week05tutsols

# week05tutsols - THE UNIVERSITY OF SYDNEY Math2968...

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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 ,α,α 2 ,α 3 ,β,αβ,α 2 β,α 3 β } where α denotes a quarter rotation and β a reflection of the square. The subgroups of G are { 1 } , G, ( α ) , ( α 2 ) , ( β ) , ( αβ ) , ( α 2 β ) , ( α 3 β ) , ( α 2 ,β ) , ( α 2 ,αβ ) . To see that this list is complete, let H be a nontrivial subgroup of G . If H contains α and a reflection, or H contains α 3 and a reflection, then H = G . Otherwise, either H ≤ ( α ) , or H ≤ ( α 2 ,γ ) , where γ 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...

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