Answer to Question #134922 in Abstract Algebra for GIRISH CHANDRA

Question

Let G be a group and H be a non empty finite subset of G.If ab belongs to H for all a,b prove that H is a subgroup of G.will the result be true if H is not finite?

Answer 1

Recall Subgroup test :

Let $G$ be a group and let $H$ be a nonempty subset of $G$ . If $ab$ is in $H$

whenever $a$ and $b$ are in $H$ ($H$ is closed under the operation), and $a^{-1}$

is in $H$ whenever $a$ is in $H$ ($H$ is closed under taking inverses), then $H$

is a subgroup of $G$ .

Solution:

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

$a\in H$ .If $a=e$ then $a^{-1}=a$ and we are done.

So assume that $a\neq e$ . Consider the sequence $a,a^2,a^3,a^4,.........$ By closer all the elements belong to $H$ .

Since $H$ is finite , not all of these elements are distinct.

Say $a^i=a^j$ where $j>i$

Therefore , $a^{j-i}=e$ ; and since $a\neq e$ , $j-i>1$ .

Thus $aa^{j-i-1}=a^{j-i}=e$ .

Therefore,$a^{j-i-1}=a^{-1}$ .

But $j-i-1\geq 1$ , implies $a^{j-i-1}\in H$ and we are done .

If $H$ is not finite then result is not true.

Consider the group $G=(\Z,+)$ and $H=\N$ . Then $a+b\in H , \forall a,b \in H$ but $\N$ is not a subgroup of $\Z$ .