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Unformatted text preview: Math 150a: Modern Algebra Homework 5 Solutions 2.4.19: Prove that if a group contains exactly one element of order 2, then that element is in the center of the group. Solution: The key fact here is that conjugation preserves order: g n = e ⇔ xg n x − 1 = xex − 1 = e , and xg n x − 1 = ( xgx − 1 ) n . Let G be a group, and g ∈ G be the unique element of order 2. Then, ∀ x ∈ G the conjugate xgx − 1 must also have order 2. Since g is the unique such element, xgx − 1 = g . Or, multiplying by x on the right, xg = gx . This exactly means that g ∈ Z ( G ) . square 2.5.3: Determine the number of equivalence relations on a set of five elements. Solution: Every equivalence relation on a set determines a partition of the set, and vice versa. So we will count the number of partitions of a set of five elements. Start by considering the largest equivalence class of the partition. If it only contains one element, then there is only one possible partition to count, { } , { } , { } , { } , { } . Next, consider the largest equivalence class to be of size 2. There are ( 5 2 ) ways to pick a set with 2 elements. Then, for each, we may split up the remaining 3 elements as follows: { , }  { } , { } , { } [in one way]; and { , }  { , } , { } [in parenleftbigg 3 2 parenrightbigg ways]; Counting in this way, we have kept track of the ordering of the two partitions of order 2. Since we don’t care about this ordering, we should divide by the number of permutations of the ordering. So there are ( 5 2 )( 3 2 ) 2 partitions of the form { , } , { , } , { } . To avoid double counting, we next consider the largest equivalence class to be of size 3. Then there are ( 5 3 ) ways to start, with two choices for each: { , , } , { } , { } , and { , , } , { , } . Finally, there are ( 5 4 ) ways to choose a partition with an equivalence class with 4 elements: { , , , } , { } . And one partition with an equivalence class with 5 elements: { , , , , } . So, all together, we have counted 1 + parenleftbigg 5 2 parenrightbigg ( 1 + ( 3 2 ) 2 )+ parenleftbigg 5 3 parenrightbigg 2 + parenleftbigg 5 4 parenrightbigg + 1 = 52 . 2.6.10: (a) Prove that every subgroup of index 2 is normal. Solution: The main ingredients of this proof are that cosets of a subgroup partition the group, and (propo sition 2.6.18) that a subgroup is normal if and only if every left coset is also a right coset. Let H be a subgroup of a group G , and let [ G : H ] = 2. Then, for any a / ∈ H we have two partitions of G (into two left or right cosets): G = H ∪ aH , and G = H ∪ Ha . This means that aH must be equal to Ha . For any x ∈ aH , x / ∈ H (because of the partition of G into left cosets). But x / ∈ H ⇒ x ∈ Ha (because of the partition of G into right cosets). Thus, aH ⊂ Ha . A similar argument shows that Ha ⊂ aH . Thus, aH = Ha , which proves that H is normal. square GK1. Set arithmetic can be interesting even when the sets involved are not subgroups or cosets. The first two parts of this problem involve the vector space...
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 Spring '03
 Kuperberg
 Algebra, Subgroup, Coset, Conjugacy class, conjugacy classes

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