Interesting. The 5/8 theorem is so easy I feel Erdos and Turan “could have” noticed it if they bothered. But I guess “could have” doesn’t mean much in math.

]]>Thank you!

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]]>(This problem was basically inspired by recalling this paper around the same time I first encountered Eberhard’s paper above…)

The obvious place to start is words in one letter, so you’re just counting elements of order dividing a fixed constant. But I have no idea if that goes anywhere. Like, it’s known that any finite group with more than 3/4 of its elements being involutions must have all elements involutions, but who knows if that’s the start of a well-ordering or not? And that’s just order 2! There’s some obvious trivial cases we can consider (the word “a”, for instance, or the identity word; or “ab”, or the analogue in any number of letters), and these aren’t counterexamples but they don’t tell us much of anything.

I’ve also wondered what might happen if you allow compact groups instead of just finite groups. Does that just add in 0, or is anything else gained?

Anyway, this is basically pure speculation on my part — I don’t actually intend to work on this problem. But it’s been bugging me all the same…

]]>Right. I’ll fix that.

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