By Gromov M.

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**Extra resources for Asymptotic invariants of infinite groups**

**Example text**

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).