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By Kondratiev A. S., Mazurov V. D.
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S / S , we have t x D ba2j v for some j and so, by the above, ha2 ix D hc 2 i. S / ¤ G (otherwise we are finished). S / S . bav/ is not possible since a semi-dihedral group is not a subgroup of a generalized quaternion group. S / such that t s D t n . S /. a2 /t D c 2k with k odd. Since y is an element of order 4 contained in ha2 i and v 0 is an element of order 4 contained in hc 2 i, we get that y t D vz l with l D 0; 1. uz l /t D uz l , recalling that yv D u. t 0 / contains a subgroup isomorphic to E8 .
8). G/ U and M are generated by involutions. Since M is not normal in G, it follows that s must invert Z. 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.
Mod p/: H 21 Remark 5. Suppose that G is a noncyclic group of order p m , m > 2 and 1 < n < m. G/ of normal subgroups N of G such that G=N is cyclic of order p n , is a multiple of p. G/ > 0. Let n D m 1. G/ of order p such that G=N is cyclic. G/ D p. Now let n < m 1. Take H 2 1 . p m 1 ; p/. The group G contains exactly p C 1 subgroups of index p n . G/ D p. mod p/, by (2). Suppose that G is a group of order p m and p n Ä jG W G 0 j. G/ of N G G such that G=N is nonabelian of order p n is a multiple of p.