Question: What are the groups (possibly infinite) $G$ satisfying the following property? $$ \forall g \in G \setminus \{ e \} \ \exists g' \in G \text{ such that } \langle g,g' \rangle = G.$$

Examples: the cyclic groups, the groups $C_p \times C_p$, and more generally $C_p \rtimes C_q$ with $p,q$ primes.

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    $\begingroup$ An immediate observation is that for any $1\ne H\le G$ you must have $G/N$ cyclic where $N=\langle H^G\rangle$. This is because for $g\in H\setminus\{1\}$ we have $\langle g,g'\rangle\le \langle g^G,g'\rangle$ so $G/N$ is a quotient of $G/\langle g^G\rangle$ which is generated by the image of $g'$. This means either $G$ has a unique minimal normal subgroup (that is $G$ is monolithic) or the socle of $G$ is a product of two cyclic groups. $\endgroup$ – Robert Chamberlain Feb 24 at 15:57
  • $\begingroup$ in the first case of course, if the monolith is central, then $G$ is abelian, so cyclic $\endgroup$ – Robert Chamberlain Feb 24 at 16:02
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    $\begingroup$ See math.stackexchange.com/questions/2263690/… for lots of information. For finite groups there’s a conjecture that the restriction in @RobertChamberlain’s comment (every proper quotient is cyclic) is necessary and sufficient. $\endgroup$ – Jeremy Rickard Feb 24 at 22:18
  • $\begingroup$ @JeremyRickard: Very nice! These groups are called $\frac{3}{2}$-generated, very chic! For $n \neq 4$, every symmetric group $S_n$ is so. This answer contains references proving that every finite simple group is so. $\endgroup$ – Sebastien Palcoux Feb 24 at 22:58
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    $\begingroup$ Something very similar holds for almost all $2$-generator groups ("random" $2$-generator groups, in the sense of Gromov): for all $g\in G$, either $\langle g, g'\rangle$ is free for all $g'\in G$, or there exists some $g'\in G$ such that $\langle g, g'\rangle=G$. This follows from Theorem B.3 of Kapovich and Schupp, Genericity, the Arzhantseva-Ol'shanskii method and the Isomorphism Problem for One-Relator Groups, Math. Ann. (2005) 331: 1. doi, arXiv. $\endgroup$ – user1729 Feb 25 at 10:55

Many such groups have a similar structure that you mentioned as examples. (This is not a complete answer to your question.)

Claim. An Artinian group $G$ with non-trivial center satisfying $$ \forall g \in G \setminus \{e\},\ \exists g' \in G \text{ such that } \langle g, g' \rangle = G \tag{$\ast$} $$ is a cyclic group or the semidirect product of cyclic groups.

Proof. Let us assume that $G$ is not cyclic and prove $G$ is the semidirect product of cyclic groups. Namely, we shall prove that there are elements $a, b \in G$ such that $$\langle a, b \rangle = G,\ \langle a \rangle \unlhd G,\ \text{ and } \langle a \rangle \cap \langle b \rangle = \{ e \}.$$ Take a random central element $a_0 \neq e$ of $G$. By $(\ast)$, there is an element $b_0$ of $G$ such that $\langle a_0, b_0 \rangle = G$. If $\langle a_0 \rangle \cap \langle b_0 \rangle = \{ e \}$ then there is nothing to do. If not, then take $e \neq a_1 \in \langle a_0 \rangle \cap \langle b_0 \rangle$. Again, by $(\ast)$, there is an element $b_1$ of $G$ such that $\langle a_1, b_1 \rangle = G$. Repeating this process, we have a descending series $$ \langle a_0 \rangle \ge \langle a_1 \rangle \ge \langle a_2 \rangle \ge \cdots $$ if $\langle a_n \rangle \cap \langle b_n \rangle \neq \{ e \}$ for all $n \ge 0$. Since $G$ is artinian, there is some index $i \ge 0$ such that $\langle a_i \rangle = \langle a_{i+1} \rangle$. Then $$ \langle a_i \rangle = \langle a_{i+1} \rangle \le \langle a_i \rangle \cap \langle b_i \rangle \le \langle a_i \rangle$$ and we have $\langle a_i \rangle \le \langle b_i \rangle $. However, as $G = \langle a_i, b_i \rangle \le \langle b_i \rangle$, we have $G$ is cyclic which condradicts to our assumption. Therefore, $\langle a_n \rangle \cap \langle b_n \rangle = \{ e \}$ for some $n \ge 0$.


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