Question

subgroup question

let G be an abelian group and H be a non-empty closed subset of G
prove that

Answer 1

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

Answer 2

@terenzreignz

Answer 3

@myko

Answer 4

what the asterics H* means?

Answer 5

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

Answer 6

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

Answer 7

i mean it is a subgroup

Answer 8

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.

Answer 9

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

Answer 10

i gues multiplication?

Answer 11

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*

Answer 12

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

Answer 13

then PQ^-1 =\[(xy ^{-1})((xy ^{-1}))^{-1}\]

Answer 14

=\[(xy ^{-1})(x ^{-1}y)=(xx ^{-1})(yy ^{-1})\]

Answer 15

@terenzreignz pls help