Let $K$ be a field and suppose that $\phi_n: K^{*} \to K^{*}$ given by $\phi_n(a) = a^n$ is a sujective group homomorphism , for all $n \in \mathbb{N}$. Show that $\frac{K^{*}}{\ker \phi_n} \cong K^{*} \iff \phi_n$ is injective.
$\textbf{My attempt:}$
I know how to solve one side, that is, if $\phi_n$ is injective then $\ker \phi_n = \{1\}$. Then $\frac{K^{*}}{\ker \phi_n} \cong \frac{K^{*}}{\{1\}} \cong K^{*}$.
But how can I solve the otherside? why $\frac{K^{*}}{\ker \phi_n} \cong K^{*}$ implies $\phi_n$ injective?
First of all the $K^*/\ker f\simeq K^*$ isomorphism always exists for surjective homomorphism. This is the first isomorphism theorem. And so your question can be restated as "if $\phi_n$ is surjective then it is injective".
And this is false. Consider $K=\mathbb{C}$ the field of complex numbers. Then $\phi_n$ is always surjective but never injective for $n>1$. Simply because the $X^n=b$ equation always has exactly $n$ solutions over $\mathbb{C}$ when $b\neq 0$.