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ThudnBlunder
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Minimum # of Conjugacy Classes
« on: Sep 30th, 2006, 2:11pm » |
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A finite group of 8k elements (k = 1,2,3,4 .......) must have at least how many conjugacy classes?
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Barukh
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Re: Minimum # of Conjugacy Classes
« Reply #1 on: Oct 3rd, 2006, 12:45am » |
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I am probably missing something...
- For every integer n, there is a cyclic group with n elements.
- Every cyclic group is abelian.
- Abelian groups have one conjugate class.
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| « Last Edit: Oct 3rd, 2006, 1:14am by Barukh » |
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Deedlit
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Re: Minimum # of Conjugacy Classes
« Reply #2 on: Oct 3rd, 2006, 1:44am » |
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Just the opposite - an abelian group of n elements has n conjugacy classes.
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The minimum number is 5, achieved by D_4, Q_8, and S_4. Note that the cardinality of every conjugacy class divides the order of the group. Also, there is a conjugacy class of size 1, namely the identity. So, if G is divided into four conjugacy classes, we have
8k/x + 8k/y + 8k/z + 1 = 8k
1/x + 1/y + 1/z + 1/8k = 1
By parity, at least one of x,y,z must be divisible by 8. So for k \geq 3, we have
1/x + 1/y + 1/z + 1/8k >= 1/2 + 1/3 + 1/8 + 1/24 = 1
with equality only when k = 3, and the conjugacy classes are of size 12, 8, 3, and 1 respectively. We see that there are 8 elements of order 3 and 15 elements of order 2 or 8, so there are 4 3-Sylow subgroups and 3 2-Sylow subgroups. There's probably a really easy way to see that this implies that the group must be S_4, but I'm not seeing it; but I classified the groups of order 24 a while back, and S_4 is the only group that is not the the direct or semidirect product of one Sylow subgroup by another. (More generally, S_4 is the only such group of order p3q for distinct primes p and q.)
But S_4 has 5 conjugacy classes.
I didn't handle the cases of fewer than 4 conjugacy classes but those are easy.
For k = 1, there are two nonabelian groups of order 8, D_4 and Q_8, each with 5 conjugacy classes. More generally, if |G| = 2n, then the above equation you can derive that you need at least n+1 conjugacy classes.
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| « Last Edit: Oct 3rd, 2006, 1:56am by Deedlit » |
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Barukh
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Re: Minimum # of Conjugacy Classes
« Reply #3 on: Oct 3rd, 2006, 1:50am » |
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on Oct 3rd, 2006, 1:44am, Deedlit wrote:| Just the opposite - an abelian group of n elements has n conjugacy classes. |
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Yes, of course! Silly me...
... And very nice demonstration, Deedlit! 
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| « Last Edit: Oct 3rd, 2006, 1:56am by Barukh » |
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ThudnBlunder
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Re: Minimum # of Conjugacy Classes
« Reply #4 on: Oct 3rd, 2006, 4:41am » |
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The number of conjugacy classes of a finite group G equals the number of irreducible representations of G. The order of G is the sum of the squares of the dimensions of these representations. At least one of these dimensions is 1 (the trivial representation). When |G|=8k, it is not possible to write |G|-1=8k-1 as a sum of fewer than four squares. Therefore the number of representations, which is also the number of conjugacy classes, is at least 4+1=5, QED.
Remarks: Equality holds for at least three groups G, namely the dihedral and quaternion groups of order 8, and the fourth symmetric group, of order 4!=24. Michael Larsen observed that for each m there is an effective upper bound on the size of a finite group with at most m conjugacy classes; when m=5, this reduces the problem, and the determination of the cases of equality, to a finite (though possibly impractical) computation. The upper bound is a consequence of the ``class equation'': |G| is the sum of the sizes of its conjugacy classes, each of which is a factor of |G|, and at least one of which equals 1.
(due to Noam Elkies)
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Michael Dagg
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Re: Minimum # of Conjugacy Classes
« Reply #5 on: Oct 9th, 2006, 9:17pm » |
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Nice solutions Deedlit and T&B!
Often in group theory one might expect the number of _something_ to evenly
divide another something, like in this case, the number m of conjugacy classes
to divide the order |G| ; but this is not one of those times, i.e. m/|G| is not the
reciprocal of any integer. But it is still an interesting quantity because it turns out
to be exactly the _probability_ that two elements of the group commute (selected
at random, with replacement, with uniform probability).
But, I am sure you already knew this nevertheless.
(Noam Elkies -- Harvard?)
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| « Last Edit: Oct 9th, 2006, 9:18pm by Michael Dagg » |
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Regards,
Michael Dagg
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ThudnBlunder
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Re: Minimum # of Conjugacy Classes
« Reply #6 on: Oct 25th, 2006, 5:08pm » |
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on Oct 9th, 2006, 9:17pm, Michael_Dagg wrote:
(Noam Elkies -- Harvard?) |
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Yes.
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