Prove intersection of two G-sets are G-set. Let $G$ be a group and $X$ be a set. Then $X$ is said to be a
$G$-set (or a set with operator $G$) if there exists a mapping $\phi : G\times X \to X$
such that for all $a, b \in G$ and $x \in X$ the following conditions are satisfied:
(i) $\phi(ab, x) = \phi(a, \phi(b, x))$,
(ii) $\phi(e, x) = x$,
where $e$ is the identity of $G$. The $G$-set $X$ defined above will be denoted by the pair $(X, \phi)$.
Now, I want to prove this lemma.
Let $X$ and $Y$ be two $G$-sets. Then $X \cap Y$ is a G-set.
Proof.
Let $z\in X\cap Y$ then $z\in X$ dan $z\in Y$. Since $X$ and $Y$ are $G$-sets, it
satisfy:
(i) $\phi(ab, z) = \phi(a, \phi(b, z))$,
(ii) $\phi(e, z) = z$.
Now I confused, how to conclude $X \cap Y$ is a G-set? Anyone please help me.
 A: I think that the important fact you are missing is $X$ and $Y$ cannot be any sets. They have to both be subsets of some larger $G$-set $Z$. For the duration of this answer, I will be writing $g \cdot x$ to mean what you call $\phi(g,x)$. This is purely to make the notation slightly cleaner, and will not change the meaning of anything.
For a proof of this, say $X = \{a,b\}$ and $Y = \{a,c\}$ where both are $\mathbb{Z}/2$-sets with $0$ acting as the identity, and $1$ swapping $a$ and $b$/$c$ respectively in $X$/$Y$. Then $X \cap Y = \{a\}$, but where should $1$ send $a$? Neither of $b$ or $c$ is in $X \cap Y$, so that doesn't make sense. Sure we could send $a$ to itself, but that is somewhat unsatisfying. If we allowed that, then no matter WHAT we do we will be left with a $G$-set by setting
$$g \cdot x = x$$ 
for every $g \in G$ and every $x \in X \cap Y$. In making this move we have "forgotten" all of the information in our original $G$-sets. This is technically correct, but isn't very interesting.

Let's say instead that we have $X, Y \subseteq Z$ with $Z$ a $G$-set and $X,Y$ both stable. That is, let's assume that $X$ and $Y$ are $G$-sets in their own right when we restrict our $G$-action on $Z$ to $X$ or $Y$. 
Formally what the means is that $g \cdot x \in X$ for every $g \in G$ for every $x \in X$. The definition for $Y$ is similar.
Notice we have defined our action on $X$ and $Y$ in terms of a pre-existing action on $Z$. Now we can ask a more insightful question -- can we put a $G$-set structure on $X \cap Y$ which is also "the same" as the action on $Z$, but restricted to $X \cap Y$? It turns out the answer is yes!
If $X$ and $Y$ are both stable in the above sense, then so is $X \cap Y$! Why might this be true? Because if $k \in X \cap Y$, then:
$g \cdot k \in X$ (because $X$ is stable and $k \in X$) and 
$g \cdot k \in Y$ (because $Y$ is stable and $k \in y$).
So we have $g \cdot k \in X \cap Y$, for any $g \in G$ and for any $k \in X \cap Y$, and thus $X \cap Y \subseteq Z$ is also a $G$-set.

Notice the difference between this proof and the earlier (contrived) example. Here we have some ambient $G$-set structure that we are trying to show is preserved under intersection. In the first example, we took the intersection and then came up with a new $G$-set structure. It is much more useful to show that some preexisting structure is preserved, but in order to do that we have to know that the structures on $X$ and $Y$ are related in some way. It turns out the correct relationship is for $X$ and $Y$ to both be sub-$G$-sets of some larger $G$-set $Z$. There is some Universal Algebra or Category Theory lurking under the surface here, but that's a discussion for another day...

I hope this helps ^_^
A: You have two conflicting definitions of a $G$-set. The first is a set where such a function exists. According to that definition, every set is a $G$-set; just choose $\phi(a,x)=x$ for all $a\in G$ and $x\in X$. Your second definition (and the one I think you actually meant) is a set $X$ equipped with a function of that type (that is, the $G$-set consist of both the set and the function, which matches denoting it with the pair (X).
Now that may be nit-picking, but I think this confusion is at the heart of the confusion your question is about.
Namely you are talking about the intersection of two $G$-sets. Taking the second definition, you'd have to define the intersection between $(X,\phi)$ and $(Y,\chi)$.
Written this way, you already see where the problem lies: While obviously the first element of the pair should be the intersection of $X$ and $Y$, we are left with the question what the second element should be.
Now one important property of the intersection of $X$ and $Y$ is that is is a subset of both $X$ and $Y$. Therefore let's first look at the concept f 4 $G$-subset.
The very obvious definition of a $G$-subset is:
Be $(X,\phi)$ a $G$-set, $S\subset X$ and $\phi|_S$ the restriction of $\phi$ to $S$. Then $S$ is a $G$-subset of $X$ if $(S,\phi|_S)$ is a $G$-set.
It is not hard to show that $S$ is a $G$-subset if and only if $\phi(a,S)=S$ for all $a\in S$, that is, for every group elements $\phi$ maps the set $S$ onto itself.
Now let's go back to the intersection of two $G$-sets $(X,\phi)$ and $(Y,\chi)$. Now as mentioned $X\cap Y$ is a subset of both $X$ and $Y$. But we need them to have $G$-subsets in order to be able to define an intersection.
So now assume that $X\cap Y$ is a $G$-subset both of $X$ and of $Y$. Then the corresponding $G$-sets are $(X\cap Y,\phi|_{X\cap Y})$ and $(X\cap Y,\chi|_{X\cap Y})$. Now in general, those will be different. But if they are different, there is no clear way to define the intersection of $G$-sets. Therefore the reasonable thing is to additionally demand that both agree.

So the intersection of two $G$-spaces $(X,\phi)$ and $(Y,\chi)$ exists if and only if $X\cap Y$ is a $G$-subset of each of the $G$-spaces and the restrictions of $\phi$ and $\chi$ agree on that subset. In that case, the intersection of the $G$-sets is $(X\cap Y,\phi|_{X\cap Y})=(X\cap Y,\chi|_{X\cap Y})$.

One way to guarantee that those conditions are fulfilled is if $X$ and $Y$ are themselves $G$-subsets of a larger $G$-set $U$. HallaSurvivor already showed a prove for this special case.
