If $\sigma : H \xrightarrow{\sim} G $ is a group isomorphism and $H = \langle S \rangle$, then does $G = \langle \sigma(S)\rangle$?

Question

Let $\sigma : \Bbb{P} \xrightarrow{\sim} \Bbb{P}$ be a permutation of the primes $\Bbb{P} = \{ 2,3,5,7,11, \dots \}$.

Then $\sigma$ extends uniquely to a surjective group hom $\Bbb{Q}^{\times} \to \Bbb{Z}$, the group of integers. We know that $\sigma$ acts as $\sigma(\dfrac{p_1 \cdots p_r}{q_1 \cdots q_s}) = \sum_{i = 1}^r \sigma(p_i) - \sum_{i=1}^s \sigma(q_i)$. For now and for simplicity of argument take $\sigma = \text{id}$ the identity permutation.

So for instance $\ker \sigma \supset \langle\dfrac{pq}{rs} : \sigma(p) + \sigma(q) = \sigma(r) + \sigma(s) \rangle$. Notice, that $\ker \sigma$ can contain no prime pair fraction $\dfrac{p}{q}$ for $p \neq q$ since $\sigma$ permutes primes and so $\sigma(p) \neq \sigma(q)$ whenever $p \neq q$.

Let $H = \langle \dfrac{p}{q}: p, q \text{ prime, and } \sigma(p) \gt \sigma(q)\rangle$ that of course includes all $\dfrac{p}{q}$ such that $\sigma(p) \lt \sigma(q)$ since generation is by $g^z$ for all $z \in \Bbb{Z}$ (including negative exponents) and $g \in $ the generating set. Notice that multiplying a finite collection of generators results in the same imbalance i.e. $\sigma(p_1) + \dots + \sigma(p_r) \gt \sigma(q_1) + \dots + \sigma(q_r)$ whenever $\sigma(p_i) \gt \sigma(q_i)$ for each $i$.

Take a look at the cosets $\dfrac{a}{b}K$ of the kernel. If $\dfrac{a}{b} \notin K$ then $\sigma(a) \gt \sigma(b)$ or $\sigma(b) \gt \sigma(a)$ which would mean $\dfrac{a}{b} \in H$.

Thus, can we conclude that $H \simeq \Bbb{Q}^{\times}/K \simeq \Bbb{Z}$? In particular, the first isomorphism is in question.

Then what can we say about prime gaps, if anything? Since $S = \{ \dfrac{p}{q} : p, q \text{ both prime}, p \neq q\}$ is essentially all non-diagonal pairs of primes $(p,q)$ and $S$ generates $H$ as a group, which then maps isomorphically onto $\Bbb{Z}$. If $H = \langle S \rangle \simeq \Bbb{Z} = \langle 1 \rangle$. Have we proved that since $\sigma(3) - \sigma(2) = 3 - 2 = 1 \implies \sigma^{-1}(1) = \dfrac{3}{2}$ so that $H = \langle \dfrac{3}{2}\rangle$? In other words, the question is:

If $\sigma : H \xrightarrow{\sim} G $ is a group isomorphism and $H = \langle S \rangle$, then does $G = \langle \sigma(S)\rangle$?

---

From the commenters we've learned that $H \cap K \neq \{1\}$. For example $H \ni \dfrac{3}{5}, \dfrac{13}{11}$ and their product is in $K$.

Answer 1

You've asked two questions and made a few claims, I'll try to answer everything.

1. Yes, if $\sigma : H \to G$ is an isomorphism and $H$ is generated by $S$, then $G$ is generated by $\langle S \rangle$. Just write any element of $H$ as $h=s_1^{\pm} \dots s_k^{\pm}$ for $s_i \in S$, then $\sigma(h) = \sigma(s_1)^{\pm} \dots \sigma(s_k)^{\pm}$ belongs to $\langle S \rangle$.
2. No, we cannot conclude that $H \simeq \mathbb{Q}^\times / K$. I do not follow your argument, and since there is a gap it is difficult to pinpoint what is wrong, exactly. If you wanted to conclude that, you would essentially need to build a surjective morphism $\mathbb{Q}^\times \to H$ with kernel $K$. You have not done that.
3. I do not get your argument that $\sigma(a) > \sigma(b) \implies a/b \in H$. You have proved that if $a/b = p_1 \dots p_r / (q_1 \dots q_r)$ is a product of generators of $H$, then $\sigma(a) > \sigma(b)$. But you have not proved the converse.