Ques-1-Let-G-be-a-group-and-b-a-fixed-element-of-G-Prove-that-the-map-G-into-G-given-by-x-bx-is-bijective-Ques-2-Let-G-be-a-group-and-g-be-an-element-of-G-Prove-that-a-g-1-1

Question Number 193711 by Mastermind last updated on 18/Jun/23

Ques . 1

Let G be a group and b a fixed element

of G . Prove that the map G into G given

by x \to bx is bijective

Ques . 2

Let G be a group and g be an element

of G . Prove that

a) {(g^{- 1})}^{- 1} = g

b) g^{m} g^{n} = g^{m + n}

Help!

Answered by Rajpurohith last updated on 19/Jun/23

(1) G b e a g r o u p, b \in G b e f i x e d .

d e f i n e f : G \to G b y f (x) = b x \forall x \in G

c l e a r l y f i s w e l l d e f i n e d .

s u p p o s e f (c) = f (d) f o r c, d \in G

\Rightarrow b c = b d, b y c a n c e l l a t i o n p r o p e r t y c = d .

s o f i s a n i n j e c t i v e f u n c t i o n f r o m G t o i t s e l f .

l e t y \in G s o f (b^{- 1} y) = {b b}^{- 1} y = y

H e n c e f i s s u r j e c t i v e .

T h u s f i s b i j e c t i v e .

(2) (a) l e t G \in g s a y h = g^{- 1} \Rightarrow h^{- 1} = {(g^{- 1})}^{- 1}

\Rightarrow h g = e

\Rightarrow h^{- 1} (h g) = h^{- 1}

\Rightarrow (h^{- 1} h) g = h^{- 1}

\Rightarrow g = h^{- 1} = {(g^{- 1})}^{- 1} ∴ {(g^{- 1})}^{- 1} = g

(b) g^{m} . g^{n} = {(g . g . g \dots g)}_{m t i m e s} {(g . g . g \dots g)}_{n t i m e s}

= {(g . g . g \dots g)}_{m + n t i m e s} = g^{m + n}

Commented by Mastermind last updated on 20/Jun/23

Thank you so much