I have a question about the explanation of wreath product in Rotman's "An introduction to the theory of groups (4th ed.)."
Let $D$ and $Q$ be groups, $\Lambda$ be a $D$-set, $\Omega$ be a finite $Q$-set. For each $q \in Q$, define a permutation $q^\ast$ of $\Lambda \times \Omega$ by $ q^\ast(\lambda, \omega) = (\lambda, q\omega) $ and define $Q^\ast = \{\, q^\ast \mid q \in Q \,\}$.
The book says that the map $Q \to Q^\ast,~q \mapsto q^\ast$ is an isomorphism. The fact that this map is homomorphism and surjective comes almost immediately from its definition. I have a trouble to show that this map is injective.
Here is my attempt. Assume that $q^\ast = \operatorname{id}$. For a fixed $\lambda \in \Lambda$ and for any $\omega \in \Omega$, $$ (\lambda, \omega) = \operatorname{id}(\lambda, \omega) = q^\ast(\lambda, \omega) = (\lambda, q\omega)$$ and conclude that $q = \operatorname{id}_\Omega$. But I think this means that $q$ belongs to the intersection of all the stabilizers and doesn't mean $q = 1_Q$. What am I missing?