Asymptotic invariants of infinite groups by Gromov M.

By Gromov M.

The articles in those volumes arose from papers given on the 1991 foreign Symposium on Geometric workforce idea, and so they symbolize a number of the most modern considering during this quarter. the various world's top figures during this box attended the convention, and their contributions conceal a large range of themes. This moment quantity comprises exclusively a floor breaking paper through Gromov, which gives a desirable examine finitely generated teams. For somebody whose curiosity lies within the interaction among teams and geometry, those books could be an important addition to their library.

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43 that |Z(G)| is divisible by p. Were it the case that |J| = p then J = H = Z(G). But then J would consist of all elements of G for which gZ(G)g −1 = Z(G), and thus would be the whole of G, which is impossible. It follows that |J| = p2 (since |J| > p and |J| divides p2 ). But then J = G, and hence H is a normal subgroup of G, as required. Example We now show that there are no simple groups of order 36. Let G be a group of order 36. Then G contains a Sylow 3-subgroup H of order 9. If this is the only Sylow 3-subgroup, then it is a normal subgroup, and therefore the group G is not simple.

It follows from this that the order of (x, y) cannot be equal to 1, p or q, and must therefore be equal to pq. Thus Np × Nq is a cyclic group generated by (x, y), and therefore G is a cyclic group, generated by xy, as required. Example Any finite group whose order is 15, 33, 35, 51, 65, 69, 85, 87, 91 or 95 is cyclic. 49 Let G be a group of order 2p where p is a prime number greater than 2. Then either the group G is cyclic, or else the group G is isomorphic to the dihedral group D2p of symmetries of a regular p-sided polygon in the plane.

Now consider the element xyx−1 of G. This must be an element of the normal subgroup N of G generated by y. Therefore xyx−1 = y k for some integer k. Moreover k is not divisible by p, since xyx−1 is not the identity element e of G. Then 2 y k = (y k )k = (xyx−1 )k = xy k x−1 = x(xyx−1 )x−1 = x2 yx−2 . But x2 = x−2 = e, since x is an element of G of order 2. It follows that 2 2 y k = y, and thus y k −1 = e. But then p divides k 2 − 1, since y is an element of order p. Moreover k 2 − 1 = (k − 1)(k + 1).

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