Section 8: Symmetric Groups

Abstract Algebra, Saracino
Second Edition
Section 8
Symmetric Groups
and Alternating
Groups

Lehman College, Department of Mathematics
Definitions and Examples (1 of 2)
Let 𝑋 be a nonempty set, then a one-to-one and onto
mapping 𝑋 → 𝑋 is called a permutation of 𝑋. The set of
all such mappings 𝑆 𝑋 forms a group (𝑆 𝑋,∘) under
composition of functions.
The group (𝑆 𝑋,∘) is called the symmetric group , or
permutation group on 𝑋.
If the set 𝑋 is finite and has, say 𝑛 elements, we can
represent 𝑋 by the set 1, 2, 3, … , 𝑛 , and we denote
(𝑆 𝑋,∘) by 𝑆 𝑛. In this case, 𝑆 𝑛 is called the symmetric
group of degree 𝑛, or the symmetric group on 𝑛 letters.
Let 𝑓 ∈ 𝑆 𝑛, then 𝑓 simply rearranges the elements of 𝑋.

Definitions and Examples (2 of 2)
We can write 𝑓 explicitly in the form:
Let us consider the group 𝑆3. What are the possible
permutations of the set 1, 2, 3 ?
The first three are obtained by cyclically permuting the
elements of the set 1, 2, 3 , while the second set are
obtained by fixing one element and rearranging the
other two. It follows that 𝑆3 = 6.
1 2
𝑓(1) 𝑓(2)
3
𝑓(3)
… 𝑛
… 𝑓(𝑛)
𝑒 =
1 2
1 2
3
3
1 2
2 3
3
1
1 2
3 1
3
2
1 2
1 3
3
2
1 2
3 2
3
1
1 2
2 1
3
3

Product of Permutations (1 of 5)
How do we determine the product of two permutations?
Recall that the group operation is function composition
and in function composition the rightmost function is
taken first:
It follows that the product of the two permutations is the
identity permutation
1 2
2 3
3
1
∘
1 2
3 1
3
2
1 = (1)
1 2
2 3
3
1
∘
1 2
3 1
3
2
2 = (2)
1 2
2 3
3
1
∘
1 2
3 1
3
2
3 = (3)
1 2
2 3
3
1
∘
1 2
3 1
3
2
=
1 2
1 2
3
3
= 𝑒

Let us look at another product
It follows that the product of the two permutations is the
permutation:
What if we reverse the order of the product?
1 2
1 3
3
2
∘
1 2
3 2
3
1
1 = (2)
1 2
1 3
3
2
∘
1 2
3 2
3
1
2 = (3)
1 2
1 3
3
2
∘
1 2
3 2
3
1
3 = (1)
1 2
1 3
3
2
∘
1 2
3 2
3
1
=
1 2
2 3
3
1
1 2
3 2
3
1
∘
1 2
1 3
3
2
=
1 2
3 1
3
2

From the previous slide, we see that
But
So the operation is not commutative and 𝑆3 is
nonabelian.
Theorem 1. If 𝒏 is a positive integer such that 𝒏 ≥ 𝟑,
then the symmetric group 𝑺 𝒏 is a nonabelian group.
Proof. Let 𝑛 ≥ 3, and let 𝛼, 𝛽 ∈ 𝑆 𝑛be defined by:
1 2
1 3
3
2
∘
1 2
3 2
3
1
=
1 2
2 3
3
1
1 2
3 2
3
1
∘
1 2
1 3
3
2
=
1 2
3 1
3
2
𝛼 1 = 2, 𝛼 2 = 1 and 𝛼 𝑥 = 𝑥 for all 𝑥 ≠ 1, 2
𝛽 1 = 3, 𝛽 3 = 1 and 𝛽 𝑥 = 𝑥 for all 𝑥 ≠ 1, 3

Proof (cont’d). 𝑛 ≥ 3, and let 𝛼, 𝛽 ∈ 𝑆 𝑛be defined by:
It follows that:
Now, we have
𝛼 1 = 2, 𝛼 2 = 1 and 𝛼 𝑥 = 𝑥 for all 𝑥 ≠ 1, 2
𝛽 1 = 3, 𝛽 3 = 1 and 𝛽 𝑥 = 𝑥 for all 𝑥 ≠ 1, 3
𝛼 =
1 2
2 1
3
3
4
4
… 𝑛
… 𝑛
𝛽 =
1 2
3 2
3
1
4
4
… 𝑛
… 𝑛
𝛼 ∘ 𝛽 =
1 2
2 1
3
3
4
4
… 𝑛
… 𝑛
∘
1 2
3 2
3
1
4
4
… 𝑛
… 𝑛
=
1 2
3 1
3
2
4
4
… 𝑛
… 𝑛

Proof (cont’d). Similarly
We note above that 𝛼 ∘ 𝛽 1 = 3, but 𝛽 ∘ 𝛼 1 = 2,
so 𝛼 ∘ 𝛽 ≠ 𝛽 ∘ 𝛼 and 𝑆 𝑛 for 𝑛 ≥ 3 is a nonabelian group.
We will introduce the following notation, suppose 𝛼 ∈ 𝑆4
is given by:
We will write 𝛼 as:
𝛽 ∘ 𝛼 =
1 2
3 2
3
1
4
4
… 𝑛
… 𝑛
∘
1 2
2 1
3
3
4
4
… 𝑛
… 𝑛
=
1 2
2 3
3
1
4
4
… 𝑛
… 𝑛
𝛼 =
1 2
1 4
3
3
4
2
𝛼 =
2 4
4 2

Cycle Notation (1 of 2)
If 𝛽 ∈ 𝑆5 is given as
then we mean the permutation:
𝛼 is an example of a 2-cycle and 𝛽 and example of a 3-
cycle. A 2-cycle is also called a transposition. Consider
the permutation 𝜎 ∈ 𝑆6:
Then 𝜎 is a 4-cycle. We will represent it by 1 3 2 5
𝛽 =
1 3
4 1
4
3
∈ 𝑆5
𝛽 =
1 2
4 2
3
1
4
3
5
5
𝜎 =
1 2
3 5
3
2
4
4
5
1
6
6

Cycle Notation (2 of 2)
Let us rewrite the elements of 𝑆3 using cycle notation:
Hence, 𝑆3 = 𝑒, 1 2 , 1 3 , 2 3 , 1 2 3 , (1 3 2)
consists of one 1-cycle, three 2-cycles and two 3-
cycles.
Non-cycles. Consider the permutation 𝛼 ∈ 𝑆4
Then 𝛼 is not a cycle.
𝑒 =
1 2
1 2
3
3
1 2
2 3
3
1
1 2
3 1
3
2
1 2
1 3
3
2
1 2
3 2
3
1
1 2
2 1
3
3
𝛼 =
1 2
3 4
3
1
4
2

Disjoint Cycles (1 of 1)
Non-cycles (cont’d). In general, the product of two
cycles may not be a cycle. Consider the permutations
𝜎, 𝛿 ∈ 𝑆6 where 𝜎 = (5 6) and 𝛿 = (3 2 4)
Then 𝜎𝛿 = 𝜎 ∘ 𝛿 is not a cycle.
Two cycles in 𝑆 𝑛 are called disjoint if no element of
1, 2, 3, … , 𝑛 is moved by both cycles. In 𝑆8, the cycles
𝛼 = 2 4 3 5 and 𝛽 = (1 6 8) are disjoint cycles. But the
cycles 𝜎 = (4 5 3 2) and 𝛿 = (1 3 8) are not disjoint.
𝜎 ∘ 𝛿 =
1 2
1 2
3
3
4
4
5
6
6
5
∘
1 2
1 4
3
2
4
3
5
5
6
6
=
1 2
1 4
3
2
4
3
5
6
6
5

Product of Cycles (1 of 3)
Let us look at the composition of the two disjoint cycles
𝛼 = 2 4 3 5 and 𝛽 = 1 6 8
Hence 𝛼𝛽 = 𝛽𝛼
Theorem 2. Let 𝜶 and 𝜷 be two disjoint cycles in 𝑺 𝒏,
then 𝜶𝜷 = 𝜷𝜶.
Conclusion: Disjoint cycles commute
𝛼𝛽 = 2 4 3 5 1 6 8
𝛽𝛼 = 1 6 8 2 4 3 5 =
1 2
6 4
3
5
4
3
5
2
6
8
7
7
8
1
=
1 2
6 4
3
5
4
3
5
2
6
8
7
7
8
1

Consider the element 𝜎 ∈ 𝑆8
Theorem 3. Any nonidentity permutation 𝝈 ∈ 𝑺 𝒏 (𝒏 ≥
𝟐) can be written as a product of disjoint cycles,
where each cycle is of length ≥ 𝟐.
= 1 3 7 ∘
1 2
1 5
3
3
4
4
5
2
6
8
7
7
8
6
𝜎 =
1 2
3 5
3
7
4
4
5
2
6
8
7
1
8
6
= 1 3 7 ∘ 2 5 ∘
1 2
1 2
3
3
4
4
5
5
6
8
7
7
8
6
= 1 3 7 ∘ 2 5 ∘ (6 8)

Now (1 3 7) ∈ 𝑆8 can be written in the form
Hence
Theorem 4. Any cycle of length ≥ 𝟐 is either a
transposition (a 2-cycle) or can be written as a
product of transpositions (not necessarily disjoint).
Theorem 5. Any nonidentity permutation 𝝈 ∈ 𝑺 𝒏 is
either a transposition or can be written as a product
of transpositions (not necessarily disjoint).
1 3 7 = 1 7 ∘ (1 3)
𝜎 =
1 2
3 5
3
7
4
4
5
2
6
8
7
1
8
6
= 1 3 7 ∘ 2 5 ∘ 6 8 = 1 7 ∘ (1 3) ∘ 2 5 ∘ 6 8

Inverse of a Permutation (1 of 2)
Let 𝜎 ∈ 𝑆3 be given by:
Then 𝜎−1
is obtained by simply reversing the orders of
the pairs in 𝜎. That is
Let 𝛼 ∈ 𝑆3 be given by:
Then
𝜎 =
1 2
3 1
3
2
= (1 3 2)
𝜎−1
=
3 1
1 2
2
3
=
1 2
2 3
3
1
= (1 2 3)
𝛼 =
1 2
3 2
3
1
= (1 3)
𝛼−1
=
3 2
1 2
1
3
=
1 2
3 2
3
1
= (1 3)

Inverse of a Permutation (2 of 2)
Let 𝛽 ∈ 𝑆4 be given by:
For 𝛽−1
we obtain
𝛽−1
=
2 3
1 2
4
3
1
4
=
1 2
4 1
3
2
4
3
= (1 4 3 2)
𝛽 =
1 2
2 3
3
4
4
1
= (1 2 3 4)

Odd and Even Permutations (1 of 1)
A permutation is called even if it can be written as the
product of an even number of transpositions.
A permutation is called odd if it can be written as the
product of an odd number of transpositions.
Theorem 6. Any permutation of 𝑺 𝒏 is either an even
permutation or an odd permutation, but never both.
Consider the element 𝛼 ∈ 𝑆8
𝛼 =
1 2
8 5
3
6
4
3
5
7
6
4
7
2
8
1
= 1 8 (2 5 7)(3 6 4)
= 1 8 (2 7)(2 5)(3 4)(3 6)

Cycles to Transpositions (1 of 1)
Consider the following 4-cycle from 𝑆4
How do we decompose it into a product of 2-cycles
(transpositions)?
Similarly for the 5-cycle (1 2 3 4 5) from from 𝑆5, we
have
Notice that cycles of odd length decompose into an
even number of 2-cycles and that cycles of even length
decompose into an odd number of 2-cycles
1 2
2 3
3
4
4
1
= (1 2 3 4)
1 2 3 4 = (1 4)(1 3)(1 2)
1 2 3 4 5 = (1 5)(1 4)(1 3)(1 2)

Alternating Group (1 of 2)
Consider the element 𝛼 ∈ 𝑆8
Hence 𝛼 is an odd permutation.
Theorem 8.5 (textbook). Let 𝑛 ≥ 2 and let 𝐴 𝑛 denote
the subset of 𝑆 𝑛 consisting of all even permutations.
Then 𝐴 𝑛 is a subgroup of 𝑆 𝑛 called the alternating group
on 𝑛 letters. In addition, 𝑆 𝑛 = 𝑛! and 𝐴 𝑛 = 𝑛!/2.
Theorem 5.3 (textbook). Let 𝐺 be a group and let 𝐻 be
a finite nonempty subset of 𝐺. Then 𝐻 is a subgroup of
𝐺 if it is closed under the group operation.
Proof. Let ℎ ∈ 𝐻. Since 𝐻 is closed under multiplication,
then ℎ, ℎ2, ℎ3, … ∈ 𝐻. But 𝐻 is finite, so all powers of ℎ
𝛼 = 1 8 (2 7)(2 5)(3 4)(3 6)

Alternating Groups (2 of 2)
Theorem 5.3 (Proof). Let ℎ ∈ 𝐻. Since 𝐻 is closed
under multiplication, then ℎ, ℎ2
, ℎ3
, … ∈ 𝐻. But 𝐻 is finite,
so all powers of ℎ cannot be distinct. Therefore ℎ has
finite order 𝑛 ≥ 1, and the distinct elements are
ℎ, ℎ2
, ℎ3
, … , ℎ 𝑛−1
, ℎ 𝑛
= 𝑒 so every power of ℎ is an
element of that set including ℎ−1
.
Proof of Theorem 8.5. 𝐴 𝑛 is a nonempty subset of a
finite set (𝑆 𝑛) and is therefore finite. Since the product of
two even permutations is even. (why?) then 𝐴 𝑛 is
closed under the group operation. By theorem 5.3, 𝐴 𝑛 is
a subgroup of 𝑆 𝑛.

Elements of 𝑺 𝟑 (1 of 4)
Consider the elements of 𝑆3
Let us denote
Then
1 2
1 2
3
3
= 𝑒 1 2
2 3
3
1
= (1 2 3)
1 2
3 1
3
2
= (1 3 2)
1 2
1 3
3
2
= (2 3)
1 2
3 2
3
1
= (1 3)
1 2
2 1
3
3
= (1 2)
𝑓 =
1 2
2 3
3
1
= (1 2 3)
𝑓2
= 1 2 3 1 2 3 =
1 2
3 1
3
2
= (1 3 2)

For 𝑓3
we have
So o 𝑓 = 3, and 𝑓 generates the cyclic subgroup
Let us denote
Then
So o 𝑔 = 2, and 𝑔 = 𝑒, 𝑔
𝑓3
= 𝑓2
𝑓 = 1 3 2 1 2 3 =
1 2
1 2
3
3
= 𝑒
𝑓 = 𝑒, 𝑓, 𝑓2
𝑔 =
1 2
1 3
3
2
= (2 3)
𝑔2
= 2 3 2 3 =
1 2
1 2
3
3
= 𝑒

Now
and
It follows that
How about the element 𝑔𝑓? Well
Similarly, 𝑔𝑓2
= 𝑓𝑔
𝑓𝑔 = 1 2 3 2 3 =
1 2
2 1
3
3
= (1 2)
𝑆3 = 𝑒, 𝑓, 𝑓2, 𝑔, 𝑓𝑔, 𝑓2 𝑔
𝑓2
𝑔 = 1 3 2 2 3 =
1 2
3 2
3
1
= (1 3)
𝑔𝑓 = 2 3 1 2 3 =
1 2
3 2
3
1
= 1 3 = 𝑓2
𝑔
𝑔𝑓2
= 2 3 1 3 2 =
1 2
2 1
3
3
= 1 2 = 𝑓𝑔

We can use the identities 𝑔𝑓 = 𝑓2
𝑔 and 𝑔𝑓2
= 𝑓𝑔
to do group multiplication without writing down the
permutations. For example
So both 𝑓𝑔 and 𝑓2
𝑔 have order 2.
Now to find the order of 𝑓2
= 1 3 2 . We have
So o 𝑓2 = 3.
𝑓𝑔 2
= 𝑓𝑔 𝑓𝑔 = 𝑓 𝑔𝑓 𝑔 = 𝑓 𝑓2
𝑔 𝑔 = 𝑓3
𝑔2
= 𝑒
𝑓2 𝑔 2 = 𝑓2 𝑔 𝑓2 𝑔 = 𝑓2 𝑔𝑓2 𝑔 = 𝑓2 𝑓𝑔 𝑔 = 𝑓3 𝑔2 = 𝑒
𝑓2 2
= 𝑓4
= 𝑓3
𝑓 = 𝑓
𝑓2 3
= 𝑓6
= 𝑓3
𝑓3
= 𝑒

Order of Permutations (1 of 2)
We may notice above that the order of the two 3-cycles
in 𝑆3 is three and the order of the 2-cycles is two.
Theorem 7. Let 𝒏 ≥ 𝟐 and let 𝝈 ∈ 𝑺 𝒏 be a cycle. Then
𝝈 is a k-cycle if and only if 𝐨 𝝈 = 𝒌.
Therefore, the order of a k-cycle is k.
How do we find the order of a permutation that is not a
cycle? Let 𝛼 ∈ 𝑆 𝑛 be a permutation. By Theorem 3, 𝛼
can be decomposed into disjoint cycles.
Theorem 8. Let 𝒏 ≥ 𝟐 and let 𝝈 ∈ 𝑺 𝒏 be a
permutation, with 𝝈 = 𝝈 𝟏 ∘ 𝝈 𝟐 ∘ 𝝈 𝟑 ∘ ⋯ ∘ 𝝈 𝒎 a product
of disjoint cycles.

Order of Permutations (2 of 2)
Theorem 8. Let 𝒏 ≥ 𝟐 and let 𝝈 ∈ 𝑺 𝒏 be a
permutation, with 𝝈 = 𝝈 𝟏 ∘ 𝝈 𝟐 ∘ 𝝈 𝟑 ∘ ⋯ ∘ 𝝈 𝒎 a product
of disjoint cycles.
Example 1. In the permutation group S8, consider the
elements 𝛼 = 1 4 7 6 and 𝛽 = 2 5 6 7 . Determine
the order of 𝛼𝛽.
Answer: Since 𝛼𝛽 is not a product of disjoint cycles, we
cannot use Theorem 8.
𝐨 𝝈 = 𝐥𝐜𝐦(𝐨 𝝈 𝟏 , 𝐨 𝝈 𝟐 , … 𝐨 𝝈 𝒎 )
𝛼𝛽 = 1 4 7 6 2 5 6 7 =
1 2
4 5
3
3
4
7
5
1
6
6
7
2
8
8
= 1 4 7 2 5

Examples (1 of 6)
Example 1 (cont’d). Since 𝛼𝛽 = 1 4 7 2 5 , then 𝛼𝛽 is
a 5-cycle. Therefore, by Theorem 7, o 𝛼𝛽 = 5.
Example 2. List all even permutations of 𝑆3. We know
that 2-cycles are odd permutations. A k-cycle with k odd
is an even permutation. Say 𝑓 = 1 2 3 = 1 3 (1 2)
and 𝑓2
= 1 3 2 = (1 2)(1 3). The remaining even
permutation is the identity 𝑒. So the alternating group
Example 3. Inverses in 𝑆3. Let 𝑆3 = 𝑒, 𝑓, 𝑓2
, 𝑔, 𝑓𝑔, 𝑓2
𝑔 .
We know that 𝑔, 𝑓𝑔 and 𝑓2
𝑔 have order 2, so they are
their own inverses. Since 𝑓3
= 𝑒, then 𝑓𝑓2
= 𝑒 and
𝑓−1 = 𝑓2. Also 𝑓2 −1 = 𝑓 by uniqueness of inverses
𝐴3 = 𝑒, 𝑓, 𝑓2

Examples (2 of 6)
Let us further examine inverses of elements in 𝑆3 in
cycle notation.
Since the 2-cycles are their own inverses, then
For the 3-cycles:
Example 4. The group 𝑆4. This group contains 4! = 24
elements. Let us list these elements starting from the
identity:
𝑆3 = 𝑒, 1 2 , 1 3 , 2 3 , 1 2 3 , (1 3 2)
1 2 −1 = (1 2) 1 3 −1 = (1 3) 2 3 −1 = (2 3)
1 2 3 −1 = (1 3 2) 1 3 2 −1
= (1 2 3)
1 2
1 2
3
3
4
4
= 𝑒

Examples (3 of 6)
Next, we list the six 4-cycles.
Now for the eight 3-cycles
1 2
2 3
3
4
4
1
= (1 2 3 4)
1 2
2 4
3
1
4
3
= (1 2 4 3)
1 2
4 3
3
1
4
2
= (1 4 2 3)
1 2
4 1
3
2
4
3
= (1 4 3 2)
1 2
3 1
3
4
4
2
= (1 3 4 2)1 2
3 4
3
2
4
1
= (1 3 2 4)
1 2
2 3
3
1
4
4
= (1 2 3 )
1 2
3 1
3
2
4
4
= (1 3 2 )
1 2
2 4
3
3
4
1
= (1 2 4 )
1 2
4 1
3
3
4
2
= (1 4 2 )

Examples (4 of 6)
More 3-cycles
Now for the six 2-cycles
1 2
3 2
3
4
4
1
= (1 3 4 )
1 2
4 2
3
1
4
3
= (1 4 3 )
1 2
1 3
3
4
4
2
= (2 3 4) 1 2
1 4
3
2
4
3
= (2 4 3)
1 2
2 1
3
3
4
4
= (1 2 )
1 2
3 2
3
1
4
4
= (1 3 )
1 2
1 3
3
2
4
4
= (2 3 ) 1 2
4 2
3
3
4
1
= (1 4 )
1 2
1 4
3
3
4
2
= (2 4 )
1 2
1 2
3
4
4
3
= (3 4 )

Examples (5 of 6)
Now for the three elements that are not cycles:
For inverses:
1 2
3 4
3
1
4
2
= (1 3)(2 4)
1 2
2 1
3
4
4
3
= (1 2)(3 4)
1 2
4 3
3
2
4
1
= (1 4)(2 3)
1 2 4 −1 = (1 4 2)
1 2 3 4 −1 = (1 4 3 2)
1 4 −1
= (1 4 )
1 2 (3 4)
−1
= 1 2 (3 4)

Examples (6 of 6)
Proof of inverses:
and vice versa.
For the element 1 2 (3 4) , this is a product of disjoint
2-cycles, so by Theorem 8:
It follows that 1 2 (3 4) is its own inverse.
1 2 3 4 1 4 3 2 =
1 2
1 2
3
3
4
4
= 𝑒
1 2 3 1 3 2 =
1 2
1 2
3
3
4
4
= 𝑒
= lcm 2 , 2 = 2
o 1 2 (3 4) = lcm(o (1 2) , o (3, 4) )

Examples (6 of 6)
Alternate (shorter) proof of Theorem 1: for 𝑛 ≥ 3, (1 2)
and (2 3) are elements of 𝑆 𝑛. But (1 2)(2 3) sends 1 to
2, while (2 3)(1 2) sends 1 to 3. It follows that the group
𝑆 𝑛 for 𝑛 ≥ 3 is nonabelian.
Example 5. Determine the center of 𝑆3.
Clearly, 𝑒 ∈ 𝑍(𝑆3). 1 2 1 3 sends 1 to 3, but 1 3 1 2
sends 1 to 2. Similarly, 1 3 2 3 sends 3 to 2, but
2 3 1 3 sends 3 to 1. So these elements do not
commute. Finally, for the two 3-cycles, 1 2 3 (1 2)
sends 1 to 3, but (1 2) 1 2 3 sends 1 to 1, Similarly,
1 3 2 (1 2) sends 1 to 1. But (1 2) 1 3 2 sends 1 to 3
𝑆3 = 𝑒, 1 2 , 1 3 , 2 3 , 1 2 3 , (1 3 2)

Examples (6 of 6)
It follows that
is trivial.
Theorem 9. For 𝑛 ≥ 3, 𝑆 𝑛 has trivial center.
Proof. Clearly
Assume that 𝑍(𝑆 𝑛) has a nonidentity element 𝜎. Then
there exists numbers 𝑖 and 𝑗 (𝑖 ≠ 𝑗), such that 𝜎 𝑖 = 𝑗.
Now, by assumption 𝑛 ≥ 3, so there exists another
number 𝑘. Consider the transposition
𝑍(𝑆3) = 𝑒
𝑒 ∈ 𝑍(𝑆 𝑛)
𝜏 = (𝑖 𝑘) ∈ 𝑆 𝑛

Examples (6 of 6)
Then we have 𝜎 𝑖 = 𝑗, and 𝜏 = (𝑖 𝑘) ∈ 𝑆 𝑛. It follows
that
Since 𝜎 𝑖 = 𝑗 and 𝜎 is one-to-one and onto. Thus
But this contradicts the assumption that 𝜎 ∈ 𝑍(𝑆 𝑛).
Therefore 𝑍(𝑆 𝑛) for 𝑛 ≥ 3 is trivial.
𝜏𝜎 𝑖 = 𝜏 𝑗 = 𝑗
𝜎𝜏 𝑖 = 𝜎 𝑘 ≠ 𝑗
𝜎𝜏 ≠ 𝜏𝜎

Section 8: Symmetric Groups

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Section 8: Symmetric Groups