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.
 A: 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
A: Simply:


*

*The neutral element $e\in A\cap B$ so $A\cap B\ne \emptyset$

*For all $x,y\in A\cap B$, $xy^{-1}\in A\cap B$ since $A$ and $B$ are subgroups of $G$
hence $A\cap B$ is a subgroup of $G$.
A: Let $x \in C$. Thus  $x \in A$ and $x \in B$. But $A$ and $B$ are subgroups of $G$ and thus $x^{-1} \in A$ and $x^{-1} \in B$. Thus, since $x^{-1}$ is an element of both $A$ and $B$, $x^{-1} \in C$. 
A: First, a quick fix to your closure argument: $$x,y\in C\Rightarrow x,y \in A \wedge x,y \in B \Rightarrow xy \in A \wedge xy \in B\Rightarrow xy\in C$$
You can in fact use the same sort of reasoning in $3$--inverses are unique--so that $$x\in C\Rightarrow x \in A \wedge x \in B \Rightarrow x^{-1} \in A \wedge x^{-1} \in B\Rightarrow x^{-1}\in C.$$
