can you rewrite this without latex urjectivity: To show that $\phi$ is surjective, we need to prove that for every $h \in H$, there exists $g \in G$ such that $\phi(g) = h$. Given $h \in H$, since $b$ generates $H$, there exists an integer $k$ such that $h = b^k$. Let $g = a^k$. Then, $\phi(g) = \phi(a^k) = b^k = h$, demonstrating surjectivity