Let $G, H$ be two abelian groups and $f : G \to H$ a group homomorphism. Suppose that $G = \langle S \rangle$ and that $f(S) = H' \leqslant H$ a subgroup. In that case we say that $S$ represents $H'$ under $f$.
Suppose this is true for $S \subset G$ and $f : G \to H$. Suppose further that $S = A^3 = A\cdot A \cdot A = \{ abc : a,b,c \in A\}$ where multiplication is done in $G$.
Then I want to conclude that $A$ represents a subgroup of $H$ under $f$ as well.
How can we go about doing that?
$\newcommand{\ZZ}{\mathbb Z}$ (Unless I'm misunderstanding the question) I don't think this is necessarily true. Suppose $G=\ZZ$, $H=\ZZ/3\ZZ$, $f$ is the quotient map, and $S=\{0,1,2,3\}$. Then $S$ generates $G$ and forms a complete residue system modulo $3$, so $f(S)=H$. Now, $S=A\cdot A\cdot A$ where $A=\{0,1\}$, but the image of $A$ is $\{0,1\}$, which is not a subgroup of $H$.