Answer to Question #105373 in Abstract Algebra for Sourav mondal

Question

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.

Answer 1

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

$ab^{-1}\in H$ when ever $a$ and $b$ in $H$ .

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

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

Say $a^i=a^j \space and \space i>j.$ Then $a^{i-j}=e$ and since $a\neq e, i-j>1.$ Thus ,$aa^{i-j-1}=a^{i-j}=e$ and ,therefore,$a^{i-j-1}=a^{-1}.$ But $i-j-1\ge1$ implies $a^{i-j-1}\in H$ and we are done.

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

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

Then $a+b\in N \space \forall \space a,b\in N$

But N is not a group under addition.