd(q,\sigma(n^2)/q)=1$ that $\sigma(n^2)/q \mid n^2$, which implies that $k=1$.
Under the assumption $$\gcd\left(q,\frac{\sigma(n^2)}{q}\right) = 1$$ we obtain $\sigma(n^2)/q \mid n^2$, which implies that $k=1$.
Hence, we actually have the biconditional $$\gcd\left(q,\frac{\sigma(n^2)}{q}\right) = 1 \iff k = 1. \tag{*}$$
Here is my question:
Is my proof of the biconditional depicted in the section marked with a (*) correct? If not, how can it be mended so as to produce a valid argument?
