can you rewrite this without latex Injectivity: To show that $\phi$ is injective, we need to demonstrate that if $\phi(a^k) = \phi(a^m)$, then $k = m$. Assume $\phi(a^k) = \phi(a^m)$, which means $b^k = b^m$ in the other group. Since $b$ generates $H$, $b^k = b^m$ implies $k \equiv m \pmod{n}$, where $n$ is the order of the cyclic groups. Hence, $k = m$, showing injectivity