week05tutsols - THE UNIVERSITY OF SYDNEY Math2968 Algebra...

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Unformatted text preview: THE UNIVERSITY OF SYDNEY Math2968 Algebra (Advanced) Semester 2 Tutorial Solutions Week 5 2008 1. (for general discussion) Give an example of a group G and subgroups H and K such that H K is not a subgroup. Find a simple necessary and sufficient condition for H K to be a subgroup of G . Solution Let G be the symmetric group on three letters, H = { 1 , (12) } and K = { 1 , (13) } . Then H G and K G , but H K negationslash G since (12)(13) = (123) negationslash H K , so that H K is not closed under multiplication. A necessary and sufficient condition for H K to be a subgroup of G is that H K or K H . This is clearly sufficient because if H K or K H then H K = K or H K = H . To see that it is necessary, suppose H K G . Suppose also H negationslash K and K negationslash H . Then there exists h H \ K and k K \ H . But H K is closed under multiplication so that hk H K . Hence hk H or hk K . If hk H , then k = h 1 ( hk ) H , since H is closed under multiplication and taking inverses, which contradicts that k negationslash H . If hk K , then h = ( hk ) k 1 K , since K is closed under multiplication and taking inverses, which contradicts that h negationslash H . This proves that H K or K H . 2. Find all subgroups of the group of symmetries of the square. Solution Let G = { 1 ,, 2 , 3 ,,, 2 , 3 } where denotes a quarter rotation and a reflection of the square. The subgroups of G are { 1 } , G, ( ) , ( 2 ) , ( ) , ( ) , ( 2 ) , ( 3 ) , ( 2 , ) , ( 2 , ) . To see that this list is complete, let H be a nontrivial subgroup of G . If H contains and a reflection, or H contains 3 and a reflection, then H = G . Otherwise, either H ( ) , or H ( 2 , ) , where is a reflection, in which case it follows quickly that H is on the above list. 3. Let G be an abelian group and H = { x g | x p = 1 } , where p is a prime number. Verify that H G . Explain why this conclusion fails if we allow G to be any group, not necessarily abelian. Solution Certainly 1 p = 1, so that 1 H negationslash = . If x,y H then x p = y p = 1 so that, by commutativity of multiplication, ( xy 1 ) p = x p ( y 1 ) p = x p ( y p ) 1 = 1(1) 1 = 1 , which proves H G , by the Subgroup Criterion. The conclusion may fail if G is not required to be abelian. For example, if G is the symmetric group on three letters and p = 2 then H = { 1 , (12)...
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week05tutsols - THE UNIVERSITY OF SYDNEY Math2968 Algebra...

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