I am confused about an exercise from Representations of Finite Groups by C.Musili (Exercise 7.10.1), which asks the reader to show that the hyperoctahedral group $B_n$ is a semi-direct product.
First I will include some definitions/notations from the book:
The hyperoctahedral group on $n$ signed letters, denoted $B_n$, is defined as the following subgroup of $S_{ \{1,...,n,-n,...,-1\} } \cong S_{2n}$.
$$ B_n = \{ \theta \in S_{2 n} \mid \theta(-i) = -\theta(i), 1 \leq i \leq n \} $$
Any element $\theta$ of $B_n$ is of the form $$ \theta=\left(\begin{matrix} 1 & \cdots & n & -n & \cdots & -1 \newline \epsilon_1 \sigma(1) & \cdots & \epsilon_n \sigma(n) & -\epsilon_n \sigma(n) & \cdots & -\epsilon_1 \sigma(1) \end{matrix}\right) \tag{1} $$ for unique $\left(\epsilon_1, \cdots, \epsilon_n\right) \in N=C_2^n$ and unique $\sigma \in S_n$, where $C_2=$ $\{ \pm 1\}$ is the cyclic group of order 2 and $S_n$ is the symmetric group on $\{1, \cdots, n\}$.
It is worth noting that permutations are composed from right to left. That is $xy(k) = x(y(k))$.
The exercise is to show that this group is a semi-direct product. (The below is not a direct quote because I have moved in the definition of the semi-direct product from another chapter)
Define the semi-direct product of $H$ by $N$ (denoted $G=N \bullet H$) as the set $G=N \times H$ along with the group multiplication given by $(a, x)(b, y)=(a(x \cdot b), x y)$ for all $a, b \in N$ and $x, y \in H$. Show that $B_n$ is the semi-direct product of $S_n$ by $N$, where $S_n$ acts on $N$ by permuting coordinates.
The only way I can show that the map described in (1) is an isomorphism is by taking the multiplication rule to instead be $(a, x)(b, y)=((y^{-1} \cdot a)b, x y)$.
Am I missing something?