# Wreath product (Rotman, p.173)

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?

• I think you are right. The map $Q \to Q^*$ is an isomorphism if and only if the action of $Q$ on $\Omega$ is faithful. Rotman is probably assuming that, but has not said so. Commented Sep 6, 2014 at 8:54
• @DerekHolt Thank you for your comment. I'm relieved to hear that from you.
– Orat
Commented Sep 6, 2014 at 9:07