Question on group mapping

**dwsmith** · September 2nd 2011, 04:40 PM

I working on a problem I can do since I have done many of this forum. However, I don't understand the group mapping:

G is a group and we are mapping G to G by $g\to g^{-1}$ .

How does this mapping work?

**FernandoRevilla** · September 2nd 2011, 05:09 PM

Originally Posted by dwsmith

How does this mapping work?

I don't understand the exact meaning of your question. The mapping $\psi:G\to G$ , $\psi(g)=g^{-1}$ is well defined (for every $g\in G$ there exists its inverse $g^{-1}$ and it is unique). Besides, $\psi(ab)=(ab)^{-1}=b^{-1}a^{-1}=\psi(b)\psi(a)$ for all $a,b\in G$ .

**dwsmith** · September 2nd 2011, 05:14 PM

Originally Posted by FernandoRevilla

The mapping $\psi:G\to G$ , $\psi(g)=g^{-1}$

Your reply helped. This is what I needed to see.

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