Sizes of Conjugacy Classes of a Group of Known Order











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Suppose G is a group of order 48 (centre consisting identity only). Show it has a conjugacy class of order 3.



I know that the size of the conjugacy classes are limited to divisors of 48: 1,2,3,4,6,8,12,16,24, and 48. These classes also partition G so their sizes must sum to the order of G (so I cannot use 48 and have to include 1). I am not sure how to continue from here.










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    Suppose G is a group of order 48 (centre consisting identity only). Show it has a conjugacy class of order 3.



    I know that the size of the conjugacy classes are limited to divisors of 48: 1,2,3,4,6,8,12,16,24, and 48. These classes also partition G so their sizes must sum to the order of G (so I cannot use 48 and have to include 1). I am not sure how to continue from here.










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      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Suppose G is a group of order 48 (centre consisting identity only). Show it has a conjugacy class of order 3.



      I know that the size of the conjugacy classes are limited to divisors of 48: 1,2,3,4,6,8,12,16,24, and 48. These classes also partition G so their sizes must sum to the order of G (so I cannot use 48 and have to include 1). I am not sure how to continue from here.










      share|cite|improve this question















      Suppose G is a group of order 48 (centre consisting identity only). Show it has a conjugacy class of order 3.



      I know that the size of the conjugacy classes are limited to divisors of 48: 1,2,3,4,6,8,12,16,24, and 48. These classes also partition G so their sizes must sum to the order of G (so I cannot use 48 and have to include 1). I am not sure how to continue from here.







      abstract-algebra group-theory permutations equivalence-relations






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      edited Nov 19 at 0:27

























      asked Nov 18 at 22:53









      J. Dawson

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          We know that $G$ partitions into conjugacy classes and the only possible orders of these classes are those that you mention. There is only one conjugacy class of order $1$ since the center is trivial. Now the sum of the occurring orders is $48$, which is even, so there is at least one conjugacy class of odd order bigger than one...






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            We know that $G$ partitions into conjugacy classes and the only possible orders of these classes are those that you mention. There is only one conjugacy class of order $1$ since the center is trivial. Now the sum of the occurring orders is $48$, which is even, so there is at least one conjugacy class of odd order bigger than one...






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              up vote
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              We know that $G$ partitions into conjugacy classes and the only possible orders of these classes are those that you mention. There is only one conjugacy class of order $1$ since the center is trivial. Now the sum of the occurring orders is $48$, which is even, so there is at least one conjugacy class of odd order bigger than one...






              share|cite|improve this answer























                up vote
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                down vote










                up vote
                4
                down vote









                We know that $G$ partitions into conjugacy classes and the only possible orders of these classes are those that you mention. There is only one conjugacy class of order $1$ since the center is trivial. Now the sum of the occurring orders is $48$, which is even, so there is at least one conjugacy class of odd order bigger than one...






                share|cite|improve this answer












                We know that $G$ partitions into conjugacy classes and the only possible orders of these classes are those that you mention. There is only one conjugacy class of order $1$ since the center is trivial. Now the sum of the occurring orders is $48$, which is even, so there is at least one conjugacy class of odd order bigger than one...







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                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 18 at 23:16









                Jef L

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