Let $(A, \cdot), (B, \ast)$ be groups, and $\phi: A \mapsto B$ a homomorphism. Then: $$ S \leq A \Rightarrow \phi(S) \leq B $$
I think this must be a false statement, because there is no guarantee that $\textrm{Ker}(\phi) \subset S$. Yet, I have written the following proof for this statement:
\begin{align*} &\quad\; S \leq A \\ &\{\text{definition of subgroup}\} \\ &\Leftrightarrow \forall s, t \in S \mid s \cdot t \in S \;\wedge\; s^{-1} \in S \\ &\{\text{definition of set inclusion}\} \\ &\Leftrightarrow \forall s, t \in S \mid (\exists u \in S \mid s \cdot t = u) \;\wedge\; (\exists v \in S \mid v = s^{-1}) \\ &\{\text{(problematic step?) $\phi$ is well-defined, so $x = y \Rightarrow \phi(x) = \phi(y)$}\} \\ &\Rightarrow \forall \phi(s), \phi(t) \in \phi(S) \mid (\exists \phi(u) \in \phi(S) \mid \phi(s \cdot t) = \phi(u)) \;\wedge\; (\exists \phi(v) \in \phi(S) \mid \phi(v) = \phi(s^{-1})) \\ &\{\text{definition of set inclusion}\} \\ &\Leftrightarrow \forall \phi(s), \phi(t) \in \phi(S) \mid\phi(s \cdot t) \in \phi(S) \;\wedge\; \phi(s^{-1}) \in \phi(S) \\ &\{\text{$\phi$ is a homomorphism, thus preserves the product, and inverse}\}\\ &\Leftrightarrow \forall \phi(s), \phi(t) \in \phi(S) \mid\phi(s) \ast \phi(t) \in \phi(S) \;\wedge\; {\phi(s)}^{-1} \in \phi(S) \\ &\{\text{definition of subgroup}\} \\ &\Leftrightarrow \phi(S) \leq B \end{align*}
I think the (weakest) correct statement should be:
$$ \textrm{Ker}(\phi) \subset S \;\wedge\; S \leq A \Rightarrow \phi(S) \leq B $$
Is this the case? If yes, how can I correct my proof to capture this issue? If not, why not?