In group theory, is it true that $f(X \vee Y) = f(X) \vee f(Y)$? ($\vee$ denotes join).
Let $G$ and $H$ denote Abelian groups, $X$ and $Y$ denote subalgebras of $G$, and let $f : G \rightarrow H$ denote a homomorphism. Then:
$$f(X \vee Y) = f(X+Y) = f(X)+f(Y) = f(X) \vee f(Y)$$
So $f(X \vee Y) = f(X) \vee f(Y)$.
Now suppose we relax our conditions, so that $G$ and $H$ are no longer assumed Abelian.

Question. Is it still provable that $f(X \vee Y) = f(X) \vee f(Y)$?


That we have $f(X\vee Y)\subseteq f(X)\vee f(Y)$ should be clear (the image of a product of elements in $X$ or $Y$ is a product of elements in $f(X)$ or $f(Y)$).
For the other direction, let $g\in f(X)\vee f(Y)$. Then $g=g_1\cdot\cdots\cdot g_n$ where $g_i\in f(X)$ or $g_i\in f(Y)$. Thus each $g_i$ is of the form $f(x_i)$ where $x_i\in X$ or $x_i\in Y$. Thus we can rewrite $g=f(x_1\cdot\cdots\cdot x_n)$. Thus $g\in f(X\vee Y)$.
This approach can be generalized for arbitrary collections of subgroups since it didn't matter that $\{X,Y\}$ was a finite collection. The only thing that was important was that elements were a finite product of other elements. Thus for any collection of subgroups $\cal S$ of $G$, we have that $f(\bigvee_{S\in\cal S} S)=\bigvee_{S\in\cal S}f(S)$.
The dual question is however false. In general, we do not have that $f(X\wedge Y)=f(X)\wedge f(Y)$ for arbitrary subgroups $X$ and $Y$, specifically the inclusion $f(X)\wedge f(Y)\leq f(X\wedge Y)$ need not hold (the other inclusion always holds). I was having trouble finding a counterexample, so I asked here where Marcin Łoś gave a counterexample in the comments. Thus in general, a group homomorphism does not induce a lattice homomorphism between the subgroup lattices.
However, if $f$ is a quotient map (or surjective by the 1st isomorphism theorem), the 4th isomorphism theorem does give us some relation between subgroup intersections.
