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

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 $$H$$?

• Given an arbitrary element $g \in G$, you wish to somehow write it as $g=x^{-1}$. I wonder what $x$ is. Apr 17 '20 at 17:28
• What defining property of G could show that arbitrary element g $\in$ H? Apr 17 '20 at 17:32

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: $$a\in G\implies a^{-1}\in G\implies (a^{-1})^{-1}\in H\implies a\in H\implies G\subseteq H$$

• I am still unable to show how $a \in H$. Apr 17 '20 at 17:40
• I added more detail. Hope it help.. Apr 17 '20 at 17:42
• So this is what @Randall was hinting at. Thanks Mostafa Apr 17 '20 at 17:48
• Yes it is. Good luck! Apr 17 '20 at 17:49

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.

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