Section 9: Equivalence Relations & Cosets

Abstract Algebra, Saracino
Second Edition
Section 9
Equivalence
Relations and
Cosets

Lehman College, Department of Mathematics
Relations (Definitions) (1 of 2)
Let 𝐴 and 𝐵 be two nonempty sets, then a relation (or
binary relation) ℛ from 𝐴 into 𝐵 is a set of ordered
pairs (𝑥, 𝑦), where 𝑥 ∈ 𝐴 and 𝑦 ∈ 𝐵. Recall the
Cartesian Product of 𝐴 and 𝐵, defined as the set:
𝐴 × 𝐵 = 𝑎, 𝑏 𝑎 ∈ 𝐴, 𝑏 ∈ 𝐵
Example: The set
ℛ1 = 1, 2 , 2, 5 , 3, −4 , 5, 2
is a relation from ℤ into ℤ, but it is not all of ℤ × ℤ.

Relations (Definitions) (2 of 2)
The set
ℛ2 = 𝑥, 𝑦 ∈ ℝ × ℝ | 𝑥2
+ 𝑦2
= 1
is a relation from ℝ into ℝ (the unit circle centered at the
origin). It is not the entire cartesian plane ℝ2
= ℝ × ℝ.
Let 𝐴 and 𝐵 be two nonempty sets and let ℛ be a
relation from 𝐴 into 𝐵. The domain of ℛ is the set
𝑥 ∈ 𝐴 | ∃ 𝑦 ∈ 𝐵, such that 𝑥, 𝑦 ∈ 𝑅
That is, the set of all first elements of ordered pairs in ℛ.
The image or range of ℛ is the set
𝑦 ∈ 𝐵 | ∃ 𝑥 ∈ 𝐴, such that 𝑥, 𝑦 ∈ 𝑅

Equivalence Relations (1 of 3)
A relation ℛ on a set 𝑆 is a relation from 𝑆 to itself. We will
define a special relation from 𝑆 to itself.
If 𝑎, 𝑏 ∈ 𝑆 such that 𝑎, 𝑏 ∈ ℛ, we write 𝑎 ∼ 𝑏
ℛ is an Equivalence Relation on the set 𝑆 if the following
three conditions hold
(1) Reflexivity:
(2) Symmetry:
(3) Transitivity:
For every 𝑎 ∈ 𝑆, we have 𝑎 ∼ 𝑎
For every 𝑎, 𝑏 ∈ 𝑆, if 𝑎 ∼ 𝑏, then 𝑏 ∼ 𝑎
For every 𝑎, 𝑏, 𝑐 ∈ 𝑆, if 𝑎 ∼ 𝑏 and 𝑏 ∼ 𝑐,
then 𝑎 ∼ 𝑐

We have already seen an example of an equivalence
relation in Homework 5 Question 5. Let 𝐺 be a group
with identity element 𝑒., and let 𝑎, 𝑏, 𝑐 ∈ 𝐺. If there exists
𝑥 ∈ 𝐺, such that 𝑎 = 𝑥𝑏𝑥−1
, we say 𝑎 is conjugate to 𝑏.
Reflexivity.
Symmetry:
For every 𝑎 ∈ 𝑆, we have 𝑎 ∼ 𝑎 since
For every 𝑎, 𝑏 ∈ 𝑆, if 𝑎 ∼ 𝑏, then 𝑏 ∼ 𝑎
𝑎 = 𝑒𝑎𝑒−1
with 𝑒 ∈ 𝐺
Since 𝑎 ∼ 𝑏 means there exists 𝑥 ∈ 𝐺 such that
𝑎 = 𝑥𝑏𝑥−1
. It follows that
𝑥−1 𝑎𝑥 = 𝑥−1 𝑥𝑏𝑥−1 𝑥 and 𝑏 = 𝑥−1 𝑎𝑥 = 𝑥−1 𝑎 𝑥−1 −1

From the last statement of the preceding slide, 𝑏 ∼ 𝑎.
Transitivity.
It follows that
For every 𝑎, 𝑏, 𝑐 ∈ 𝑆, if 𝑎 ∼ 𝑏 and 𝑏 ∼ 𝑐,
then 𝑎 ∼ 𝑐
Since 𝑎 ∼ 𝑏 means there exists 𝑥 ∈ 𝐺 such that
𝑎 = 𝑥𝑏𝑥−1
. And 𝑏 ∼ 𝑐 means there exists 𝑦 ∈ 𝐺
such that 𝑏 = 𝑦𝑐𝑦−1
𝑎 = 𝑥𝑏𝑥−1 = 𝑥 𝑦𝑐𝑦−1 𝑥−1 = 𝑥𝑦𝑐𝑦−1 𝑥−1
= (𝑥𝑦)𝑐 𝑥𝑦 −1

Modular Arithmetic (1 of 2)
Recall modular arithmetic
Because 8 − 2 is divisible by 3 (integer division). In
general for integers 𝑎, 𝑏 and for positive integers 𝑛
if and only if 𝑛 divides the difference 𝑎 − 𝑏. We will show
that modular arithmetic forms an equivalence relation
on the set of integers.
(1) Reflexivity. Let 𝑛 be a positive integer and let 𝑎 be
any integer. Then 𝑎 ≡ 𝑎 (mod 𝑛) since 𝑎 − 𝑎 = 0 is
divisible by any positive integer 𝑛
8 ≡ 2 (mod 3)
𝑎 ≡ 𝑏 (mod 𝑛)

Modular Arithmetic (2 of 2)
(2) Symmetry. Let 𝑛 be a positive integer and let 𝑎 and
𝑏 be integers, such that 𝑎 ≡ 𝑏 (mod 𝑛). Then 𝑎 − 𝑏 is
divisible by 𝑛. It follows that 𝑏 − 𝑎 is also divisible by 𝑛
and
(2) Transitivity. Let 𝑛 be a positive integer and let 𝑎, 𝑏
and 𝑐 be integers, such that 𝑎 ≡ 𝑏 (mod 𝑛) and 𝑏 ≡
𝑐 (mod 𝑛). Then both 𝑎 − 𝑏 and 𝑏 − 𝑐 are divisible by 𝑛.
It follows their sum is also divisible by 𝑛. That is
Since 𝑎 − 𝑐 is divisible by 𝑛, then by definition
𝑏 ≡ 𝑎 (mod 𝑛)
𝑎 − 𝑏 + 𝑏 − 𝑐 = 𝑎 − 𝑐
𝑎 ≡ 𝑐 (mod 𝑛)

Equivalence Classes (1 of 5)
Let 𝑆 be a set and let ∼ be an equivalence relation on 𝑆.
Then the set of all 𝑥 ∈ 𝑆, such that 𝑥 ∼ 𝑎 is called the
equivalence class of the element 𝑎, and denoted 𝑎 or
sometimes 𝑎 . In set notation:
Theorem 9.1 (textbook): Let 𝑆 be a set and let ∼ be an
equivalence relation 𝑆. Then every element of 𝑆 is in
exactly one equivalence class under ∼. That is, the
equivalence classes partition 𝑆 into a family of mutually
disjoint, nonempty subsets. Conversely, given any
partition of a set 𝑆 into mutually disjoint, nonempty
subsets, then there is an equivalence relation on
𝑎 = 𝑥 ∈ 𝑆 | 𝑥 ∼ 𝑎

Theorem 9.1 (cont’d): Conversely, given any partition
of a set 𝑆 into mutually disjoint, nonempty subsets, then
there is an equivalence relation on 𝑆 whose equivalence
classes are precisely the subsets in the given partition
of 𝑆.
Proof. By reflexivity 𝑎 ∈ 𝑎 for any 𝑎 ∈ 𝑆. So, we see that
every element of 𝑆 is contained in some equivalence
class and that each equivalence class is nonempty. We
will now show that if 𝑎 ∼ 𝑏, then 𝑎 = 𝑏, otherwise their
respective equivalence classes are disjoint. Recall
From 𝑎 ∼ 𝑏 then 𝑎 ∈ 𝑏. Now, let 𝑦 ∈ 𝑎 then 𝑦 ∼ 𝑎, and
𝑏 = 𝑥 ∈ 𝑆 | 𝑥 ∼ 𝑏

Theorem 9.1 (cont’d): From 𝑎 ∼ 𝑏 then 𝑎 ∈ 𝑏. Now, let
𝑦 ∈ 𝑎 then 𝑦 ∼ 𝑎, and 𝑦 ∼ 𝑏. Hence 𝑎 ⊆ 𝑏. Similarly, we
can show that 𝑏 ⊆ 𝑎. Therefore, 𝑎 = 𝑏. If not, suppose
the intersection is nonempty,
It follows that 𝑥 ∈ 𝑎 and 𝑥 ∈ 𝑏. Hence 𝑥 ∼ 𝑎 and 𝑥 ∼ 𝑏.
By symmetry then by transitivity we have 𝑎 ∼ 𝑏, a
contradiction.
We can show that 𝑎 = 𝑏 if and only if 𝑎 ∼ 𝑏. We have
shown one part already. We only need to show that if
𝑎 = 𝑏 then 𝑎 ∼ 𝑏. Since 𝑎 ∈ 𝑎 and 𝑎 = 𝑏 then 𝑎 ∈ 𝑏
and 𝑎 ∼ 𝑏. That is each element is a representative.
𝑥 ∈ 𝑎 ∩ 𝑏

Recall congruence modulo 𝑛as an equivalence class.
There 𝑎 ≡ 𝑏 (mod 𝑛) if and only if 𝑛 divides 𝑎 − 𝑏. By
analogy in groups 𝑎 − 𝑏 would be 𝑎𝑏−1
.
Theorem 9.2. Let 𝐺 be a group and let 𝐻 be a subgroup
of 𝐺. Define a relation ≡ 𝐻 on 𝐺 by the following
Then ≡ 𝐻 is an equivalence operation.
(1) Reflexivity. 𝑎 ≡ 𝐻 𝑎 for all 𝑎 ∈ 𝐺 since 𝑎𝑎−1
= 𝑒 ∈ 𝐻.
2) Symmetry. Let 𝑎, 𝑏 ∈ 𝐺, such that 𝑎 ≡ 𝐻 𝑏. That is
𝑎𝑏−1
∈ 𝐻. But 𝐻 is closed under inverses, so
𝑥 ≡ 𝐻 𝑦 if and only if 𝑥𝑦−1
∈ 𝐻
𝑎𝑏−1 −1
= 𝑏−1 −1
𝑎−1
= 𝑏𝑎−1
∈ 𝐻

(2) Symmetry. Let 𝑎, 𝑏 ∈ 𝐺, such that 𝑎 ≡ 𝐻 𝑏. That is
𝑎𝑏−1
∈ 𝐻. But 𝐻 is closed under inverses, so
It follows that 𝑏 ≡ 𝐻 𝑎.
2) Transitivity. Let 𝑎, 𝑏, 𝑐 ∈ 𝐺, such that 𝑎 ≡ 𝐻 𝑏 and
𝑏 ≡ 𝐻 𝑐. That is 𝑎𝑏−1
∈ 𝐻 and 𝑏𝑐−1
∈ 𝐻. But 𝐻 is closed
under products, so
It follows that 𝑎 ≡ 𝐻 𝑐
𝑎𝑏−1 −1
= 𝑏−1 −1
𝑎−1
= 𝑏𝑎−1
∈ 𝐻
𝑎𝑏−1
𝑏𝑐−1
= 𝑎 𝑏−1
𝑏 𝑐−1
= 𝑎𝑐−1

Right Cosets (1 of 8)
Let G be a group and let H be a subgroup of G. A Right
Coset of H in G is a subset of the form 𝐻𝑎 for some
element 𝑎 ∈ 𝐺. That is
Think of these as right-translates of 𝐻 by the element 𝑎.
Every element of 𝐻 gets translated by 𝑎 on the right
(especially in additive notation).
Example 1. Let 𝐺 be the group ℤ6 under addition
modulo 6. That is ℤ6 = 0, 1, 2, 3, 4, 5 , and let 𝐻 = 0, 3 .
Let us construct all right cosets of 𝐻 in ℤ6.
𝐻𝑎 = ℎ𝑎 | ℎ ∈ 𝐻
𝐻 + 0 = 0 + 0, 3 + 0 = 0, 3 = 𝐻

Next
Observe that the right cosets are in set notation, so
there are only three distinct right cosets:
𝐻 + 1 = 0 + 1, 3 + 1 = 1, 4
𝐻 + 2 = 0 + 2, 3 + 2 = 2, 5
𝐻 + 3 = 0 + 3, 3 + 3 = 3, 0
𝐻 + 4 = 0 + 4, 3 + 4 = 4, 1
𝐻 + 5 = 0 + 5, 3 + 5 = 5, 2
𝐻 + 0 = 𝐻 + 3 = 0, 3 = 𝐻
𝐻 + 1 = 𝐻 + 4 = 1, 4
𝐻 + 2 = 𝐻 + 5 = 2, 5

Note above than any two right corset are either equal or
disjoint.
Example 2. Let 𝐺 be the group 𝑆3 and let 𝐻 = 𝑒, (1 2) .
For the right cosets:
𝑆3 = 𝑒, 1 2 , 1 3 , 2 3 , (1 2 3), (1 3 2)
𝐻𝑒 = 𝑒𝑒, 1 2 𝑒 = 𝑒, (1 2) = 𝐻
𝐻(1 2) = 𝑒(1 2), 1 2 (1 2) = (1 2), 𝑒)
𝐻(1 3) = 𝑒(1 3), 1 2 (1 3) = (1 3), (1 3 2)
𝐻(2 3) = 𝑒(2 3), 1 2 (2 3) = (2 3), (1 2 3)
𝐻(1 2 3) = 𝑒(1 2 3), 1 2 ( 1 2 3) = (1 2 3), (2 3)
𝐻(1 3 2) = 𝑒(1 3 2), 1 2 ( 1 3 2) = (1 3 2), (1 3)

Note again that of the six right cosets only three are
distinct:
Theorem 9.3 (textbook). Let 𝐺 be a group and let 𝐻 be
a subgroup of 𝐺. Define an equivalence relation ≡ 𝐻 on
𝐺 by the following
Let 𝑎 denote the equivalence class of 𝑎 ∈ 𝐺 under ≡ 𝐻.
Then
𝐻𝑒 = 𝐻(1 2) = 𝑒, (1 2) = 𝐻
𝐻(1 3) = 𝐻(1 3 2) = (1 3), (1 3 2)
𝐻(2 3) = 𝐻(1 2 3) = (2 3), (1 2 3)
𝑥 ≡ 𝐻 𝑦 if and only if 𝑥𝑦−1
∈ 𝐻
𝑎 = 𝐻𝑎

Theorem 9.3 (cont’d). Then 𝑎 = 𝐻𝑎.
And the equivalence classes of ≡ 𝐻 are precisely the
right cosets of 𝐻 in 𝐺.
Proof. We will show that if 𝑥 ∈ 𝐺, then 𝑥 ∈ 𝑎 if and only
if 𝑥 ∈ 𝐻𝑎. Now 𝑥 ∈ 𝑎 iff (by definition) 𝑥 ≡ 𝐻 𝑎 iff (by
definition) 𝑥𝑎−1
∈ 𝐻 iff 𝑥𝑎−1
= ℎ for some ℎ ∈ 𝐻 if and
only if 𝑥 = ℎ𝑎 for some ℎ ∈ 𝐻 iff 𝑥 ∈ 𝐻𝑎.
Corollary 9.4. Let 𝐻 be a subgroup of a group 𝐺, and
let 𝑎, 𝑏 ∈ 𝐺. Then 𝐻𝑎 = 𝐻𝑏 iff 𝑎𝑏−1 ∈ 𝐻
Proof. From theorem 9.3, 𝐻𝑎 = 𝑎, and 𝐻𝑏 = 𝑏. From
Theorem 9.1, 𝑎 = 𝑏 iff 𝑎 ≡ 𝐻 𝑏 iff (definition) 𝑎𝑏−1
∈ 𝐻.

Theorem 1. Let 𝐻 be a subgroup of a group 𝐺. If 𝑎 ∈ 𝐺,
then 𝐻𝑎 = |𝐻|. That is, the cardinality of every right
coset is the same.
Proof. To show that 𝐻 = 𝐻𝑎 , we will show that there
is a one-to-one and onto function from 𝐻 to 𝐻𝑎. Define
𝑓: 𝐻 → 𝐻𝑎 by 𝑓 ℎ = ℎ𝑎 for all ℎ ∈ 𝐻. To show that 𝑓 is
one-to-one, let ℎ1, ℎ2 ∈ 𝐻. Then, 𝑓 ℎ1 = ℎ1 𝑎 and we
also have 𝑓 ℎ2 = ℎ2 𝑎. If 𝑓 ℎ1 = 𝑓 ℎ2 , then ℎ1 𝑎 = ℎ2 𝑎.
By right-cancellation of the element 𝑎, we have ℎ1 = ℎ2
and 𝑓 is one-to-one. To show 𝑓 is onto the set 𝐻𝑎, pick
an arbitrary element ℎ𝑎 ∈ 𝐻𝑎, but there exists ℎ ∈ 𝐻,
such that ℎ𝑎 = 𝑓(ℎ). It follows that 𝑓 is onto the set 𝐻𝑎.

Example 3. Let 𝐺 be the group 𝑆3 and let
Now
For the right cosets:
All six right cosets have cardinality 3 from Theorem 1.
𝐻 = 𝐴3 = 𝑒, 1 2 3 , (1 3 2)
𝑆3 = 𝑒, 1 2 , 1 3 , 2 3 , (1 2 3), (1 3 2)
𝐻𝑒 = 𝑒𝑒, 1 2 3 𝑒, 1 3 2 𝑒 = 𝑒, 1 2 3 , (1 3 2 = 𝐻
𝐻(1 2) = 𝑒 1 2 , 1 2 3 1 2 , (1 3 2)(1 2) = 1 2 , 1 3 , (2 3)
𝐻(1 3) = 𝑒 1 3 , 1 2 3 1 3 , (1 3 2)(1 3) = 1 3 , 2 3 , (1 2)
𝐻(2 3) = 𝑒 2 3 , 1 2 3 2 3 , (1 3 2)(2 3) = 2 3 , 1 2 , (1 3)
𝐻(1 2 3) = 𝑒 1 2 3 , 1 2 3 1 2 3 , (1 3 2)(1 2 3) = 1 2 3 , 1 3 2 , 𝑒)
𝐻(1 3 2) = 𝑒 1 3 2 , 1 2 3 1 3 2 , (1 3 2)(1 3 2) = 1 3 2 , 𝑒, (1 2 3)

Note that of the six right cosets only two are distinct:
Let us check on Corollary 9.4. Let 𝐻 be a subgroup of
a group 𝐺, and let 𝑎, 𝑏 ∈ 𝐺. Then 𝐻𝑎 = 𝐻𝑏 iff 𝑎𝑏−1 ∈ 𝐻
We see that 𝐻(1 2) = 𝐻(1 3), so 𝑎 = (1 2) and 𝑏 = (1 3)
Then 𝑎𝑏−1
= 1 2 1 3 −1
= 1 2 1 3 =
1
3
2
1
3
2
= (1 3 2)
and 1 3 2 ∈ 𝐴3 = 𝐻.
𝐻𝑒 = 𝐻(1 2 3) = 𝐻(1 3 2) = 𝑒, 1 2 3 , (1 3 2 = 𝐻
𝐻(1 2) = 𝐻(1 3) = 𝐻(2 3) = 1 2 , 1 3 , (2 3)

Left Cosets (1 of 6)
Let G be a group and let H be a subgroup of G. A Left
Coset of H in G is a subset of the form 𝑎𝐻 for some
element 𝑎 ∈ 𝐺. That is
Think of these as left-translates of 𝐻 by the element 𝑎.
Every element of 𝐻 gets translated by 𝑎 on the left
(especially in additive notation).
Example 4. Let 𝐺 be the group ℤ6 under addition
modulo 6. That is ℤ6 = 0, 1, 2, 3, 4, 5 , and let 𝐻 = 0, 3 .
Let us construct all right cosets of 𝐻 in ℤ6.
𝑎𝐻 = 𝑎ℎ| ℎ ∈ 𝐻
0 + 𝐻 = 0 + 0, 0 + 3 = 0, 3 = 𝐻

Next
Observe that there are only three distinct left cosets:
Note also that the set of right cosets of 𝐻 in 𝐺 coincide
with the set of left cosets of of 𝐻 in 𝐺 from Example 1.
1 + 𝐻 = 1 + 0, 1 + 3 = 1, 4
2 + 𝐻 = 2 + 0 + 2, 2 + 3 = 2, 5
3 + 𝐻 = 3 + 0, 3 + 3 = 3, 0
4 + 𝐻 = 4 + 0, 4 + 3 = 4, 1
5 + 𝐻 = 5 + 0, 5 + 3 = 5, 2
0 + 𝐻 = 3 + 𝐻 = 0, 3 = 𝐻
1 + 𝐻 = 4 + 𝐻 = 1, 4
2 + 𝐻 = 5 + 𝐻 = 2, 5

Example 5. Let 𝐺 be the group 𝑆3 and let 𝐻 = 𝑒, (1 2) .
For the left cosets:
𝑆3 = 𝑒, 1 2 , 1 3 , 2 3 , (1 2 3), (1 3 2)
𝑒𝐻 = 𝑒𝑒, 𝑒 1 2 = 𝑒, (1 2) = 𝐻
1 2 𝐻 = 1 2 𝑒, 1 2 (1 2) = (1 2), 𝑒)
1 3 𝐻 = 1 3 𝑒, (1 3) 1 2 = (1 3), (1 2 3)
2 3 𝐻 = 2 3 𝑒, (2 3) 1 2 = (2 3), (1 3 2)
(1 2 3)𝐻 = 1 2 3 𝑒, ( 1 2 3) 1 2 = (1 2 3), (1 3)
(1 3 2)𝐻 = 𝑒(1 3 2), ( 1 3 2) 1 2 = (1 3 2), (2 3)

Note again that of the 6 right cosets only 3 are distinct:
Observe that 𝐻 is both a left coset and a right coset, but
no other left coset is a right coset and vice versa.
We will later look at subgroups with the property that
right cosets are left cosets. Such subgroups will be
called normal. As is Example 4, this is true of subgroups
of Abelian groups, but we will be particularly interested
in normal subgroups of nonabelian groups
𝑒𝐻 = (1 2)𝐻 = 𝑒, (1 2) = 𝐻
(1 3)𝐻 = 1 2 3 𝐻 = (1 3), (1 2 3)
(2 3)𝐻 = 1 3 2 𝐻 = (2 3), (1 3 2)

Example 6. Let 𝐺 be the group of unit quaternions 𝒬8
and let 𝐻 = 1, −1 be a subgroup.
The group is given by
For the left cosets of 𝐻 in 𝐺, we have
ℍ = 𝒬8 = 1, −1, 𝑖, −𝑖, 𝑗, −𝑗, 𝑘, −𝑘
1𝐻 = 1(1), 1(−1) = 1, −1 = 𝐻
(−1)𝐻 = −1 1, −1 (−1) = −1, 1
𝑗𝐻 = 𝑗(1), 𝑗(−1) = 𝑗, −𝑗
(−𝑖)𝐻 = −𝑖 1, −𝑖 (−1) = −𝑖, 𝑖
𝑖𝐻 = 𝑖(1), 𝑖(−1) = 𝑖, −𝑖
(−𝑗)𝐻 = −𝑗 1, −𝑗 (−1) = −𝑗, 𝑗
𝑘𝐻 = 𝑘(1), 𝑘(−1) = 𝑘, −𝑘
(−𝑘)𝐻 = −𝑘 1, −𝑘 (−1) = −𝑘, 𝑘

Note that of the 8 right cosets only 4 are distinct:
and let 𝐻 = 1, −1, 𝑖, −𝑖 be a subgroup.
1𝐻 = −1 𝐻 = 1, −1 = 𝐻
𝑖𝐻 = −𝑖 𝐻 = 𝑖, −𝑖
𝑗𝐻 = −𝑗 𝐻 = 𝑗, −𝑗
𝑘𝐻 = −𝑘 𝐻 = 𝑘, −𝑘
1𝐻 = 1 1 , 1 −1 , 1 𝑖 , 1(−𝑖) = 1, −1, 𝑖, −𝑖 = 𝐻
𝑖𝐻 = 𝑖 1 , 𝑖 −1 , 𝑖 𝑖 , 𝑖(−𝑖) = 𝑖, −𝑖, −1, 1
(−𝑖)𝐻 = −𝑖 1 , −𝑖 −1 , −𝑖 𝑖 , −𝑖(−𝑖) = −𝑖, 𝑖, 1, −1
(−1)𝐻 = −1 1 , −1 −1 , −1 𝑖 , −1(−𝑖) = −1, 1, −𝑖, 𝑖

and let 𝐻 = 1, −1, 𝑖, −𝑖 be a subgroup.
Note that of the 8 right cosets only 2 are distinct:
Note that only the coset 𝐻 is a subgroup of 𝐺.
𝑗𝐻 = 𝑗 1 , 𝑗 −1 , 𝑗 𝑖 , 𝑗(−𝑖) = 𝑗, −𝑗, −𝑘, 𝑘
(−𝑗)𝐻 = −𝑗 1 , −𝑗 −1 , −𝑗 𝑖 , −𝑗(−𝑖) = −𝑗, 𝑗, 𝑘, −𝑘
𝑘𝐻 = 𝑘 1 , 𝑘 −1 , 𝑘 𝑖 , 𝑘(−𝑖) = 𝑘, −𝑘, 𝑗, −𝑗
(−𝑘)𝐻 = −𝑘 1 , −𝑘 −1 , −𝑘 𝑖 , −𝑘(−𝑖) = −𝑘, 𝑘, 𝑗, −𝑗
1𝐻 = −1 𝐻 = 𝑖𝐻 = −𝑖 𝐻 = 1, −1, 𝑖, −𝑖 = 𝐻
𝑗𝐻 = −𝑗 𝐻 = 𝑘𝐻 = −𝑘 𝐻 = 𝑗, −𝑗, 𝑘, −𝑘

Section 9: Equivalence Relations & Cosets

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Section 9: Equivalence Relations & Cosets