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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 2:11

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What is $\Omega$? –  Mark Sapir Aug 28 '11 at 22:31
    
It's the set of points on which G is acting. I've updated the question to make it clearer. –  Steve Aug 28 '11 at 22:42
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Why not $k=n$, all the distinct elements of $G$? If you mean that no proper subset of $\langle g_1 \cdots g_k$ generates $G$ then $n = 2^k$ no? If you mean something else what? –  Daniel Mehkeri Aug 28 '11 at 22:53
    
You are right, it didn't make much sense. I've uptaded the question. –  Steve Aug 28 '11 at 23:12
    
Assuming $k$ is fixed, the possibilities for $|G|$ are all the powers of 2 in the range $k\leq |G|\leq 2^k$. (This might or might not answer the question, depending on what the question actually is.) –  Andreas Blass Aug 29 '11 at 1:21