Normal subgroups of even index in an NEC group

Large Automorphism Groups of Hyperelliptic Klein Surfaces

“…Then we have an epimorphism 0i from F"' onto G] with kernel fi. If Ti has signature (1), 0i is defined by #i(ci0) = x, 0i(cn) = !> 0i(ci2) = 1, 0i(ci2,) = y, 0i(cii) = z, 0i(cis) = x; if y = z = 1, then Gi = Z2, and so 2(p -1) = 2, p = 2; if y = 1, z ^ 1, then we call r the order of xz and the number of period-cycles of ker 0lt 1, equals 2(p -l)/2r [7, §3]; so r -p -1; and the number of periods of the period-cycle, 2k, equals Ar [7], and so 2r = fc. Hence 2(p -1) = fc and p = fc/2 -f 1; since in case (1) p = fc -1, we obtain p = 3.…”

“…As then 6 would have to divide 16 [7], this is impossible. If Ti has signature (2), 0i is given by 0i(ciO) = x, 0i(cn) = 1, 0i(ci2) -y, 9i(ciz) = z, 0i(ci4) = x.…”

“…The kernel of 0 is orientable [8], and the signature of T = ker 0 is (0, +, [-],{(-), -k-, (-)}) [3,7].…”

On Nilpotent Groups of Automorphisms of Klein Surfaces

“…Without loss of generality we can assume that 9(cx) = 1 . Then 9(c2) ^ 1, since otherwise Y has nontrivial period cycle by [5]. As a result 9(c0) = x, 9(cx) = l, 9(c2)=y, 9(c}) = z, where x, y, and z are elements of order 2 such that yz and xz have orders 2 and k , respectively.…”

“…First notice that 6(c0) £ 1 and 9(c3) / 1, since otherwise Y would have nonempty period cycle by [5]. Moreover since Y has a boundary it contains one of the remaining reflections.…”