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 $f(X \vee Y) = f(X) \vee f(Y)$?


Here is how I see it (without reference to "messy" product representations).
If we partially order the lattice of subgroups of $G$ by inclusion, then $H \vee K = \langle H,K\rangle$, the subgroup generated by the set $H \cup K$. By definition this is the smallest subgroup (via our partial order) of $G$ that contains both $H$ and $K$.
Now for any function $f:A \to B$, if $X \subseteq Y \subseteq A$, we certainly have: $f(X) \subseteq f(Y)$.
Homomorphisms are functions, so we have $f(H) \subseteq f(\langle H,K\rangle)$, and similarly for $f(K)$ (since $H,K \subseteq \langle H,K\rangle$).
This shows that $\langle f(H),f(K)\rangle \subseteq f(\langle H,K\rangle)$, by minimality (there is a "hidden appeal" here to the fact $f$ is a homomorphism, so we may conclude $f(\langle H,K\rangle)$ is a group, as our property of "smallest subgroup that contains the generating set" only applies to elements of the subgroup lattice, not arbitrary subsets of $f(G)$).
Now suppose $L$ is ANY subgroup of $f(G)$ containing $f(H)$ and $f(K)$. It follows that $f^{-1}(L)$ is a subgroup of $G$ containing $H$ and $K$ (this is again where we invoke the fact that $f$ is a homomorphism), so:
$\langle H,K\rangle \subseteq f^{-1}(L)$
Thus $f(\langle H,K\rangle) \subseteq f(f^{-1}(L)) = L$.
Since $\langle f(H),f(K)\rangle$ is such a subgroup, we have: $f(\langle H,K\rangle) \subseteq \langle f(H),f(K)\rangle$, and equality is established.
