# Complement of the kernel of a group epimorphism of a semidirect product

Currently, I am going through Chapter 25 of Character theory of finite groups by Bertram Huppert.

Here the authors introduce the concept of semidirect product $G \wr H$ and the kernel of its epimorphism as follows.

$$G \wr H = \{(f, h) | h \in H, f:\Omega \to H\}$$

and the product is

$$(f_1, h_1) (f_2, h_2) = (f, h_1 h_2)$$

where

$$f (j) = f_1(j) f_2(jh_1)$$ for $j \in \Omega$.

Now, $\alpha$ with $(f, h) \alpha = h$ is an epimorphism of $G \wr H$ onto $H$ and $$\ker \alpha = B = \{(f, 1) | f: \Omega \to G\}$$ is isomorphic to $G \times \underset{n}{\ldots} \times G$.

$B$ has a complement $$\overline{H} = \{(e, h) | h \in H\}$$ in $G \wr H$, where $e(j) = 1$ for all $j \in \Omega$ and $\overline{H} \cong H$.

My question:

Why is the complement of $B$ is not $\overline{B}$?