Given a finite group $G$, let $a,b\in G$ be of order $p$, and denote by $A:=\langle a \rangle, B:=\langle b \rangle$ the generated cyclic subgroups, respectively. We demand that they have trivial intersection.
Does $A\vee B = \langle a, b \rangle$ have order $p^2m$ for some $m$?
My first idea was as follows: If $H=A\vee B$ is our generated subgroup, $A$ acts by left multiplication on $H/B$, meaning we have $\alpha\colon A\to Aut(A/B)\simeq S_{[H:B]}$. The image of a prime cyclic group is either the group itself or trivial. It cannot be trivial, because that would mean $A$ maps $B$ to itself, contradicting trivial intersection. Because $S_n$ starts to admit $C_p$-subgroups at n=p, we know that $[H:B]\geq p$. Even more, we know that $p=\lvert \alpha(A) \rvert$ divides $[H:B]! = \lvert Aut(H/B)\rvert$ – however, we would like to have something like $p\mid [H:B]$, which would yield the desired statement.
Is there any way to finish that proof, or is the assumption wrong?
Best Answer
No, take $S_3$, it is generated by two elements of order $2$, but it has order $6$.