Question

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})

Abstract algebra
ANSWERED
asked 2021-02-25

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

Answers (1)

2021-02-26

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
\(x\in X\otimes Y\Leftrightarrow x\in (\frac{X}{Y})\cup (\frac{Y}{X}) \Leftrightarrow (x\in X \bar{\wedge x} \not \in Y)\bar{\vee} (x\in Y\bar{\wedge x} \not\in X),\)
and similarly
\(x\in Y\otimes Z  \Leftrightarrow  x\in (\frac{Y}{Z})\cup (\frac{Z}{Y}) \Leftrightarrow  (x\in Y\wedge{x} \not \in Z)\vee\)

\((x\in Z\wedge x \not \in Y).\)

(2) \(x\in Y\otimes Z \Leftrightarrow x\in (\frac{Y}{Z})\cup (\frac{Z}{Y}) \Leftrightarrow (x\in Y\wedge x \not \in Z)\vee(x\in Z\wedge x \not \in Y).\)
(2)Hence

\(x \not \in X\otimes Y  \Leftrightarrow  (x \not \in X\vee x\in Y)\wedge (x \not \in Y\vee x\in X) \)

\( \Leftrightarrow  (x \not \in X\wedge x \not \in Y)\vee (x \not \in X\wedge x\in X)\vee\)

\( (x\in Y\wedge x \not \in Y)\vee (x\in Y\wedge x\in X)  \)

\(\Leftrightarrow  (x\not\in X\wedge x \not \in Y)\vee 0\vee 0\vee (x\in Y\wedge x\in X) \)

\( \Leftrightarrow  (x \not \in X\wedge x \not \in Y)\vee (x\in Y\wedge x\in X).\)

(3) \(x \not \in X\otimes Y \Leftrightarrow (x \not \in X\vee x\in Y)\wedge (x \not \in Y\vee x\in X) \Leftrightarrow\)

\((x \not \in X\wedge x \not \in Y)\vee (x \not \in X\wedge x\in X) \vee (x\in Y\wedge x \not \in Y)\vee\)

\((x\in Y\wedge x\in X) \Leftrightarrow (x\in / X\wedge x \not \in Y)\)

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

\(\Leftrightarrow (x \not \in X\wedge x \not \in Y)\vee (x\in Y\wedge x\in X).\)

(3) And \(x \not \in Y\otimes Z  \Leftrightarrow  (x \not \in Y\vee x\in Z)\wedge (x \not \in Z\vee x\in Y)\)

\( \Leftrightarrow  (x \not \in Y\wedge x \not \in Z)\vee (x \not \in Y\wedge x\in Y)\vee\)

\((x\in Z\wedge x \not \in Z)\vee (x\in Z\wedge x\in Y)  \Leftrightarrow\)

\((x \not \in Y\wedge x \not \in Z)\vee 0\vee 0\vee (x\in Z\wedge x\in Y) \)

\( \Leftrightarrow  (x \not \in Y\wedge x \not \in Z)\vee (x\in Z\wedge x\in Y).\)

(4) \(x \not \in Y\otimes Z \Leftrightarrow (x \not \in Y\vee x\in Z)\)

\(\wedge (x \not \in Z\vee x\in Y) \Leftrightarrow (x \not \in Y\wedge x \not \in Z)\vee\)

\((x \not \in Y\wedge x\in Y) \vee (x\in Z\wedge x \not \in Z)\vee (x\in Z\wedge x\in Y) \Leftrightarrow\)

\((x \not \in Y\wedge x \not \in Z)\vee 0\vee 0\vee (x\in Z\wedge x\in Y)\)

\( \Leftrightarrow (x \not \in Y\wedge x \not \in Z)\vee (x\in Z\wedge x\in Y). \)

(4) Thus \(x\in (X\otimes Y)\otimes Z  \Leftrightarrow  (x\in X\otimes Y\wedge x \not \in Z)\)

\(\vee (x\in Z\wedge x \not \in X\otimes Y)  \Leftrightarrow  (((x\in X\wedge x \not \in Y)\vee\)

\((x\in Y\wedge x \not \in X))\wedge x \not \in Z)\vee (x\in Z\)

\(\wedge ((x \not \in X\wedge x \not \in Y)\vee (x\in Y\wedge x\in X)))  \Leftrightarrow\)

\((x\in X\wedge x \not \in Y\wedge x \not \in Z)\vee (x \not \in X\wedge x\in Y\wedge x \not \in Z)\)

\(\vee (x \not \in X\wedge x \not \in Y\wedge x\in Z)\vee (x\in X\wedge x\in Y\wedge x\in Z)\)

(5) \(x\in (X\otimes Y)\otimes Z \Leftrightarrow (x\in X\otimes Y\wedge x \not \in Z)\vee (x\in Z\wedge x \not \in X\otimes Y) \Leftrightarrow\)

\((((x\in X\wedge x \not \in Y)\vee (x\in Y\wedge x \not \in X))\wedge x \not \in Z) \)

\(\vee (x\in Z\wedge ((x \not \in X\wedge x \not \in Y)\vee (x\in Y\wedge x\in X)))\)

\(\Leftrightarrow (x\in X\wedge x \not \in Y\wedge x \not \in Z)\vee (x \not \in X\wedge x\in Y\wedge x \not \in Z)\)

\( \vee (x \not \in X\wedge x \not \in Y\wedge x\in Z)\vee (x\in X\wedge x\in Y\wedge x\in Z)\)

(5) And \(x\in X\otimes (Y\otimes Z)  \Leftrightarrow  (x\in X\wedge x \not \in Y\otimes Z)\)

\(\vee (x \not \in X\wedge x\in Y\otimes Z)  \Leftrightarrow\)

\(((x\in X\wedge ((x \not \in Y\wedge x \not \in Z)\vee (x\in Y\wedge x\in Z)))\)

\(\vee (x \not \in X\wedge ((x\in Y\wedge x \not \in Z)\vee (x\in Z\wedge x \not \in Y)))\)

\(\Leftrightarrow  (x\in X\wedge x \not \in Y\wedge x \not \in Z)\vee (x\in X\wedge x\in Y\wedge x\in Z)\)

\(\vee (x \not \in X\wedge x\in Y\wedge x \not \in Z)\vee (x \not \in X\wedge x \not \in Y\wedge x\in Z)\)

(6) \(x\in X\otimes (Y\otimes Z) \Leftrightarrow (x\in X\wedge x \not \in Y\otimes Z)\vee (x \not \in X\wedge x\in Y\otimes Z) \Leftrightarrow\)

\(((x\in X\wedge ((x \not \in Y\wedge x \not \in Z)\vee (x\in Y\wedge x\in Z)))\)

\( \vee (x \not \in X\wedge ((x\in Y\wedge x \not \in Z)\vee\)

\((x\in Z\wedge x \not \in Y))) \Leftrightarrow (x\in X\wedge x \not \in Y\wedge x \not \in Z)\)

\(\vee (x\in X\wedge x\in Y\wedge x\in Z) \vee\)

\((x \not \in X\wedge x\in Y\wedge x \not \in Z)\vee (x \not \in X\wedge x \not \in Y\wedge x\in Z)\)

(6) By (5)(5) and (6)(6)

\(x\in (X\otimes Y)\otimes Z  \Leftrightarrow  x\in X\otimes (Y\otimes Z) x\in (X\otimes Y)\otimes Z\Leftrightarrow x\in X\otimes (Y\otimes Z)\)

Hence \((X\otimes Y)\otimes Z=x\in X\otimes (Y\otimes Z)\).

Hence GG is a group. Abelian part of \(\otimes \otimes\) is trivial by definition.

0
 
Best answer

expert advice

Need a better answer?
...