Let $G$ be an abelian permutation group acting on a set $\Omega$ such that $|G| = n$ and all its elements are of order at most 2. Let $g_1, \ldots, g_k \in G$ be distinct elements and suppose that $G = \langle g_1, \ldots, g_k\rangle$.

Question: Is $k$ necessarily small (logarithmic) in $n$ or can it be large (polynomial)?

Let $g_1, \ldots, g_k$ be distinct permutations on a set $\Omega$. Suppose that $G = \langle g_1, \ldots, g_k \rangle$ is an abelian permutation group with only elements of order at most 2. Is it possible that $|G| = k$? If not, can $|G|$ be polynomial in $k$?

Let $g_1, \ldots, g_n$ be permutations on $\Omega$ such that $$g_i^2 = 1 \quad \forall i \in [n]$$ $$g_i \not= g_j \quad \forall i, j \in [n] \text{ s.t. } i \not = j.$$

Let $G = \langle g_1, \ldots, g_n\rangle$ and suppose that $G$ is an abelian transitive group.

Question: Can $|G|$ be exponential in $n$ or is it always polynomial? More precisely, what is the largest $|G|$ possible?

Let $G$ be an abelian permutation group acting on a set $\Omega$ such that $|G| = n$ and all its elements are of order at most 2. Let $g_1, \ldots, g_k \in G$ be distinct elements and suppose that $G = \langle g_1, \ldots, g_k\rangle$.

Question: Is $k$ necessarily small (logarithmic) in $n$ or can it be large (polynomial)?