## How large can a symmetric generating set of a finite group be?

Let $$G$$ be a finite group of order $$n$$ and let $$\Delta$$ be its generating set. I’ll say that $$\Delta$$ generates $$G$$ symmetrically if for every permutation $$\pi$$ of $$\Delta$$ there exists $$f:G\rightarrow G$$ an automorphism of $$G$$ such that $$f\restriction\Delta=\pi$$.

How large can $$\Delta$$ be with respect to $$n$$? Specifically what’s the asymptotic behaviour? Is there a class of groups (along with their symmetric generating sets) of unbounded order such that $$n$$ is polynomially bounded by $$|\Delta|$$. Has any other research been done on these generating sets?

## Is there any difference between these two solutions for finding generating function?

Consider the following question.

Find the generating function for the number of integer solutions to the equation $$c_{1}+c_{2}+c_{3}+c_{4}=20$$ where $$-3\leq c_{1}, -3 \leq c_{2}, -5\leq c_{3}\leq 5,$$ and $$0 \leq c_{4}$$.

• The first solution: $$(c_{1} + 3)+(c_{2}+3)+(c_{3}+5)+c_{4}=20 + 3 + 3 + 5 \Rightarrow c_{1}’+c_{2}’+c_{3}’+c_{4}’=31$$ where $$0\leq c_{1}’, 0 \leq c_{2}’, 0\leq c_{3}’\leq 10,$$ and $$0 \leq c_{4}’$$. The generation function is $$f(x) = (x^{0}+x^{1}+…)^{3}(x^{0}+x^{1}+…+x^{10})$$, and the number of integer solutions is the coefficient of $$x^{31}$$ in $$f(x)$$.

• The second solution (My solution): $$g(x) = (x^{-3}+x^{-2}+…+x^{3})^{2}(x^{-5}+x^{-4}+…+x^{5})(x^{0}+x^{1}+…)$$, and the number of integer solutions is the coefficient of $$x^{20}$$ in $$g(x)$$.

My question: In the solution manual book, the presented solution is the first solution, but I think that the second solution is correct, and it is simpler than the first solution. Is my solution correct? What is the difference between these two solutions?