I'm making some notes for Number Theory in Function Fields by Rosen, and for the life of me I cant figure out why the following is true (in the notation below $A = \mathbb{F}_{q}[T]$, $P \in A$ is irreducible, and $|P| = q^{\deg(P)}$):
...To deal with the remaining cases, we employ a little group theory. The natural map (Le., reduction modulo $P$) is a homomorphism from $(A/P^{e}A)^{\ast}$ onto $(A/PA)^{\ast}$ and the kernel is a $p$-group as follows from Proposition 1.6. Since the order of $(A/PA)^{\ast}$ is $|P|-1$ which is prime to $p$ it follows that $(A/P^{e}A)^{\ast}$ is the direct product of a $p$-group and a copy of $(A/PA)^{\ast}$.
Why is the statement in bold true? Clearly this would follow if the general fact $G \cong \ker\varphi \oplus \text{im}\varphi$ were true where $\varphi:G \to H$ is a surjective homomorphism. But there are a bounty of examples where this isomorphism does not hold. I'm also not sure why the relative primality of $|P|-1$ and $p$ is important. I'm probably missing something trivial...