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

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 $$\displaystyle{X}\oplus{Y}={\left({\frac{{{x}}}{{{y}}}}\right)}\cup{\left({\frac{{{y}}}{{{x}}}}\right)}$$

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.