what are all finite subgroups of $\mathbb{Z}^n \rtimes \mathbb{Z}_2$? $\mathbb{Z}^n \rtimes \mathbb{Z}_2$ := $\{(u_1,u_2....,u_n,t), u_iu_j=u_ju_i, tu_jt=u_j^{-1},t^2=1\}
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*what are all finite subgroups of $\mathbb{Z}^n \rtimes \mathbb{Z}_2$? (in terms of the generators $u_1,u_2....,u_n,t$)

*what are all finite maximal subgroups up to conjugacy for the same $\mathbb{Z}^n \rtimes \mathbb{Z}_2$? (in terms of the generators $u_1,u_2....,u_n,t$)
 A: It's easy to check that the nontrivial elements in $\Bbb Z^n$ have infinite order, any element outside of it has order two, and any two elements outside of it multiply to get a nontrivial element of $\Bbb Z^n$.
Therefore, the only finite subgroups are of the form $\langle g\rangle=\{e,g\}$ with $g$ not in $\Bbb Z^n$.
Explicitly, given your presentation of $\Bbb Z^n\rtimes C_2$, every element can be written in the standard form as $u_1^{e_1}\cdots u_n^{e_n}t^\epsilon$ with $e_1,\cdots,e_n\in\Bbb Z$ and $\epsilon\in\{0,1\}$. The $g$s mentioned are those with $\epsilon=1$.
Since all nontrivial finite subgroups have order $2$, they are all maximal among finite subgroups, but when are they conjugate? Note $\langle g\rangle\sim\langle h\rangle$ iff $g\sim h$ in this case, so we can instead ask when two elements in standard form are conjugate. For convenience, we can write elements as $u$ and $ut$, where $u\in\Bbb Z^n$ is a vector. Then $x(ut)x^{-1}=xuxt$ and $(xt)ut(xt)^{-1}=xtuttx^{-1}=xu^{-1}xt$. Thus the conjugates of $ut$ are $vt$ with $v$ in the same coset of $2\Bbb Z^n$ as $u$. Therefore, the subgroups
$$\{{\rm identity},u_1^{\epsilon_1}\cdots u_n^{\epsilon_n}t\}$$
for $\epsilon_1,\cdots,\epsilon_n\in\{0,1\}$ chosen independently are pairwise nonconjugate and every finite subgroup is conjugate to one of them. This is the full classification for question 2.
