By L.C. Grove, C.T. Benson

This isn't my evaluate; yet i've got consciously learn this booklet (chapters 1,2,3,4,5) for getting ready my thesis, and that i used to be considering the interpretation (from English to Spanish)of this booklet.

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Let X < G/N. Put H = {y & G : Ny e X}. For n&N,Nn = Ne e X. Hence, N C H. If y, z € H, then Ny, Nz e X. Thus, Nyz~l = (Ny)(Nz)~1 e X so that yz~1 e H. Therefore, by the subgroup test theorem, H < G. By definition, H/N = X. If H < G, let TT : G/N -> G/H be the factor map, n(Ng) = Hg. Since N C H it is easy to check that TT is well defined and, if well defined, TT is plainly a homomorphism. We have Ker TT = H/N = X, so X < G/N . Suppose X = H/N < G/N and let g e G. Then (Ng)-l(H/N)(Ng) C F/AT. Thus, (g-1Hg)/N C H/N.

Thus, s~, = 1 or ,ss = 6. If 85 = 1, then G has a unique-hence normal-Sylow ^-subgroup, and we're done. Suppose that 55 = 6. Since each Sylow ^-subgroup is isomorphic to Z(5), the pairwise intersection of any two distinct ones is < e > . Thus, the Sylow 5-subgroups account for 25 = 6 • 4 + 1 distinct elements. Next, look at 33. If 33 = 1, we're done. So, suppose that 53 = 10, the only other possibility. In this case, reasoning as above, we account for 20 = 10-2 additional elements. But we've already used more elements than are in the group!

If we number the elements of 5(3), we obtain a corresponding homomorphism c : G —> 84. The homomorphism c sends each g 6 G to the permutation induced by conjugation by g. Since G = 108 < 24 =| 84 |, the kernel of the map c will be a proper normal subgroup of G. Note that c cannot map each element g & G to the identity of 84, since any two 3-Sylow subgroups are conjugate. 4 Our final example has a different flavor. We construct a nonabelian group of G of order 21. Consider a group G with \ G \~ 3 • 7.