Let's say I have a noncyclic multiplicative group $G_m = \{j < m, \gcd(j,m) = 1\}$, with multiplication $\bmod m$. This has order $\varphi(m)$, and suppose I can determine its factorization.
Now suppose I have some prime $p_1 \in G_m$ which has order $n_1$ such that $n_1 < \varphi(m)$ (since $G_m$ is noncyclic) and $n_1 \mid \varphi(m)$ (Lagrange's theorem).
Given another prime $p_2 \in G_m$, is there a way to tell whether $p_2 \in \left<p_1\right>$, that is, $p_2 = {p_1}^k \pmod m$ for some $k$ --- without trying all possible values of $k$?
If $p_2 \notin \left<p_1\right>$, is there a way to determine the order of the multiplicative group $H_{p_1,p_2} = \{{p_1}^{k_1}{p_2}^{k_2} \mod m\}$ without exhaustively trying to enumerate all elements?
I know I can determine the order of $p_1$ or $p_2$ by computing $p_1^{\varphi(m)/j}$ for various primes $j \mid \varphi(m)$; if the result is 1 then I should be able to confirm that $K$ = some product of powers of those primes $j$ is the largest possible value such that $p_1^{\varphi(m)/K} =1$, and therefore the order of $p_1$ is $\varphi(m)/K$.
Not sure how to determine the order of a group that has a generating set of more than one element, though.