Cyclicity is 2-local for finite groups

Statement

Suppose $G$ is a finite group with the property that for any two elements $x,y\in G$, the subgroup $\langle x,y\rangle$ is a Cyclic group (?). Then, $G$ itself is a cyclic group.

The proof generalizes to $G$ a Finitely generated group (?).

Related facts

• Abelianness is 2-local
• Solvability is 2-local for finite groups
• Nilpotency is 2-local for finite groups

Proof

Given: A finite group $G$ with the property that for any $x,y\in G$, the subgroup $\langle x,y\rangle$ is cyclic.

To prove: $G$ is cyclic.

Proof: Let $A=\{a_{1},a_{2},\dots ,a_{r}\}$ be a generating set of minimum cardinality for $G$ (such a generating set exists because $G$ is finite).

Now, if $r\geq 2$, consider the subgroup $\langle a_{1},a_{2}\rangle$. By hypothesis, there exists $b$ such that $\langle a_{1},a_{2}\rangle$ is generated by $b$. Then, the set $B=\{b,a_{3},\dots ,a_{r}\}$ also generates $G$. Thus, there is a set of size $r-1$ that generates $G$, contradicting the minimality of $r$.

Thus, $r=0$ or $r=1$. In either case, $G$ is cyclic.