Suppose you are working with an abelian Galois group $G=G(L/K)$ of the Galois extension $L/K$. You know the Decomposition group is:
$D=D(Q,P) = \{ \sigma \in G : \sigma(Q) = Q \}$
where $Q$ (in $L$) is a prime lying over $P$ (in $K$).
I'm going to prove that $D$ does not depend on $Q$.
Proof: Since we are working with Galois extensions, then, if $Q$ and $Q'$ are two primes lying over $P$, then there exists $\tau \in G$ s.t. $\tau(Q) = Q'$. Then $D(Q',P)=D(\tau(Q),P)$. Moreover, $\sigma \in D(\tau(Q),P)$ iff $\sigma(\tau (Q)) = \tau(Q)$. Applayin $\tau^{-1}$ to both side of the identity yields $(\tau^{-1} \sigma \tau) Q= Q$ iff $(\tau^{-1} \sigma \tau) \in D(Q,P)$. The inverse inclusion is similar. Finally we have
$ D(Q',P) = \tau^{-1} D(Q,P) \tau$.
Since $G$ is abelian, then
$ D(Q',P)=D(Q,P)$
(I wanted to stress the fact that in general $ D(Q',P) = \tau D(Q,P) \tau^{-1}$).
Can this work?
You know that the Galois group of the extension acts transitively on the primes lying over a fixed prime $P$. Then the decomposition group is the stabilizer of this action. We conclude that (since we have just one orbit) the decomposition group remains the same whenever we take different prime lying over $P$. Obviously this is not true if we pick a prime lying over another prime different from $P$.