ahw4 - Math 4124 Wednesday February 16 Fourth Homework...

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Math 4124 Wednesday, February 16 Fourth Homework Solutions 1. 2.3.6 on page 60. In Z / 48 Z write out all elements of a for every a . Find all inclu- sions between subgroups in Z / 48 Z . Since 48 = 3 *16, 3 has two divisors and 16 has five divisors, we see that there are 2*5 = 10 subgroups. 0 = { 0 } Order 1 1 = { 0 , 1 , 2 ,..., 47 } order 48 2 = { 0 , 2 , 4 ,..., 46 } order 24 3 = { 0 , 3 , 6 ,..., 45 } order 16 4 = { 0 , 4 , 8 ,..., 44 } order 12 6 = { 0 , 6 , 10 ,..., 42 } order 8 8 = { 0 , 8 , 12 ,..., 40 } order 6 12 = { 0 , 12 , 24 , 36 } order 4 16 = { 0 , 16 , 32 } order 3 24 = { 0 , 24 } order 2 Then we have 1 = 5 = 7 = 11 = 13 = 17 = 19 = 23 = 25 = 29 = 31 = 35 = 37 = 41 = 43 = 47 2 = 10 = 14 = 22 = 26 = 34 = 38 = 46 3 = 9 = 15 = 21 = 27 = 33 = 39 = 45 4 = 20 = 28 = 44 6 = 18 = 30 = 42 8 = 40 12 = 36 16 = 32 0 , 24 are on their own. The lattice of subgroups looks like (where we have labeled each subgroup with its order)
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1 2 4 8 16 3 6 12 24 48 2. 2.4.3 on page 65. Prove that if H is an abelian subgroup of a group G , then H , Z ( G ) is abelian. Give an explicit example of an abelian subgroup H of a group G such that H , C G ( H ) is not abelian. Set Z = Z ( G ) . Then from the ungraded HW of February 15, every element of H , Z ( G ) can be written in the form hx where h H and x Z ( G ) . Suppose we have two el- ements hx and ky of H ,
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