herstein

# herstein - Abstract We sketch the solutions of just a few...

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Unformatted text preview: Abstract We sketch the solutions of just a few problems in Herstein’s book. 1 Sketches of some problems in Herstein’s abstract algebra Ching-Lueh Chang March 12, 2008 2.5.19 If H is a subgroup of finite index in G, prove that there is only a finite number of distinct subgroups of the form aHa- 1 . Proof. Observe aHa- 1 = aHHa- 1 = aH- 1 Ha- 1 = ( Ha- 1 )- 1 ( Ha- 1 ) . The second inequality follows from H = H- 1 . 2.5.20 If H is of finite index in G prove that there is a subgroup N of G, contained in H, and of finite index in G such that aNa- 1 = N for all a ∈ G. Can you give an upper bound for the index of this N is G ? Proof. The set N = ∩ g ∈ G g- 1 Hg provably forms a normal subgroup of G. Clearly, N ⊆ H. Let a ∈ G be arbitrary and Hg 1 ,...,Hg n be all the distinct right cosets of H. Then Na = ( ∩ g ∈ G g- 1 Hg ) a = ( ∩ g ∈ G g- 1 HHg ) a = ( ∩ g ∈ G g- 1 H- 1 Hg ) a = ( ∩ g ∈ G ( Hg )- 1 ( Hg )) a = ( ∩ n i =1 ( Hg i )- 1 ( Hg i )) a = ( ∩ n i =1 ( Hg i )- 1 ( Hg i a ) . We observe that Hg 1 a,...,Hg n a are right cosets of H and are distinct, hence they form a permutation of Hg 1 ,...,Hg n . This and the above equation show that N has at most n ! right cosets. 2.5.24 Let G be a finite group whose order is not divisible by 3 . Suppose that ( ab ) 3 = a 3 b 3 for all a,b ∈ G. Prove that G must be abelian. Proof. We assume o ( G ) > 1 for otherwise the statement is trivial. Let g ∈ G be any non-identity element. We show that g 3 6 = e where e is the identity of...
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herstein - Abstract We sketch the solutions of just a few...

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