@BalarkaSen, Here it goes.
Proof of the statement (you mentioned):
If $a$ be a unique element of order $m$ in $J$, then $a \in Z(J)$.
Proof: Consider the element $jaj^{-1}$.
In this case $x \in Z(G)$. So, for any element $g \in G$
$g\{x,e\}=\{gx,g\}=\{xg,g\}=\{x,e\}g$.
Evidently, $<x>$ is normal in $G$. Here, $Z(G)=<x>$ [since the only possible order of subgroups are $3$ and $2$. If any other element, of order $3$, say $p$ belongs to $Z(G)$, then $p^2=x \implies p^3=e=xp \implies p=x^{-1} $ But, $o(x^{-1}=2$, contradiction ] The quotient group $G/<x>$ exists. But, $o(G/<x>)=o(G)/o(<x>…