Is $\sigma(n)$ injective in set $A=\left\{n\in\mathbb{N}: \mbox{$n$ is odd and $\omega(n)=1$} \right\}$? Some time ago I asked if the sum function of divisors $σ(n)$ was injective, to which the answer was no and I was presented with several counterexamples, then I began to wonder if by restricting $σ(n)$ to a certain $A\subset\mathbb{N}$, it could be injective. The first one I found was the set of prime numbers, and from there I tried to see a more general set, for example, $A=\left\{n\in\mathbb{N}: \mbox{$n$ is odd and $\omega(n)=1$}  \right\}$ where $\omega(n)$ represents the number of prime divisors other than $n$. In this way, if I take $a,b\in A$ such that $a\neq b$, so we want to prove that $\sigma(a)\neq \sigma(b)$. Note that $a,b\in A$ implies that $a=p^{\alpha}$ and $b=q^\beta$ with $\alpha,\beta\in\mathbb{N}$ and $p,q$ odd primes. Now, since $a\neq b$, then suppose without loss of generality that $a<b$. We have the cases:
Case 1: If $p=q$, then mandatory $\alpha<\beta$ and $\sigma(a)< \sigma(b)$.
Case 2: If $p\neq q$, then
Case 2.1: If $p<q$ and $\alpha\le\beta$, then $\sigma(a)< \sigma(b)$
Case 2.2: If $p<q$ and $\beta<\alpha$, then $\tau(b)=\beta +1<\tau(a)=\alpha+1$.
And there stop, could someone give me an idea how to continue the test? or tell me if it is false that sigma is not injective in A?
If I have an error in the test, please let me know.
Note: $\tau(n)$ represents the number of positive divisors of $n$.
Thanks in advance.
 A: I think this is the proof, let's first look at the following proposition:
Proposition:
$I\left( p^{n}q^{m}\right) <2$ for any $ p, q $ different odd primes and $ n, m $ positive integers.
Where $I$ denotes the abundance index
Proof:
Note that $p,q$ are different odd primes, so $(p, q)=1$ implying in turn $\left(p^{n}, q ^{m}\right) = 1 $ for any $n, m \in\mathbb{N}$ and since the abundance index is multiplicative, we have that
\begin{eqnarray*}
I\left( p^{n}q^{m}\right)=I\left( p^{n}\right)I\left(q^{m}\right)=\left(\cfrac{\sigma\left( p^{n}\right)}{p^{n}} \right) \left(\cfrac{\sigma\left( q^{m}\right)}{q^{m}} \right)
\end{eqnarray*}
But,
\begin{eqnarray*}
  \cfrac{\sigma\left( p^{n}\right)}{p^{n}}&=&\cfrac{1}{p^{n}}+\cfrac{p}{p^{n}}+\dots+\cfrac{p^{n}}{p^{n}}\\ &=&\cfrac{1}{p^{n}}+\cfrac{1}{p^{n-1}}+\dots+1\\
  &=&\sum_{k=0}^{n}\left({1}/{p}\right)^{k}=\cfrac{1-{\left( {1}/{p}\right)}^{n}}{1-{\left({1}/{p}\right)}}<\cfrac{1}{1-{\left({1}/{p}\right)}}
 \end{eqnarray*}
Similarly for $ q $ we get
\begin{eqnarray*}\cfrac{\sigma\left( q^{m}\right)}{q^{m}}<\cfrac{1}{1-{\left({1}/{q}\right)}}\end{eqnarray*}
On the other hand, as $ p $ and $ q $ are different odd primes, then we can assume without loss of generality that $3\le p<q$, this is $p\ge3$ and $q\ge5$, from here
\begin{eqnarray*}
\cfrac{1}{1-{\left({1}/{p}\right)}}\le\cfrac{1}{1-{\left({1}/{3}\right)}}\quad and\quad\cfrac{1}{1-{\left({1}/{q}\right)}}\le\cfrac{1}{1-{\left({1}/{5}\right)}}
\end{eqnarray*}
So,
\begin{eqnarray*}
 I\left( p^{n}q^{m}\right)=\left(\cfrac{\sigma\left( p^{n}\right)}{p^{n}} \right) \left(\cfrac{\sigma\left( q^{m}\right)}{q^{m}} \right)&<&\left( \cfrac{1}{1-{\left({1}/{p}\right)}}\right)\left( \cfrac{1}{1-{\left({1}/{q}\right)}}\right) \\
 &\le&\left( \cfrac{1}{1-{\left({1}/{3}\right)}}\right) \left(\cfrac{1}{1-{\left({1}/{5}\right)}}\right)=\cfrac{15}{8}<2
\end{eqnarray*}
Now, let's go with the proof of $σ(n)$ is injective in set $A=\left\lbrace n\in\mathbb{N}:\mbox{$n$ is odd and $\omega(n)=1$}\right\rbrace $
Proof:
Given $a,b∈A$ such that $a≠b$, so we want to prove that $σ(a)≠σ(b)$. Note that $a,b∈A$ implies that $a=p^α$ and $b=q^β$ with $α,β∈N$ and $p,q$ odd primes. Now, from $a\neq b$ the following cases emerge:
Case 1: If $p=q$, then obligatorily $\alpha\neq\beta$ and $\sigma(a)\neq\sigma(b)$.
Case 2: If $p\neq q$, then suppose $\sigma(a)=\sigma(b)$, in consecuense $I\left( ab\right)=I\left( a\right)I\left( b\right)<2$, where do we get \begin{eqnarray*}\left(\cfrac{\sigma\left( p^{\alpha}\right)}{p^{\alpha}} \right) \left(\cfrac{\sigma\left( q^{\beta}\right)}{q^{\beta}} \right)<2\end{eqnarray*} or equivalently \begin{eqnarray*}\sigma\left( p^{\alpha}\right) \sigma\left( q^{\beta}\right)<2p^{\alpha}q^{\beta}\end{eqnarray*} but how $\sigma(a)=\sigma(b)$, then \begin{eqnarray*}\left( \sigma\left( p^{\alpha}\right)\right)^2<2p^{\alpha}q^{\beta}\end{eqnarray*} and the latter is valid for any $ p, q $ different odd primes such that $\sigma(a)=\sigma(b)$ and $ \alpha, \beta $ positive integers. \end{eqnarray*}
I had made a mistake in the previous test, that's why I edit it, I thank @shibai for making me notice. The current test is incomplete, but perhaps a clue to the full test.
