Answer to Question #105373 in Abstract Algebra for Sourav mondal

Question #105373
Let G be a group and H be a non-empty finite subset of .G If ab∈H∀ b,a ∈ H then show that H ≤ .G Will the result remain true if H is not finite? Give reasons for your answer.
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Expert's answer
2020-03-13T10:36:07-0400

We known that , A nonempty subset H of a group G is a subgroup if

ab1Hab^{-1}\in H when ever aa and bb in HH .

In view of the above theorem ,we need only prove that a1H whenever aHa^{-1}\in H \space whenever \space a\in H .

If a=e,a=e, then a1=aa^{-1}=a and we are done.If aea\neq e consider the sequence. a,a2,a2,........a,a^2,a^2,........ By closure ,all of these elements belongs to HH .Since HH is finite ,not all of these elements are distinct.

Say ai=aj and i>j.a^i=a^j \space and \space i>j. Then aij=ea^{i-j}=e and since ae,ij>1.a\neq e, i-j>1. Thus ,aaij1=aij=eaa^{i-j-1}=a^{i-j}=e and ,therefore,aij1=a1.a^{i-j-1}=a^{-1}. But ij11i-j-1\ge1 implies aij1Ha^{i-j-1}\in H and we are done.

If H is not finite then we have a counter example.

Let G=ZG=Z (set of integer) group under addition and H=NH=N (set of natural number).

Then a+bN  a,bNa+b\in N \space \forall \space a,b\in N

But N is not a group under addition.


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