- #1
Lebnm
- 31
- 1
In general, the textbooks says that, if the set ##G## is a group, so to every element ##g \in G## there is other element ##g^{-1} \in G## such that ##g g^{-1} = g^{-1}g = e##, where ##e## is the identity of the group. But I am reading a book where this propriete is write only as ##g^{-1} g = e##, and the book says that ##g g^{-1} = e## follows from this. The proof it gives is: Applaying ##g## on the left on the both sides, we have $$g(g^{-1} g) = (g g^{-1}) g = g e = g,$$and of this the book concludes that ##g g^{-1}## is equal to ##e##, because its action in ##g## gives ##g##.
Is it correct? To me, it was necessary to proof also that ##g(g g^{-1}) = g##, because with this I could use the fact that the identity of a group is unique.
Is it correct? To me, it was necessary to proof also that ##g(g g^{-1}) = g##, because with this I could use the fact that the identity of a group is unique.