Checking understanding on proving uniqueness of identity and inverse elements of a group. Sorry for such a trivial question, but just wanted to check my understanding.
When proving a statement, for example, that the inverse of a group element is unique (in elementary group theory) one starts by supposing that there exists two inverses $h$ and $k$ for a given element $g \in G$, where $G$ is some group with binary operation $\ast:G\times G \rightarrow G$, such that $h\ast g = g\ast h = e$  and $k\ast g = g\ast k = e$ where $e \in G$ is the unique identity for the group $G$. From this, one can show that $h=k$ in the following manner: $$ h=h\ast e =h\ast\left( g\ast k\right) = \left(h\ast g\right)\ast k = e\ast k =k$$ and as such $h=k$. 
Now, is the reason we can from this state that the inverse of an element $g\in G$ is unique because $h$ and $k$ were chosen arbitrarily, apart from the requirement that they are inverses of $g$, and as such, if we know the value of one of the inverses, say $h$, then we know that for any other value $k$ to be an inverse of $g$ it must be equivalent to the known inverse $h$, and thus $h$ is the unique inverse of $g$. Is this reasoning correct? 
Sorry for the wordiness of this question, but just wanted to check my understanding explicitly. Also, although I understand that, logically, I should have proven this first (but I'd already written out the inverses part before thinking of this, so apologies for that), but is the reasoning the same for arguing that the identity element of a group is unique? (i.e. If we assume that there are two identity elements $e,g \in G$ and subsequently show that $e=g$, then this implies that if we know the form of one of the identities, say $e$, then for any other value $g$ to be an identity of the group $G$ it must be equivalent to $e$ and thus the identity element of a group is unique). 
 A: Group axioms tell you that an identity element must exist, and also that every element has an inverse. They don't tell you that there's only one identity element, and they don't tell you that an element can have only one inverse: these are things that you have to prove.
So, if for example you want to prove that there is only one identity element, suppose instead that there are more than one. If so, you can choose two of them that are not the same element: yet as the proof goes on you see that those elements are instead the same one, as you know. This means that supposing that there are lots of identity elements yields a contradiction: hence, since you can't have more than one of them, and since at least one has to exist because of the group axioms, you see that the only possible option is that there's exactly only one identity element.
The same happens when you're trying to show that every element of the group has only one inverse.
A: Suppose $x\in G$ has two inverses $y$ and $z$ then $$zx=yx\Rightarrow (zx)y=(yx)y\Rightarrow z(xy)=y(xy)\Rightarrow ze=ye$$ So $y=z.$ 
