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  1. does any subset of symmetric group generated by repeated multiplication of 3 or more permutations can be generated by only 2 of them? (trivial cases of identity and single transpositions aside)

  2. is there a way for finding such set of two permutations from bigger sets? (naive brute-force search aside)

(and, well, what is correct name and terminology for this problem?)

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    $\begingroup$ What do you mean by "repeated multiplication of three of more permutations"? $\endgroup$ – Pedro Tamaroff Aug 2 '16 at 11:02
  • $\begingroup$ result of (1)(2 3 4) and (1 2 3)(4) is alternating group. result of (1 2)(3 4) and (1 4)(2 3) is klein group rotation. $\endgroup$ – pb7 Aug 2 '16 at 11:03
  • $\begingroup$ does all subgroups can be generated only by two of them? these seems trivial, but i do not know any proof of it, or of possibilities of edge-cases in here. $\endgroup$ – pb7 Aug 2 '16 at 11:07
  • $\begingroup$ yes, right. counter-example for first question is (1 2), (3 4), (5 6). and answer to second and third is "dig into k generators and group presentations". $\endgroup$ – pb7 Aug 3 '16 at 4:38
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If I understood your question (1) correctly the answer is certainly no. For example, in $\;S_6\;$ , we have that

$$\;H:=\langle\,(12),\,(34),\,(56)\,\rangle\cong\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_2$$

and this group cannot be generated by less than three elements. You can easily generalize the above to any finite number of generators.

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    $\begingroup$ This is an example of something, perhaps counter intuitive in group theory: A minimally $k$-generated group can have subgroups that cannot be generated by any set of $k$ elements (i.e. are minimally $k+l$ generated for some natural number $l>0$). I believe the smallest example is an order 16 group that is minimally $2$-generated but contains a copy of the elementary abelian group of order 8 in it, which is minimally $3$-generated. $\endgroup$ – Justin Benfield Aug 2 '16 at 11:33

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