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Intersection of 3 Proper Subgroups


Welcome back to the world of magic Math tricks! Today we shall be covering an interesting problem that I found a few days ago on AoPS regarding the intersection of 3 proper subgroups (duh...). Apparently the problem was first proposed in the 9th number of 'Gazeta Matematica' from 1985. I surely do love problems that are older than me.

Let G be a group and H1,H2,H3 be proper subgroups of G such that G=H1H2H3
Prove that x2H1H2H3 for all xG.


Solution:

First of all, let us prove that none of this subgroups includes another. Let's presume by absurd that H2 is contained in H1. This would imply that H1H3=G, which, in turn, would imply that either H1 or H3 is G, which contradicts the fact that the subgroups are proper. This argument works if one group is included in the reunion of the others, too.

If xH1H2H3, then it is obvious that x2H1H2H3.

If xH1H2H3, then we shall do a magic trick and write x2=xaa1x, where aH3H1H2. xa obviously doesn't belong to H1H2, so it must be in H3. This argument works for a1x, too, so x2H1H2H3. The other possible intersections of these groups can be treated the same.

If xH1H2H3, then we write x2=xaa1x again, this time with aH2H1H3. Obviously, xa can't be in either H2 or H1, so it is in H3, so x2H3. By choosing a suitable bH3H1H2, we get that x2H2, so x2H1H2H3. All similar cases are treated as above.

This little exercise can work as a Mathematical 'Creativity Test' if you think about it, because it's about discerning a certain pattern and coming up with an idea that solves the problem instantly. More posts might be coming this week, as I am working on 2 more articles atm: the psychology article (finally) and another one with partial derivatives / Lagrange multipliers. Tune in next time for one of these articles (probably the psychology article will come first)!

Cris.





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