By Mazurov V. D.

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**Extra info for 2-Transitive permutation groups**

**Sample text**

He proved the ambivalency of G~S 2 assuming that G is ambivalent. Using this and the associativity of the wreath product multiplication he showed, that the 2-Sylow-eubgroups of symmetric groups are ambivalent. And this implies, that every 2-group can be embedded in an ambivalent 2-group. Cyclic groups Cp of order p are not ambivalent in general, therefore that not every p-Sylow-subgroup plied by the following: of S n is ambivalent is im- 51 The ambivalency ambivalency Proof: of G~H implies the ambivalency of H.

15 If G is ambivalent, then G~S n is ambivalent. Proof: There is obviously a 1-1-correspondence between the cycles 5O (j... r(j)) of ~ and (j... -r(j)) of - 1 . Let g be the cycle product to (j... r(j)) with respect to f. Then the oycleproduct to (j... -r(j)) with respect to f-~1 is (recall that (f;~)-1 = (f-11|~-I)) " If now G is an ambivalent group, then g ~ g-l, y g E G, what implies, that in this case And this proves the assertion since two elements of the same type are conjugates (of. 7). d.

The a k cyclic factors of ~ which are of length k can be distributed into the s conJugacy classes of G in alkl ~ 1 "'" \ .... ask = . ask ! ways which are in accordance with the considered type (aik). T,et f:G ~ G be a mapping which yields such a distribution cycleproducts. It remains to show, what freedom of choice is left for choosing the values of f. , f(-k+2(j)) at will and can choose an f(~-k+1(j)) E G so that the complete product is an element of O i ~ G. ask ! = (lelk-llcil) ~ mappings f:~ ~ G which distribute the ak k-cycles of ~ as the considered type (aik) prescribes.

### 2-Transitive permutation groups by Mazurov V. D.

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