Intersection of subgroups Let $A,B$ be subgroups of a group $G$. We define:
$$C=\{x; x \in A \wedge x \in B \}$$
Prove that $C$ is a subgroup of $G$
My attempt at a proof:
$1)$ Closure: Since A and B are subgroups:
$$x \in A \wedge x \in B \Rightarrow xy \in A \wedge xy \in B$$
$2)$ Associativity: Since the elements of $C$ are also elements of $G$ it follows that they are associative.
$3)$ Existence of an identity element: Since there is only one identity element in any group, and since $e \in A \wedge e \in B$ it follows $e \in C$
I wanted to apply the same logic to the inverse element proof as I did in $3)$, but if I recall correctly groups can have distinct identity elements for distinct elements. Could someone help me with the inverse element proof? Thanks in advance.

Your first step in proving closure needs to consider two elements, the second implication following from the fact that $A, B$ are subgroups and hence closed under the group operation:
$$\begin{align} x \in C, y\in C & \implies x \in A \land x \in B,\; y \in A \land y \in B\; \\ \\
& \implies \;xy \in A \land xy\in B \tag{closure: A, B}\\ \\ & \implies xy \in C\tag{Def. Intersection} \end{align}$$
You need to show closure under taking inverses: If $x \in C$, then $x \in A \land x\in B$. Because $A, B$ are subgroups, they are closed under taking inverses. Hence $x \in A \implies x^{-1} \in A$, and $x^{-1} \in B$. And so $x^{-1} \in A \cap B = C$. Therefore, $C$ is closed under taking inverses.
N.B. Each element has a unique inverse, that is correct (but it is not correct to say each element has a unique/distinct identity: there is one and only one identity element in a group.)  each
