Let G be a finite group,f be an automorphism of G such that f(x)=x ⇒x=e . Then prove that, (i)∀g∈G, ∃x∈G such that g=x^(−1) f(x). (ii)If ∀x∈G , f(f(x))=x ⇒ G is Abelian.


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Question Number 193796 by Rajpurohith last updated on 20/Jun/23

$L e t G b e a f i n i t e g r o u p, f b e a n a u t o m o r p h i s m o f G$ $s u c h t h a t f (x) = x \Rightarrow x = e .$ $T h e n p r o v e t h a t,$ $(i) \forall g \in G, \exists x \in G s u c h t h a t g = x^{- 1} f (x) .$ $(i i) I f \forall x \in G, f (f (x)) = x \Rightarrow G i s A b e l i a n .$
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