By Wilson J.S.

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**Example text**

G=D/0 is cyclic. G/ D 2. N (GN is defined in (ii1)). G/j D 25 . G/g; then jN0 j is odd (Sylow) and N0 Â 1 M. 1 M (otherwise, jMj D j1 j jN0 j is even). M [ N0 /; then GN D G=U1 is not metacyclic. G/j D 12 25 D 24 . If V < H1 is a member of the set 1 , then G=V is nonmetacyclic. Let N1 D fU0 2 1 N0 j U0 < H1 g. Then jN0 [ N1 j D jN0 j C jN1 j, the number of members of the set 1 contained in H1 , is odd (Sylow). It follows that the number 2 so jN1 j is even since the number jN0 j is odd. Since U1 2 N1 , we obtain jN1 j jN0 [N1 j 1C2 D 3.

S /. S /; which contradicts the maximality of S . bav/ so that hP i D S and P Â G T . t 0 /j D 2m ; a contradiction. S /; a contradiction. bav/. x0 /j D 2m , a contradiction. S / does not fuse bv and bav. bav/. x0 /j D 2m , a contradiction. S / D G and so jG W S j D 2, as required. S / to t c 2 L but S ¤ G. S / must fuse t to an involution in S L U . S / to bv. bv/ Š C2 Q2n . S /. bav/ so that hP i D S and P Â G T . S / S and t 0 acts faithfully on U . Suppose at first that m D n 4. S / S we have Lx D K and so hcix D hai.

A/ and so hz; u; si Š E8 . G/ so normal in G, which contradicts the maximal choice of S since S < AM (see (i–iv)). 2). A/ D A0 . A/, as we saw, acts faithfully on L). We have m 4 since GN Š M2m in view of the fact that GN is not of maximal class, N by the assumption of this paragraph; we conclude that jAj 8. G/. a/ D1 Š D8 . A/ centralizes L. Thus, that A=ha2 i C m 1 D n and so i D n m C 1. Since i 2, we must have n m C 1. Because m 4, we have here n 5. It remains to determine the commutator Œa; s.