# Classification of groups whose any non-trivial element can be associate to another to generate all

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.

• 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. – Robert Chamberlain Feb 24 at 15:57
• in the first case of course, if the monolith is central, then $G$ is abelian, so cyclic – Robert Chamberlain Feb 24 at 16:02
• 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. – Jeremy Rickard Feb 24 at 22:18
• @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. – Sebastien Palcoux Feb 24 at 22:58
• 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. – user1729 Feb 25 at 10:55

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 $$n \ge 0$$.