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

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[H ^{*}=\left\{ xy ^{1}:x,y \in H \right\}\le G\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what the asterics H* means?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it sort of another H (it can be any later ie Q)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and by ≤ you mean that H* is a subset of G?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i mean it is a subgroup

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it'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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by the way, which operation respect to is this an abelian group?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i gues multiplication?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes 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*

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.01 H*\[\neq \] 2 let P,Q \[\in H ^{*}\]we want to show that PQ^1 element H*

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then PQ^1 =\[(xy ^{1})((xy ^{1}))^{1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0=\[(xy ^{1})(x ^{1}y)=(xx ^{1})(yy ^{1})\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@terenzreignz pls help
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