Let $A,B$ be two subgroups of $G$. Show that $g(A\cap B)=gA \cap gB$ for all $g \in G$ Let $A,B$ be two subgroups of $G$. Show that $$g(A\cap B)=gA \cap gB$$ for all $g \in G$. I think I have to use inclusion to show it but I have no idea how to start. Can I let $g(a\cap b) \in g(A\cap B)$? But then I don know how to proceed. 
 A: To show that $gA \cap gB \subseteq g(A \cap B)$.  Let $x \in gA \cap gB$.  Then $x \in gA$, $x \in gB$  which implies $x = gw_1 = gw_2$, where $w_1 \in A$ and $w_2 \in B$.  Multiply on the left side by $g^{-1}$ and see that $w_1 = w_2 \in B$, so $x \in g(A \cap B)$.
A: For any fixed $g$, the function $f(x)=gx$ is injective (because inverses exist). It is a general fact about injective functions that $f(A\cap B)=f(A)\cap f(B)$ for any subsets $A$, $B$ of the domain.
A: If $x\in g(A\cap B)$ then there is an element $w\in (A\cap B)$ such that $x=gw$. $$w\in (A\cap B)\Longrightarrow w\in A,~~~w\in B$$ Then  $x=gw$ and $w\in A$ means that $x\in g A$ and  $x=gw$ and $w\in B$ means that $x\in gB$. So $x\in gA\cap gB$. Follow the inverse path from the last deduction. Try to show that we have a common element in $A$ and $B$ which $x=gw,~~x=gw',~~w\in A, w'\in B$
A: $g (A \cap B)$ are elements of the form $gx$ for some $x \in A$, $x \in B$.
$g A \cap g B$ are elements of the form $gx$ for some $x \in A$, $x \in B$.
therefore they are the same.
