Index of Intersection of Subgroups with Finite Index

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Homework Help Overview

The problem involves subgroups of finite index within a possibly infinite group, specifically examining the relationship between the indices of these subgroups and their intersection. The original poster attempts to prove a bound involving the least common multiple of the indices of the subgroups.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss using Lagrange's theorem to establish divisibility conditions for the index of the intersection of the subgroups. There is uncertainty about the applicability of Lagrange's theorem in the context of infinite groups.

Discussion Status

Some participants have offered insights regarding the divisibility of the index of the intersection by the indices of the individual subgroups. However, there is a lack of consensus on the validity of using Lagrange's theorem in this scenario, leading to further exploration of assumptions regarding the finiteness of the group.

Contextual Notes

There is a noted concern regarding the applicability of Lagrange's theorem, particularly in relation to the finiteness of the group involved in the problem.

Bashyboy
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Homework Statement


Suppose that ##H## and ##K## are subgroups of finite index in the (possibly infinite) group ##G## with ##|G : H|m## and ##|G:K|=n##. Prove that ##lcm(m,n) \le |G : H \cap K | < mn##.

Homework Equations

The Attempt at a Solution



I was able to get the upper bound on ##|G : H \cap K|##, but am having difficulty showing that the lower bound is ##lcm(m,n)##. I tried showing that ##m## and ##n## both divide ##|G : H \cap K |##, but I couldn't get anywhere. I could use some hints!
 
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I think you can use the theorem of Lagrange to get ##|G : H \cap K|=|G : K| \cdot |K : H \cap K|## so ##n\,\vert \,|G : H \cap K|## and similar ##m\,\vert \,|G : H \cap K|##.
 
fresh_42 said:
I think you can use the theorem of Lagrange to get ##|G : H \cap K|=|G : K| \cdot |K : H \cap K|## so ##n\,\vert \,|G : H \cap K|## and similar ##m\,\vert \,|G : H \cap K|##.

Unfortunately, Lagrange's theorem only applies to finite groups.
 
Bashyboy said:
Unfortunately, Lagrange's theorem only applies to finite groups.
I haven't checked the proof, but on the Wiki page it has been first stated (## |G|=|G : H| \cdot |H| ##) and then appended "Especially for ##|G|<\infty \; \ldots ##" so I assumed that finiteness of ##G## isn't really required.
 

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