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Let $G$ be a group and let $A$ be a non empty subset of $G$. Let $H$ be a set defined by $H = \{ x \in G \mid \text{For all }a \in A,\text{ we have }xa \in A\text{ and }x^{-1}a \in A \}$

Show that $H$ is a subgroup of $G$.

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    @anon And that is contains the identity.2012-07-08

1 Answers 1

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Note that $H$ contains $e$, since for all $a\in A$, $ea = a\in A$ and $e^{-1}a=a\in A$.

Now assume that $x,y\in H$. That means that for each $a\in A$, $xa\in A$ and $x^{-1}a$, and $ya\in A$ and $y^{-1}a\in A$. We want to show that $xy^{-1}\in H$.

What do we need in order for $xy^{-1}$ to be in $H$? We need it to be the case that if $a\in A$, then both $xy^{-1}a$ and $(xy^{-1})^{-1}a\in A$.

I'll do the first one: why is $(xy^{-1})a\in A$? Well, we know $y^{-1}a=a'\in A$ because $y\in H$. And since $y^{-1}a=a'\in A$, then $xa'\in A$, because $x\in H$. So $xy^{-1}a = xa'\in A$.

I'll let you do the second one.