# How do we bound the cardinality of this group

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 $g$?

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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 $\# \operatorname{Sp}(2g,\mathbb{Z}/15\mathbb{Z})$.

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