If $N = {q^k}{n^2}$ is an odd perfect number with Euler prime $q$,

then Sorli's conjecture predicts that $k = 1$.

From previous posts, we have the implication "$n < q \rightarrow k = 1$".

Here, we show other statements that imply $n < q$.

**Claim #1: If $q = \sigma(n)$, then $n < q$.**

Proof of Claim #1:

Suppose that $q = \sigma(n)$. Then

$$\frac{\sigma(n)}{q} = 1 < \frac{\sigma(n)}{n} = \frac{q}{n} < \frac{\sigma(q)}{n}$$

From previous posts, $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n} \rightarrow n < q$.

**Claim #2: If $\sigma(q) = \sigma(n)$,**

**then $n < q$.**

Proof of Claim #2:

Suppose that $\sigma(q) = \sigma(n)$. Then

$$\frac{\sigma(n)}{q} = \frac{\sigma(q)}{q} < \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n}$$

From previous posts, $\frac{\sigma(n)}{q} < \frac{\sigma(q)}{n} \rightarrow n < q$.