Generating sets of semi-direct products with $\mathbb{Z}_2$

Question

Suppose a group $G$ splits as a semidirect product $N\rtimes\mathbb{Z}_2$, and let $\phi:G\to\mathbb{Z}_2$ the the associated quotient map. If I have a subset of elements $\{g_1,\dots,g_n,h\}$ of $G$ such that $g_i\in N$ for $1\le i\le n$, and $\phi(h)=[1]_2$ which I know a priori generates $G$, can I conclude that $\{g_1,\dots,g_n\}$ generates $N$ as a subgroup of $G$?

I am particularly interested in the case the $G$ is a Coxeter group, so $N$ can be thought of (morally at any rate) as the subgroup of orientation preserving transformations. I can't see how I might prove this, but I also haven't been able to think of a counter example.

Answer 1

A counterexample is the Coxeter Group $$G =W(A_3) = \langle x,y,z \mid x^2,y^2,z^2,(xy)^3,(yz)^3,(xz)^2 \rangle.$$

Then $G$ is isomorphic to $S_4$ with $x,y,z \mapsto (1,2),(2,3),(3,4)$, and the orientation preserving subgroup $N$ corresponds to $A_4$.

Now $G = \langle x,yz \rangle$ (because $S_4$ is generated by $(1,2)$ and $(2,3,4)$), but $N$ is not generated by $yz$.

Answer 2

No, you can't.

Consider $G = \mathbb{Z}^2 \rtimes \mathbb{Z}_2$ where $1 \in \mathbb{Z}_2$ corresponds to the automorphism $\varphi \in \operatorname{Aut}(\mathbb{Z}^2, +)$ such that $\varphi(x, y) = (y, x)$. Then it's easy to see that $\{ g, h \} = \{ ((1, 0), 0), ((0, 0), 1) \}$ generate $G$ but $(1, 0)$ does not generate $\mathbb{Z}^2$.