If U is a set, let G = {X| X subseteq U}. Show that G is an abelian group under the operation oplus defined by X oplus Y = (frac{x}{y}) cup (frac{y}{x})

Here, $G=X:X\subset U,$ where U is any set. Here binary operation defined as follows
$X\oplus Y=(\frac{X}{Y})\cup (\frac{Y}{X}).$
Also, let $\varnothing$ be the empty set and correspond to 0. Suppose $X,Y,Z\in G$. Since $U\in G,$ So, G is non-empty.
Additive identity: $X\oplus \oslash =(\frac{x}{\oslash })\cup \frac{\oslash }{x}=X\cup \oslash =X=\oslash \otimes X.$
Additive inverse: $X\otimes X=(\frac{x}{x})\cup (\frac{x}{x})=\oslash \cup \oslash =\oslash .$
So additive inverse of X is X itself. 4. Associativity of addition: By definition
$x\in X\otimes Y\iff x\in (\frac{X}{Y})\cup (\frac{Y}{X})\iff (x\in X\overline{\wedge x}\notin Y)\overline{\vee }(x\in Y\overline{\wedge x}\notin X),$
and similarly
$x\in Y\otimes Z \iff x\in (\frac{Y}{Z})\cup (\frac{Z}{Y})\iff (x\in Y\wedge x\notin Z)\vee$

$(x\in Z\wedge x\notin Y).$

(2) $x\in Y\otimes Z\iff x\in (\frac{Y}{Z})\cup (\frac{Z}{Y})\iff (x\in Y\wedge x\notin Z)\vee (x\in Z\wedge x\notin Y).$
(2)Hence

$x\notin X\otimes Y \iff (x\notin X\vee x\in Y)\wedge (x\notin Y\vee x\in X)$

$\iff (x\notin X\wedge x\notin Y)\vee (x\notin X\wedge x\in X)\vee$

$(x\in Y\wedge x\notin Y)\vee (x\in Y\wedge x\in X)$

$\iff (x\notin X\wedge x\notin Y)\vee 0\vee 0\vee (x\in Y\wedge x\in X)$

$\iff (x\notin X\wedge x\notin Y)\vee (x\in Y\wedge x\in X).$

(3) $x\notin X\otimes Y\iff (x\notin X\vee x\in Y)\wedge (x\notin Y\vee x\in X)\iff$

$(x\notin X\wedge x\notin Y)\vee (x\notin X\wedge x\in X)\vee (x\in Y\wedge x\notin Y)\vee$

$(x\in Y\wedge x\in X)\iff (x\in /X\wedge x\notin Y)$

$\vee 0\vee 0\vee (x\in Y\wedge x\in X)$

$\iff (x\notin X\wedge x\notin Y)\vee (x\in Y\wedge x\in X).$

(3) And $x\notin Y\otimes Z \iff (x\notin Y\vee x\in Z)\wedge (x\notin Z\vee x\in Y)$

$\iff (x\notin Y\wedge x\notin Z)\vee (x\notin Y\wedge x\in Y)\vee$

$(x\in Z\wedge x\notin Z)\vee (x\in Z\wedge x\in Y) \iff$

$(x\notin Y\wedge x\notin Z)\vee 0\vee 0\vee (x\in Z\wedge x\in Y)$

$\iff (x\notin Y\wedge x\notin Z)\vee (x\in Z\wedge x\in Y).$

(4) $x\notin Y\otimes Z\iff (x\notin Y\vee x\in Z)$

$∧(x∉Z∨x∈Y)⇔(xDo you have a similar question?Recalculate according to your conditions!RecalculateNew Questions in Abstract algebraHow to find out the mirror image of a point? Generators of a free groupIf G is a free group generated by n elements, is it possible to find an isomorphism of G with a free group generated by n-1 (or any fewer number) of elements? How many 3/4 Are in 1Convert 10 meters to feet. Round your answer to the nearest tenth 6. Reduce the following matrix to reduced row echelon form:                                                                                                                                                               Let v be a vector over a field F with zero vector  0 and let s,T be a substance of V .then which of the following statements are falseDescribe Aut(Zp), the automorphism group of the cyclic group Zp where p is prime. In particular find the order of this group. (Hint: A generator must map to another generator)Ask your question. Get an expert answer.Let our experts help you. Answer in as fast as 15 minutes.Ask a QuestionQuestionAnswer & ExplanationNew QuestionsDidn't find what you were looking for?ProductHigh School QuestionsCollege QuestionsMath SolverTop QuestionsTop Questions 2CompanyAbout usTerm of ServicePrivacyPayment PolicyHonorConnect with usContact usYoutubeFacebookGet Plainmath AppGoogle PlayE-mail us: [email protected]Our Service is useful for:Plainmath is a platform aimed to help users to understand how to solve math problems by providing accumulated knowledge on different topics and accessible examples.2023 Plainmath. All rights reserved $