Exercise on Conjugate classes

1. How many cyclic strucutres of type (_,_,_,_,_)(_,_)(_) are there in 𝓢₈?

2. How many cyclic strucutres of type (_,_,_)(_,_,_)(_,_,_)(_,_,_)(_,_,_) are there in 𝓢₁₅?

3. Determine the cardinality of every conjugacy class in 𝓢₅.

4. Prove the group 𝓢_n is generated by the following n − 1 transpositions.

   (1 2) (2 3) (3 4), ..., (n − 1 n)

5. Prove that 𝓢_n is generated by the two permutations (1 2), (1 2 3 ... n).

6. Prove ∀n ≥ 3 the center of 𝓢_n (see also definition 7.2.3) is made by the identity only.

Solutions

1. 
   1^st method: C_8,5 ⋅ (5 − 1)! C_3,2 ⋅ (2 − 1)! ⋅ C_1,1 = 4032

   2^nd method: Fill in the blanks in one of 8! = 40320 ways. Each r-cycle can be restarted anywhere:

   (1,6,3,2,5) = (6,3,2,5,1) = (3,2,5,1,6) = (2,5,1,6,3) = (5,1,6,3,2)

   We overcounted each cycle of length r a total of r times, so divide by the product of the cycle lengths:

   8!/(5·2·1) = 40320/10 = 4032

2. 
   

   1^st method: [C_15,3 ⋅ (3 − 1)! ⋅ C_12,3 ⋅ (3 − 1)! ⋅ C_9,3 ⋅ (3 − 1)! ⋅ C_9,3 ⋅ (3 − 1)! ⋅ C_9,3 ⋅ (3 − 1)!]/5! = 44844800. We divided by 5! to avoid overcounting the same permutations (a b c)(c d e) = (c d e) (a b c).

   Remark. (a b)(c d) and (c d)(a b) (with a,b,c,d distinct) are the same permutation, so we divide by 2!. In other cases e.g. when filling in the blanks in (_ _)(_ _ _), while you still have (a b)(c d e) = (c d e)(a b), the latter would have shape (_ _ _)(_ _) which is different from (_ _)(_ _ _), so there's no overcounting here.

   2^nd method: Fill in the blanks in one of 15! ways. Each cycle has 3 representations matching this format (by restarting at any of 3 places), so divide by 3⁵.

   The order of the whole cycles can be changed while keeping the pattern, e.g., (1,2,3)(4,5,6) = (4,5,6)(1,2,3). Divide by 5! ways to reorder the cycles.

   15!/(3⁵ · 5!) = 44844800

3. 𝓢₅

   Cyclic structure	Number of distinct conjugates
   (−)	1
   (−,−)	C5,2 = 10 (To construct a transposition (a,b) we choose 2 of 5 elements, and this can be done in 10 ways.
   (−,−,−)	C5,3 ⋅ (3 − 1)! = 20 (To construct a 3-cycle, we choose 3 of 5 elements, but then each choice can be arranged in two different ways).
   (−,−)(−,−)	[C5,2 ⋅ (2 − 1)! ⋅ C3,2⋅ (2 − 1)!]/ 2! = 15 or equivalently as (5!)/(4 ⋅ 2!) = 15
   (−,−,−,−)	C5,4 ⋅ (3 − 1)! = 30
   (−,−,−)(−,−)	C5,3 ⋅ (3 − 1)! ⋅ C2,2 ⋅ (2 − 1)! = 20
   (−,−,−,−,−)	C5,5 ⋅ (5 − 1)! = 24

4. Let H = ⟨(m, m+1), m = 1, ...,n − 1⟩. H contains (1 2)(2 3)(1 2) = (1 3), (1 3)(3 4)(1 3) = (1 4), etc. Thus H contains all the transpositions (1 r), (r = 2,...n). Then it contains every transpositions (r s), (r ≠ s), because (1 s) (1 r) (1 s) = (r s). Hence H = 𝓢_n.

5. Let H = ⟨(1 2) (1,2,...,n)⟩ . In a "group generated by elements" we're allowed to take products and inverses of the elements generating the group.

   (12)(123… n) = (13)
   (1 3)(1 2 3 … n) = (14)
   (14) (123…n) = (15)
   ⋮
   (1 n -1)(123 … n) = (1 n)

   Next we do,

   (12) (13) (12)⁻¹ = (23)
   (13) (14) (13)⁻¹ = (34)
   ⋮
   (1 n-1) (1 n) (1 n − 1)⁻¹ = (n −1 n)

   Hence H contains all the traspositions of type (m m+1) and by what we seen in the previous exercise H = 𝓢_n.

6. An element g belongs to the group's center iff its conjugate class is reduced to {g}: say a is in center and b conjugate to a so b = g⁻¹ag, a commutes with g (it's in the center) so b = g⁻¹g a = a, therefore, if a ∈ Z(G), cl(A) = {a}. We just proved any "other" element that is conjugate to a, must be a itself.

   Now, in 𝓢_n the only element whose conjugate class is reduced to {g} is the one containing id, thus Z(𝓢_n) = {id}.

   Remark: We specified ∀n >3, because in 𝓢₂ = {id, (1 2)}, the non-identity element (1 2) commutes with everything. So here the center is not just {id} .

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