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Apr 13, 2017 at 12:19 history edited CommunityBot
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Jan 30, 2015 at 13:56 comment added Sean Eberhard You can prove $k\geq n$ by adapting the argument here: mathoverflow.net/questions/20471/…
Jan 25, 2015 at 22:03 comment added Geoff Robinson Another small remark is that you might as well suppose that $c_{1} = 1.$
Jan 25, 2015 at 20:56 comment added Milo Brandt @Jim It has no particular motivation, beyond the thought that a proof that $k<\text{lcm}(1,\ldots,n)$ for infinitely many $n$ would likely reveal something interesting about some "additional structure" allowing that to happen; in the other case, if $k=\text{lcm}(1,\ldots,n)$, it might be interesting to see why the exponent of a group is "irreducible"; I ask about the symmetric group in particular, because it has exactly one normal subgroup - so any approach applicable to it is likely to extend well to simple groups, but an approach could also leverage that $S_n$ is not itself simple.
Jan 25, 2015 at 20:43 comment added Jim Humphreys The question looks quite nontrivial, but is there some specific motivation for it?
Jan 25, 2015 at 20:13 comment added Geoff Robinson Note that $k$ must be even.
Jan 25, 2015 at 20:07 review First posts
Jan 25, 2015 at 20:16
Jan 25, 2015 at 19:59 history asked Milo Brandt CC BY-SA 3.0