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walters
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subgroup question
let G be an abelian group and H be a nonempty closed subset of G
prove that
 one year ago
 one year ago
walters Group Title
subgroup question let G be an abelian group and H be a nonempty closed subset of G prove that
 one year ago
 one year ago

This Question is Closed

walters Group TitleBest ResponseYou've already chosen the best response.0
\[H ^{*}=\left\{ xy ^{1}:x,y \in H \right\}\le G\]
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.0
@terenzreignz
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.0
what the asterics H* means?
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.0
it sort of another H (it can be any later ie Q)
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.0
and by ≤ you mean that H* is a subset of G?
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.0
i mean it is a subgroup
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.0
by the way, which operation respect to is this an abelian group?
 one year ago

myko Group TitleBest ResponseYou've already chosen the best response.0
i gues multiplication?
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.0
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*
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.0
1 H*\[\neq \] 2 let P,Q \[\in H ^{*}\]we want to show that PQ^1 element H*
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.0
then PQ^1 =\[(xy ^{1})((xy ^{1}))^{1}\]
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.0
=\[(xy ^{1})(x ^{1}y)=(xx ^{1})(yy ^{1})\]
 one year ago

walters Group TitleBest ResponseYou've already chosen the best response.0
@terenzreignz pls help
 one year ago
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