Let $g$ be a positive integer.
How do I bound the number of elements of the group $Sp(2g,\mathbb{Z}/15)$?
Is there a polynomial bound in $g$, or can we not do better than exponential in $g$?
Let $g$ be a positive integer.
How do I bound the number of elements of the group $Sp(2g,\mathbb{Z}/15)$?
Is there a polynomial bound in $g$, or can we not do better than exponential in $g$?
For any odd prime power $q$,
$\# \operatorname{Sp}(2g,\mathbb{F}_q) = (q^{2g} - 1)(q^{2g-2} - 1) \cdots (q^2 - 1) q^{g^2}$.
This formula is proved, for instance, here.
Since $\operatorname{Sp}(2g,\mathbb{Z}/15\mathbb{Z})$ has $\operatorname{Sp}(2g,\mathbb{F}_3)$ and $\operatorname{Sp}(2g,\mathbb{F}_5)$ as homomorphic images, we certainly have $\# \operatorname{Sp}(2g,\mathbb{Z}/15\mathbb{Z}) \geq \max \# \operatorname{Sp}(2g,\mathbb{F}_3), \# \operatorname{Sp}(2g,\mathbb{F}_5)$, which is enough to answer your question: the growth is indeed at least exponential in $g$.
In fact, using the Chinese Remainder Theorem one can see that
$\operatorname{Sp}(2g,\mathbb{Z}/15\mathbb{Z}) \cong \operatorname{Sp}(2g,\mathbb{F}_3) \times \operatorname{Sp}(2g,\mathbb{F}_5)$,
so now you know an exact formula for $\# \operatorname{Sp}(2g,\mathbb{Z}/15\mathbb{Z})$.