By D. Arnold, R. Hunter, E. Walker

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**Example text**

Its order is equal to jM22 j 100. Proof. /. / . / D M22 . N / is the union of the orbits of the group M22 . N / is equal to 1 or 100. The length of every N -orbit cannot be equal to 1 since N acts faithfully on 0 . / . / D NM22 . Since N \ M22 M22 and M22 is simple, either N \ M22 D M22 or N \ M22 D f1g. / D N , in the second jN j D 100. But any group of order 100 has a unique Sylow subgroup of order 25. One can take it as N and obtains a contradiction. Thus the group HS is simple. 9. 9, we prove several auxiliary lemmas.

C2 ; : : : ; c7 ; c1 /. Using this code, one can construct the Steiner As usual standard means that the i -th coordinate of the vector ei is equal to 1 and the other coordinates are equal to zero. 8 18. 7; 3; 2/: the set of points of this system is f1; 2; : : : ; 7g and the set of blocks is the set of supports of words of this code of weight 3. 2/. 2/ of order 168. 7; 4/code. We denote it by Cx . 4 Theorem. 8; 4; 3/. Proof. Writing out the 16 words of this code, one can verify that it contains the words 0, 1 and 14 words of weight 4.

Moreover, any pair of points of k can be carried by an appropriate element h 2 H to one of the pairs fxN 2 ; xN 3 g, fxN 2 ; x2 C x3 g, fx2 C x3 ; ax2 C a 1 x3 g. Changing P by the oval hP , we may assume that k intersects P in one of these pairs of points. 12, ' carries the oval P to some block. M /. However, ' … M22 since ' is an odd permutation of the points of M : ' has order 2 and fixes 8 points of 22. The automorphism f , as it was mentioned above, is also odd. Therefore f ' 2 M22 . 1/ D xN 1 .