Suppose that I have $A$ is a subset of $\mathbb{Z_p}$. Let $|A|=a\cdot b $ where $\gcd(a,b)=1$ and let me attache a multiplication modulus $p$. Are the two following requirements
- There is a cyclic group of order $a$ in $A$
- There is a cyclic group of order $b$ in $A$
sufficient to insure that $A$ is cyclic under multiplication modulus $p$, that is, there exist an element $g\in A$ with order $a\cdot b$.
Even if you exclude $0$ (which would ruin any hope of multiplication being a group operation), there is a counterexample for $p=13$. In that case, the multiplicative group of invertible elements has $12$ elements. Take $A=\{1,12,3, 9,2,5\}$, which has size $6=2\cdot3$, and contains the order $2$ group $\{1,12\}$ and the order $3$ group $\{1,3,9\}$. But it's not a group, since $2\in A$ but $2\cdot 2=4\notin A$.