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Let $G$ be a group and $H = \{x^{-1}\mid x\in G\}$. Show $G=H$.

I have showed that $H \subseteq G$.

Can somebody give me a hint to show how an arbitrary $x \in G$ also belongs in $H$?

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    $\begingroup$ Given an arbitrary element $g \in G$, you wish to somehow write it as $g=x^{-1}$. I wonder what $x$ is. $\endgroup$
    – Randall
    Apr 17 '20 at 17:28
  • $\begingroup$ What defining property of G could show that arbitrary element g $\in$ H? $\endgroup$ Apr 17 '20 at 17:32
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Based on the Axiom of Inverse Element, if $a\in G$, then $b\in G$ if $a* b=b* a=e$, where $*$ represents the groupg operation. Hence the rest of the proof goes as follows: $$a\in G\implies a^{-1}\in G\implies (a^{-1})^{-1}\in H\implies a\in H\implies G\subseteq H$$

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  • $\begingroup$ I am still unable to show how $a \in H$. $\endgroup$ Apr 17 '20 at 17:40
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    $\begingroup$ I added more detail. Hope it help.. $\endgroup$ Apr 17 '20 at 17:42
  • $\begingroup$ So this is what @Randall was hinting at. Thanks Mostafa $\endgroup$ Apr 17 '20 at 17:48
  • $\begingroup$ Yes it is. Good luck! $\endgroup$ Apr 17 '20 at 17:49
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Hint: Conside the map $G \to G$ given by $x \mapsto x^{-1}$. Prove that this map is an involution, that is, is its own inverse. Conclude that the map is a bijection and so is a surjection.

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  • $\begingroup$ I have proved that the map is a bijection but I am unable to see your guidance. $\endgroup$ Apr 17 '20 at 17:58
  • $\begingroup$ @KevinDudeja, if the map is a bijection then it is a surjection. $\endgroup$
    – lhf
    Apr 17 '20 at 18:10

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