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Mar 11, 2017 at 0:08 comment added Ehud Meir I see now that I made too much haste with this question. We can take the collection of groups $P=\mathbb{Z}/2^r$ and the group $G=S_3$. Then the number of homomorphisms will be $3\cdot 2^r - 2$. Then all the prime numbers $p$ for which $3\in\langle 2 \rangle$ in $(\mathbb{Z}/p)^{\times}$ will appear as prime divisors, and there are infinitely many such.
Mar 11, 2017 at 0:00 vote accept Ehud Meir
Mar 10, 2017 at 23:47 comment added Ehud Meir Thanks for this example. Do you have any insight about the second question?
Mar 10, 2017 at 23:42 history answered Jeremy Rickard CC BY-SA 3.0