Math 3560 HW Set 4
Kara
September 24, 2013
(6.1) Note that we first apply the column element and then we apply the row element.
id
(12)
(23)
(13)
(123)
(132)
id
id
(12)
(23)
(13)
(123)
(132)
(12)
(12)
id
(132)
(123)
(13)
(23)
(23)
(23)
(123)
id
(132)
(12)
(13)
(13)
(13)
(132)
(123)
id
(23)
(12)
(123)
(123)
(23)
(13)
(12)
(132)
id
(132)
(132)
(13)
(12)
(23)
id
(123)
(6.2)
1. (1734)(26)(58)=(14)(13)(17)(26)(58)
2. (4568)(1245)=(125)(468)=(15)(12)(48)(46)
3. (624)(253)(876)(45)=(25687)(34)=(27)(28)(26)(25)(34)
Only the element in (b) is an even permutation i.e. in A8 .
(6.4) By the previous exercise, the subset H1 consisting of all permutations sending 1, 2, 3
among themselves (that is fixing 4) is a subgroup and its order is 3! = 6. Similarly
Hi = {permutations fixing i} for i = 1, 2, 3 are subgroups of order 6.
To prove. These are the only subgroups of order 6.
Proof. H needs to contain an element of order 3, i.e. 3-cycle, and some element of order 2
as these are prime divisors of 6. H cannot contain any element of order 4, as 4 does not
divide 6. Up to relabeling we can assume the 3-cycle is σ = (1 2 3). Then H also contains
σ 2 = σ −1 = (1 3 2). Can we have more 3-cycles? Any other 3-cycle in S4 will move element
4. Then together with σ, σ 2 they would generate all A4 which already has 12 elements.
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(check that. hint - ex 6.9). Therefore there are precisely two 3-cycles in H: σ, σ 2 . Now
analyze the element of order 2. It cannot be a transposition involving 4 as then its product
with σ would give a 4-cycle:
(123)(34) = (1234)
(123)(24) = (231)(24) = (21)(23)(24) = (3124)
(123)(14) = (13)(12)(14) = (1423).
Also it cannot be a product of disjoint transpositions, as then its product with σ gives an
even nontrivial permutation (so in S4 it has to be a 3-cycle) moving 4 and we cannot have
such elements in H as explained above. Therefore H can only contain transpositions not
moving 4. Counting the elements of H we see it has to contain all such transpositions.
Thus H is a subgroup fixing element 4. Relabeling the elements we obtain all 4 subgroups
found in the first part of this exercise.
(6.9) We want to show that (123) and (12 · · · n) generate An when n is odd and (123) and
(23 · · · n) generate An if n is even. Note that the claim is true for n = 3 as (123) generates
A3 . The proof consists of a few steps. We know that An is generated by 3-cyles if n ≥ 3 by
Theorem 6.5 in the book. Actually, it proves An is generated by the 3-cycles of the form
(1ba) if n ≥ 3.
Step 1: An is generated by the 3-cycles of the form (1 k k + 1).
In order to show this, we will prove that 3-cycles of the form (1ba) can be written as a
word in the 3-cycles of the form (1 k k + 1). Assume a > b. Then
(1ba) = (1 a − 1 a)(1 a − 2 a − 1) · · · (1 b + 1 b + 2)(1 b b + 1)
(e.g (127) = (167)(156)(145)(134)(123))
If b > a, then (1ba) = (1ab)(1ab) and we can write (1ab) as a word in the 3-cycles of the
form (1 k k + 1).
Step 2: An is generated by the 3-cycles of the form (k k + 1 k + 2).
In order to show this, we will prove that 3-cycles of the form (1 a a + 1) can be written
as a word in the 3-cycles of the form (k k + 1 k + 2). First observe that we can write
(134) = (234)(123)(234)−1 . Let us generalize this to a > 1 as follows:
(1 a a+1) = (a−1 a a+1)(a−2 a−1 a) · · · (234)(123)(234)−1 · · · (a−2 a−1 a)−1 (a−1 a a+1)−1
Step 3: Assume n is odd, then (12 · · · n) is in An . We will prove that a 3-cycle of the form
(k k + 1 k + 2) can be written as a word in terms of (123) and (12 · · · n). First observe that
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we can write (234) = (12 · · · n)(123)(12 · · · n)−1 . We can generalize this to
(k k + 1 k + 2) = (12 · · · n)k−1 (123)(12 · · · n)1−k
for 2 ≤ k ≤ n − 2.
Now assume n is even, then (23 · · · n) is in An . We will show that a 3-cycle of the form
(1 k k + 1) can be written as a word in terms of (123) and (23 · · · n) since we know that
An is generated by the 3-cycles of the form (1 k k + 1) by Step 1. First observe that
(134) = (23 · · · n)(123)(23 · · · n)−1 . We can generalize this to
(1 k k + 1) = (23 · · · n)k−2 (123)(23 · · · n)2−k
for 3 ≤ k ≤ n − 1.
(6.11) We will use exercise (6.12) to find the orders of permutations listed in (6.2). The
order of (1734)(26)(58) is 4, the order of (125)(468) is 3 and the order of (25687)(34) is
10.
(6.12)
To prove. The order of α ∈ Sn is the least common multiple of the lengths of the cycles
which are obtained when α is written as a product of disjoint cyclic permutations.
Proof. Let α ∈ Sn , then we know that we can write it as a product of disjoint cycles.
Suppose α = τ1 τ2 · · · τk , where each τi is a disjoint cycle of length `i . Let ` be the least
common multiple of all the `i ’s. Since the τi ’s are disjoint cycles, they commute. Therefore,
αj = τ1j τ2j · · · τkj for any j ∈ Z. Since the τi ’s are disjoint, then the τij ’s are also disjoint.
Let n be the order of α, then αn = id. This means τ1n τ2n · · · τkn = id, which implies τin = id
for all i since the τin ’s are disjoint. Then we get `i divides n for all i, which implies ` divides
n since ` is the least common multiple of `i ’s. (note that the order of a cycle is equal to
its length and if order of x ∈ G, where G is a group, is n and if xm = id, then n divides
m). Also, we have α` = τ1` τ2` · · · τk` = id since τi`i = id for each i and ` is the least common
multiple of `i for each i. So order of α divides `. Thus, α has order `.
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