Noether’s problem for the groups with a cyclic subgroup of index 4

Let G be a finite group and k be a field. Let G act on the rational function field k(xg : g ∈ G) by k-automorphisms defined by g ∙ xh = xgh for any g, h ∈ G. Noether's problem asks whether the fixed field k(G) = k(xg : g ∈ G)G is rational (i.e., purely transcendental) over k. Theorem 1. If G is a group of order 2n (n ≥ 4) and of exponent 2e such that (i) e ≥ n − 2 and (ii) \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ {\zeta_{{{2^{{e - 1}}}}}} \in k $$\end{document}, then k(G) is k-rational. Theorem 2. Let G be a group of order 4n where n is any positive integer (it is unnecessary to assume that n is a power of 2). Assume that (i) char k ≠ 2, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$ {\zeta_n} \in k $$\end{document}, and (ii) G contains an element of order n. Then k(G) is rational over k, except for the case n = 2 m and G ≃ Cm ⋊ C8 where m is an odd integer and the center of G is of even order (note that Cm is normal in Cm ⋊ C8); for the exceptional case, k(G) is rational over k if and only if at least one of −1; 2;−2 belongs to (k×)2.