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Thus x y = 1 = y 2 x = y ( y x ) = y, so also x = 1, and hence G = 1. = ( a , b 1 am = b2 = 1, ub = ba '). In this case all elements of G can be written in the form b'aj, with 0 5 i I1 and 0 I j I m - 1 , so [GII 2m. 2 that IGl = 2m. We suggest two proofs that such a group exists. Consider first the group 0,of symmetries of a regular m-gon in the plane (assume here that m 2 3). It has a generating rotation o through angle 2n/m and it has m reflections. If 7 is one of the reflections then 0 and 7 generate 0, and satisfy the relations for G, and ID,l = 2m, so G 2 0,.

InducGk for all k, so G(“) = 1 and G is solvable. 4. Suppose K 4C. Then G is solvable if and only if both K and G / K are solvable. Proof. * : We observed earlier that any subgroup of a solvable group is solvable. Since the canonical quotient map q: G + G / K is an epimorphism every commutator [ x K , yK] in G / K is the image q ( [ x , y]) = [qx, qy] of a commutator in G. Thus (GjK)‘ = q(G’), and likewise ( G / K ) ‘ k= ) v(G(~)), all k, so G / K is solvable. -=: Choose subnormal series with abelian factors for K and for G / K , say K = K,,2 K , 2 ...

Exercise I I . 1. Prove 2(c) above. 2. Classify all groups of orders 12 and 20. 12. FURTHER EXERCISES 1. If G is a group and f : G -+ G is defined by f ( x ) = x-', all x E G, show that f is a homomorphism if and only if G is abelian. 2. If a group G has a unique element x of order 2 show that x E Z ( G ) . 3. Suppose G is finite, K 4G, H I G, and I KI is relatively prime to [ G : H ] . Show that K I H . 42 1 Group 4. If G is not abelian show that Z(G)is properly contained in an abelian subgroup of G.