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The s u b g r o u p elements powers a fourth- R, one has -+I glgB=gl E zeZ(R) , g in- so that one has the d e f i n i n g -I gl =g2 =i, of with the same g2 do not commute, R g3 is a third R = QE g2=i, (Io) as the q u a t e r n i o n For and Z(R), Such a grou p is k n o w n which g4=e suppose be in Q= 2 gl =g2 =e, g3 from square and, since these 2 -2 g =g that they are 2 2 2 = g z = iz , a subgroup 4 the same it follows cannot is an involution. , g~Z(R) all e l e m e n t s qcQ, if R is of the an i n v o l u t i o n q2=i, an index zeZ(R); and in this two s u b g r o u p of C(g)= R latter R- since 52 the quaternion group contains exactly six different fourth-order elements, C(g) is one of the three Abelian subgroups of cI=®K(2 q) , c2=eK (2q), R: c 3=®K (2q) , (16) with C1C1C2=C1cLC3=C2rlC3=C1Rc2Nc3=Z(Q)@K(2q) =K(2q+l ).

Since eq. all togther group it is e a s i l y form of eq. (15) : Any checked imply that from R o__rrit is a p r o d u c t g=qz with C(g)=sK(2q) , w h i c h (6) holds is either is indeed R K(2q) : (15i that eq. g Hence, (6) shows (e,i}sK(2q). Z(R) = Conversely, case, (I 2) from eq. R m u s t be normal, to b e l o n g is a d i r e c t holds 8 (1). N o w if group of order -1 consists that (11) -+I of those e l e m e n t s Q. It follows glfg2 this 3 =g2g I (IB) -I For they , of g2 then relations g2g3=g2 four p o s s i b i l i t i e s , of any e l e m e n t g3 this equation.

21) 53 Step in eq. (b). Let R now be a regular group with an automorphism (4). The permutation Tg (I) =g -+1 (I) iff T ~ as defined by ~ (g) = g -+1 (22) is such that TgT = ~(g) holds; (23) this permutation keeps all graphs of the MI of R invariant: TGk=T{g (I) ,ggk(1) }={e(g) (I) ,~(ggk ) (I) }={~(g) (I) ,~(g)~(gk ) (I) }= ={g(1) ,g~(gk) (I) }={g(1),gg k NOW a subgroup T T = {glgsR, s~T; ±I (24) (I) }=Gk={g(1),ggk(1)}. is defined by (25) ~(g)=g}. Let tsT, then the product not. In the first case, either belongs ~(ts)=e(t)e(s)=ts -I volut ion, which must be in second case applies, ts to T holds and T- This contradiction ~(ts)=s-lt-l=e(t)~(s)=ts-l; or it does s shows, is an inthat only the this leads to the re- quirement tst = S Now let for all s I and te T, s~ T.

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