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 $k$?
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closed as off topic by Alex Bartel, Andreas Blass, Felipe Voloch, Gerry Myerson, S. Carnahan♦ Aug 29 '11 at 2:11Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question. 

$k\leq G\leq 2^k$
. (This might or might not answer the question, depending on what the question actually is.) – Andreas Blass Aug 29 '11 at 1:21