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No need to mark "edit" (see https://cs.meta.stackexchange.com/q/657/755)

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edit approved Sep 6, 2022 at 5:10

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This is an answer to problem 1 (edit: and problem 3):

As noted in the comments, $\operatorname{Poly}(G)$ with pointwise multiplication is a group for finite $G$. Consider the homomorphism $\operatorname{Poly}(G)\to G$ given by evaluating at $1$. It is surjective (as seen by constant polynomials), and the kernel contains an element of order $\operatorname{exp}(G)$, the identity polynomial. So the kernel has order divisible by $\operatorname{exp}(G)$, and thus $\operatorname{Poly}(G)$ has order divisible by $\operatorname{exp}(G)\cdot |G|$.

Edit: Let me also answer problem 3! If $|\operatorname{Poly}(G)|$ is divisible by $p$, then there is an element $f\in \operatorname{Poly}(G)$ of order $p$. This means all values of $f$ have order dividing $p$, and at least one is nontrivial. So $G$ must have an element of order $p$, thus $p$ divides $|G|$.

This is an answer to problem 1 (edit: and problem 3):

As noted in the comments, $\operatorname{Poly}(G)$ with pointwise multiplication is a group for finite $G$. Consider the homomorphism $\operatorname{Poly}(G)\to G$ given by evaluating at $1$. It is surjective (as seen by constant polynomials), and the kernel contains an element of order $\operatorname{exp}(G)$, the identity polynomial. So the kernel has order divisible by $\operatorname{exp}(G)$, and thus $\operatorname{Poly}(G)$ has order divisible by $\operatorname{exp}(G)\cdot |G|$.

Edit: Let me also answer problem 3! If $|\operatorname{Poly}(G)|$ is divisible by $p$, then there is an element $f\in \operatorname{Poly}(G)$ of order $p$. This means all values of $f$ have order dividing $p$, and at least one is nontrivial. So $G$ must have an element of order $p$, thus $p$ divides $|G|$.

This is an answer to problem 1 and problem 3:

As noted in the comments, $\operatorname{Poly}(G)$ with pointwise multiplication is a group for finite $G$. Consider the homomorphism $\operatorname{Poly}(G)\to G$ given by evaluating at $1$. It is surjective (as seen by constant polynomials), and the kernel contains an element of order $\operatorname{exp}(G)$, the identity polynomial. So the kernel has order divisible by $\operatorname{exp}(G)$, and thus $\operatorname{Poly}(G)$ has order divisible by $\operatorname{exp}(G)\cdot |G|$.

Let me also answer problem 3! If $|\operatorname{Poly}(G)|$ is divisible by $p$, then there is an element $f\in \operatorname{Poly}(G)$ of order $p$. This means all values of $f$ have order dividing $p$, and at least one is nontrivial. So $G$ must have an element of order $p$, thus $p$ divides $|G|$.

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created Sep 5, 2022 at 7:13

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This is an answer to problem 1 (edit: and problem 3):

As noted in the comments, $\operatorname{Poly}(G)$ with pointwise multiplication is a group for finite $G$. Consider the homomorphism $\operatorname{Poly}(G)\to G$ given by evaluating at $1$. It is surjective (as seen by constant polynomials), and the kernel contains an element of order $\operatorname{exp}(G)$, the identity polynomial. So the kernel has order divisible by $\operatorname{exp}(G)$, and thus $\operatorname{Poly}(G)$ has order divisible by $\operatorname{exp}(G)\cdot |G|$.

Edit: Let me also answer problem 3! If $|\operatorname{Poly}(G)|$ is divisible by $p$, then there is an element $f\in \operatorname{Poly}(G)$ of order $p$. This means all values of $f$ have order dividing $p$, and at least one is nontrivial. So $G$ must have an element of order $p$, thus $p$ divides $|G|$.