I Is the identity of a group unique?

[Lebnm]

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.

 

[fresh_42]

To be exact, one first proves that left and right identity are the same: $e_lg=ge_r=g \Longrightarrow e_l=e_r$ and that there cannot be two identities, i.e. we have to show $e=e'$. Then you can conclude that the right inverse is equal to the left inverse. (You wrote $eg=ge$ which you cannot know from the start, then you concluded $gg^{-1}=e$ which is also not known until the uniqueness of $e$ has been proven.)

Another way to define a group is by demanding that the equations $ax=b$ and $xa=b$ can uniquely be solved (IIRC).

 

[WWGD]

A nice exercise in this area is to show that the identity of a subgroup must be the same as that of the supergroup. And that a subset S is a group iff for all a,b in S, $ab^{-1}$ is also in S.