Suppose $G$ is a group and $g\in G$ be such that $\{ g^6,g^{10},g^{15}\}$ generates all of $G$. Which of the following is not a possible order of $G$:
$(a)1$
$(b)7 $
$(c)12$
$(d) 30$
$(e)$ none of the above
Clearly choosing $g=e$, $(a)$ is possible. And since $7$ is prime therefore $(b)$ is also possible. Since lcm$(6,10,15)=30$, I guess $(d)$ is also possible but I’m not sure. Any suggestions to do it properly?
The crucial observation is that the following are equivalent for any group $G$:
$(i)$ $G$ is generated by $\{g^6, g^{10}, g^{15}\}$ for some $g \in G$
$(ii)$ $G$ is cyclic.
It's obvious that $(i) \to (ii)$.To see that $(ii) \to (i)$, suppose $G$ is cyclic and let $g_0$ be a generator of $G$. Now, note that since $\gcd(6, 10, 15) = 1$, we have that there exist integers $a,b,c$ such that $6a+10b+15c=1$, so that $g_0=g_0^{6a}g_0^{10b}g_0^{15c}$.
It's clear that $\{6, 10, 15\}$ is not special, and can be replaced with any set of coprime integers.