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By Howie J.M. (ed.)

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G/; then D 0. Suppose that s > t D 1. If D < hai hbi, then G=D is nonmetacyclic. p r ; p s / so metacyclic, and in that case, D p 2 . Now suppose that r > s D 1. G/ is the unique normal subgroup of order p in G such that G=D is not metacyclic so D p 2 C p. Thus, p divides in all cases. 3. Suppose that G D ha; b j a4 D b 4 D c 2 D 1; c D Œa; b; Œa; c D Œb; c D 1i is a nonmetacyclic minimal nonabelian group of order 25 . 1(a) for n D 4. The group G has exactly seven central subgroups of order 2: X1 D ha2 i; X2 D hb 2 i; X3 D ha2 b 2 i; X5 D ha2 ci; X6 D hb 2 ci; X7 D ha2 b 2 ci: X4 D hci; Then G=Xi is metacyclic for i D 3; 4; 5; 6 and nonmetacyclic for i D 1; 2; 7.

If x D a u, then we replace x with xt . xt /2 D xtxt D x 2 t x t D a2 ut ut D a2 . Hence, we may assume from m 3 the start that x 2 D a2 and so the structure of G is uniquely determined in this m 4 case. G/ D ht i ha2 i D G 0 and G D ha; b; xi. Also, xa2 t is an involution in G ha; b; t i. b/. 2 Now let b x D bat . bat /x D bat at u D ba2 u. On the other m 3 1C2m 3 hand, by setting v D a2 , we see that b a D b av D ba2 u. Hence, we may set 2 2 x D av or x D avu. Note that the group hai=hui produces by conjugation exactly 2m 3 distinct conjugates of b.

L/ D A0 is of index 2 in A. a/ D1 . This yields at once that b a D bz and so n 4. b Hence we also have a D az. L=hzi/ Š A hb 0 i, where hb 0 i is a cyclic subgroup of order 4 in L. Hence A hb 0 i is normal in G since L=hzi is G-invariant. If s normalizes A, then as D a 1 z , D 0; 1, since Ahsi=hzi is of maximal class. But from st s D t b we get acting on a: asts D atb which yields az D a. a 1 z /s D az z D a, and this is a contradiction. Hence s does not normalize A. Ahb 0 i=hb 0 i/hsi is of maximal class, one may set (3) as D a (4) as Da 1 0 b or 1C2m 2 b 0.

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