This is a tangent of this question. I wanted to remark in my answer that it is not generally true that given a group homomorphism $f:G\rightarrow H$ and two subgroups $X,Y\leq G$ that $$f(X\cap Y)=f(X)\cap f(Y)$$
However, I am having difficulty coming up with a counter-example.
We know it's true that $f(X\cap Y)\subseteq f(X)\cap f(Y)$. So I want two subgroups that don't satisfy $f(X)\cap f(Y)\subseteq f(X\cap Y)$. I cannot have that either $X$ or $Y$ is contained in the kernel of $f$ else I'll have equality. I also cannot have that $X\subseteq Y$ or $Y\subseteq X$. Further, in a finite group, I cannot have that $X$ and $Y$ have coprime order.
The smallest group that gave me subgroups which satisfied these three conditions was $S_3$. But those quotients gave equality. The next were $Q_8$ and $D_8$, but also had no luck. This is becoming tedious.
Any help in coming up with a counter-example (or proving it) is appreciated. Also, outlining other observations that could cut down the computation is appreciated.
Let $G = \Bbb Z_2 \times \Bbb Z_2$ and $H = \Bbb Z_2$ with:
$f: G \to H$ given by $f(a,b) = a$.
Let $X = \{(0,0),(1,0)\}$ and $Y = \{(0,0),(1,1)\}$.
Then $X \cap Y = \{(0,0)\}$ and $f(X \cap Y) = \{0\}$.
Since $f(X) = \{f((0,0)), f((1,0))\} = \{0,1\}$ and
$f(Y) = \{f((0,0),f((1,1))\} = \{0,1\}$, we have:
$f(X) \cap f(Y) = \Bbb Z_2 \neq \{0\} = f(X \cap Y)$.
I believe this is the minimal counter-example.