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Unformatted text preview: Math 113 Homework 5 – Solutions IMPORTANT REMINDER: While collaboration is encouraged, you are expected to write up your solutions in your own words, without looking at someone else’s work or any other source. From now on, solutions that are identical to another source may receive no credit. 1. (a) Find the left and right cosets of the subgroup H = { r , s } of D 4 . Are they the same? Recall the group table for D 4 : r r 1 r 2 r 3 s s 1 s 2 s 3 r r r 1 r 2 r 3 s s 1 s 2 s 3 r 1 r 1 r 2 r 3 r s 1 s 2 s 3 s r 2 r 2 r 3 r r 1 s 2 s 3 s s 1 r 3 r 3 r r 1 r 2 s 3 s s 1 s 2 s s s 3 s 2 s 1 r r 3 r 2 r 1 s 1 s 1 s s 3 s 2 r 1 r r 3 r 2 s 2 s 2 s 1 s s 3 r 2 r 1 r r 3 s 3 s 3 s 2 s 1 s r 3 r 2 r 1 r The left cosets are: r H = s H = { r , s } ; r 1 H = s 1 H = { r 1 , s 1 } ; r 2 H = s 2 H = { r 2 , s 2 } ; and r 3 H = s 3 H = { r 3 , s 3 } . The right cosets are: Hr = Hs = { r , s } ; Hr 1 = Hs 3 = { r 1 , s 3 } ; Hr 2 = Hs 2 = { r 2 , s 2 } ; and Hr 3 = Hs 1 = { r 3 , s 1 } . The left and right cosets are different (e.g., the right coset Hr 1 is not a left coset aH for any a ∈ D 4 ). (b) Same question for H ′ = { r , r 2 } . The left cosets are: r H ′ = r 2 H ′ = { r , r 2 } ; r 1 H ′ = r 3 H ′ = { r 1 , r 3 } ; s H ′ = s 2 H ′ = { s , s 2 } ; and s 1 H ′ = s 3 H ′ = { s 1 , s 3 } . The right cosets are: H ′ r = H ′ r 2 = { r , r 2 } ; H ′ r 1 = H ′ r 3 = { r 1 , r 3 } ; H ′ s = H ′ s 2 = { s , s 2 } ; and H ′ s 1 = H ′ s 3 = { s 1 , s 3 } . Hence the left and right cosets are the same: in fact aH ′ = H ′ a for all a ∈ D 4 . 2. Fraleigh section 10, exercises 34, 39, 44. # 34: Let G be a group of order pq , where p and q are prime numbers. Show that every proper subgroup of G is cyclic. Let H be a proper subgroup of G : then  H  divides  G  = pq , and  H  negationslash = pq . Since p, q are prime, we conclude that  H  ∈ { 1 , p, q } . The only subgroup of order 1 is the trivial subgroup { e } , which is cyclic (generated by e ). Otherwise,  H  is a prime number (either p or q ), and so H is cyclic by Corollary 10.11 (in fact, as shown in the proof of that corollary, H = ( a ) for any element a ∈ H such that a negationslash = e ). # 39: Show that if H is a subgroup of index 2 in a finite group G , then every left coset of H is also a right coset of H . By definition of the index,  H  =  G  / 2 and there are exactly two left cosets, which form a partition of G . Since of the cosets is eH = H , the other coset must consist of all those elements of G which are not in H , i.e. it is the complement { x ∈ G, x negationslash∈ H } . Likewise, there are two right cosets, forming a partition of G ; one of these is He = H , and therefore the other one must be the complement of H . Hence every left coset ( H or its complement) is also a right coset....
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 Spring '08
 OGUS
 Math, Cyclic group, Z4

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