Let $D_{2^{n-1}}$ be the Dihedral group of order $2^n$. Letting $s_1:D_{2^{n-1}}\rightarrow \mathbb{Z}_2$ be determined by one of its subgroups isomorphic to $D_{2^{n-2}}$. The book I'm reading says that the pullback given by this map has a normal subgroup $N$ such that when quotiend by it, the resulting group is isomorphic to the semi-dihedral group $QD_{2^{n-1}}$.
What I'm assuming this means, is that we get a fiber product $D_{2^{n-1}}\times_{\mathbb{Z}_2} \mathbb{Z}_4$ where $\mathbb{Z}_4\rightarrow \mathbb{Z}_2$ is the canonical epimorphism.
(Here $\mathbb{Z}_n$ denotes the cyclic group of order $n$.)
What I am wondering is how to find this subgroup $N$. I think one could try to construct an explicit surjection from the fiber product to $QD_{2^{n-1}}$, and then we have the result by the first isomorphism theorem. But this seems like a lot of work. The book usually does some calculations in e.g. Maple. Is there an easy way to see this? My main problem being that while I know explicitly how the fiber product looks like, I find it computationally hard to work with.
The direct product can be defined by the presentation $$\langle x,y,z \mid x^{2^{n-1}}=y^2=z^4=1, y^{-1}xy=x^{2^{n-2}-1}, [x,z]=[y,z]=1 \rangle,$$ and the fibre product is the subgroup $\langle u,y \rangle$ of index $2$, where $u=xz$.
There are three normal subgroups of index $2$, and (as you might expect) factoring out either of the two contained in one of the original direct factors does not give what you want, so it must the third of these, whichn is $\langle x^{2^{n-2}}z^2 \rangle = \langle u^{2^{n-2}}z^2\rangle$.
Now $y^{-1}uy = x^{-1}z$, whereas $$u^{2^{n-2}-1} = (xz)^{2^{n-2}-1} = x^{-1}u^{2^{n-2}}z^{-1} = x^{-1}z(u^{2^{n-2}}z^2) = y^{-1}uy(u^{2^{n-2}}z^2),$$ we see that the quotient group is indeed isomorphic to ${\rm QD}_{2^{n-1}}$.