## anonymous 3 years ago subgroup question let G be an abelian group and H be a non-empty closed subset of G prove that

1. anonymous

$H ^{*}=\left\{ xy ^{-1}:x,y \in H \right\}\le G$

2. anonymous

@terenzreignz

3. anonymous

@myko

4. anonymous

what the asterics H* means?

5. anonymous

it sort of another H (it can be any later ie Q)

6. anonymous

and by ≤ you mean that H* is a subset of G?

7. anonymous

i mean it is a subgroup

8. anonymous

it's vean long time since I did this, but I would try to prove the axioms for the elements of H* to be a abelian group: Closure: For all a, b in H*, the result of the operation a • b is also in H*. Associativity For all a, b and c in H*, the equation (a • b) • c = a • (b • c) holds. Identity element There exists an element e in H*, such that for all elements a in H*, the equation e • a = a • e = a holds. Inverse element For each a in H*, there exists an element b in H* such that a • b = b • a = e, where e is the identity element. CommutativityFor all a, b in H*, a • b = b • a.

9. anonymous

by the way, which operation respect to is this an abelian group?

10. anonymous

i gues multiplication?

11. anonymous

yes that one i get i want to show that 1 H* is not empty ie e element of H* 2 pq^-1 element H* for every p,q element H*

12. anonymous

1 H*$\neq$ 2 let P,Q $\in H ^{*}$we want to show that PQ^-1 element H*

13. anonymous

then PQ^-1 =$(xy ^{-1})((xy ^{-1}))^{-1}$

14. anonymous

=$(xy ^{-1})(x ^{-1}y)=(xx ^{-1})(yy ^{-1})$

15. anonymous

@terenzreignz pls help