COSETS
Left and Right Coset
Assume that G is a multiplicative group
Definition: Let H be a subgroup of group G and 𝑎 ∈ 𝐺, then the set
𝑎𝐻 = {𝑎ℎ ∶ ℎ ∈ 𝐻} is called the left coset of H in G determined by a, and the set
𝐻𝑎 = {ℎ𝑎 ∶ ℎ ∈ 𝐻} is called the right coset of H in G determined by a.
It is obvious that the cosets are subsets of G, since the closure axiom holds.
Note: For an additive group we just take ∀ 𝑎 ∈ 𝐺, 𝑎 + 𝐻 = {𝑎 + ℎ ∶ ℎ ∈ 𝐻} and 𝐻 +
𝑎 = {ℎ + 𝑎 ∶ ℎ ∈ 𝐻} as the left and right cosets of H in G determined by a
respectively.
Also corresponding to every element of G we have a right and left coset of H in
G. It is clear that H is itself a right as well as left coset of H in G, corresponding to the
identity element 𝑒 ∈ 𝐺, since
𝑒𝐻 = {𝑒ℎ ∶ ℎ ∈ 𝐻 }
= {ℎ ∶ ℎ ∈ 𝐻 } = 𝐻
= {ℎ𝑒 ∶ ℎ ∈ 𝐻 }
= 𝐻𝑒
But in general, the right coset of H for a may not be equal to the left coset left coset of
H for a, i.e. 𝑎𝐻 ≠ 𝐻𝑎 ∀ 𝑎 ∈ 𝐺.
Example 1:The set of integers equipped with addition is a subgroup of real numbers
equipped with addition, i.e. (ℤ , +) ⊂ (ℝ , +)
1
Let 𝑎 = ∈ (ℝ , +)
2
then,
1
1
2
1
2
ℤ + = { + 𝑧 ∀ 𝑧 ∈ ℤ}
1
1
1
1
1
2
2
2
3
2
2
1 1 3 5
2
2 2 2 2
ℤ + = {… , −2 + , −1 + , 0 + , 1 + , 2 + , … }
2
1
ℤ + = {… , − , − , , , … }
2
1
is a right coset of H for .
2
Example 2: Let 𝐺 = {1, −1, 𝑖, −𝑖} be the multiplicative group and 𝐻 = {1, −1} be a
subgroup of G. Find all the cosets of H in G.
Sol: We start by finding the left cosets of H in G.
Left cosets of H are:
1) (1)𝐻 = {1(1), 1(−1)} = {1, −1} = 𝐻
2) (−1)𝐻 = {−1(1), −1(−1)} = {−1,1} = 𝐻
3) (𝑖)𝐻 = {𝑖 (1), 𝑖(−1)} = {𝑖, −𝑖}
4) (−𝑖)𝐻 = {−𝑖(1), −𝑖(−1)} = {−𝑖, 𝑖} = (𝑖)𝐻
Right cosets of H are:
1) 𝐻 (1) = {1(1), −1(1)} = {1, −1} = 𝐻
2) 𝐻 (−1) = {1(−1), −1(−1)} = {−1,1} = 𝐻
3) 𝐻 (𝑖) = {1(𝑖), −1(𝑖)} = {𝑖, −𝑖}
4) 𝐻 (−𝑖) = {1(−𝑖), −1(−𝑖)} = {−𝑖, 𝑖} = 𝐻 (𝑖)
Here 𝐻 (𝑖) = (𝑖)𝐻 i.e. the left coset is equal to the right coset.
It follows from the fact that the group G is Abelian.
Notice that:
(i) Only two of the cosets are distinct as sets; we have called these 𝐻 and 𝐻(𝑖).
(ii) The set  = {H , H (i )} partition the four elements of G, into two subsets each containing
H = 2 elements.
Note that a coset is not essentially a subgroup. If G be the Quaternion group then
𝐻 = {1, −1} is a subgroup of G. Take a=i, then 𝐻𝑎 = {𝑖, −𝑖} which is not a subgroup
of G. (it does not contain identity).
Example 3: Consider 𝐺 = (ℤ, +) the additive group of integers. If 𝐻 = {3ℤ} then we
know, 𝐻 + 0, 𝐻 + 1, 𝐻 + 2 are the only three distinct right cosets (or left cosets) of
H in G.
In general mZ will have m distinct cosets in Z.
Also it can be seen that Z = H  ( H + 1)  ( H + 2) .
Note: In general if m is a positive integer then the subgroup 𝐻 = {𝑚ℤ} of (ℤ, +) has a
finite index m.
Example 4: Find the distinct left (or right) cosets of E = {0,  2,  4,....} = {2n : n  z} in
the group (Z, +)
Solution. The only two distinct left cosets (or right cosets) of E in Z are 0+E and 1+E
Here 0 + E = {0 + 2n : n  z} = E and 1 + E = {1 + 2n : n  z} = {  1,  3,  5....}
and E  {1 + E} = Z and E  {1 + E} = 
Example 5: Consider the subgroup H={I, (1 2 3), (1 3 2)} of the subgroup.
S 3 . The complete set of left cosets of H in S 3 are
IH = {I , I (1 2 3), I (1 3 2)} = X 1
(1 2) H = {(1 2), (2, 3), (1 3)} = X 2
(2 3) H = X 2
(1 3) H = X 2
(1 2 3) H = X 1
(1 3 2) H = X 1
Notice that:
(i) Only two of the cosets are distinct as sets; we have called these X1and X 2
(ii) The set  = { X1 , X 2 } partition the six elements of S 3 , into two subsets each containing H = 3
elements
Theorem: Let G be a group and H be a subgroup of the group G,
then ∀ 𝒙, 𝒚 ∈ 𝑮:
i. 𝑯𝒙 = 𝑯𝒚 𝒊𝒇𝒇 𝒙𝒚−𝟏 ∈ 𝑯
ii. 𝒙𝑯 = 𝒚𝑯 𝒊𝒇𝒇 𝒚−𝟏 𝒙 ∈ 𝑯
Proof:(i) The Necessary Condition:
Suppose that 𝑥𝐻 = 𝑦𝐻. To prove that 𝑥𝑦 −1 ∈ 𝐻;
We know that ∈ 𝐻𝑥(∵ 𝐻𝑥 = {ℎ𝑥 ∶ ℎ ∈ 𝐻 } 𝑎𝑛𝑑 𝑒 ∈ 𝐻).
Also 𝑦 ∈ 𝐻𝑦 ⇒ ∃ ℎ ∈ 𝐻 such that for some h we have 𝑥 = ℎ𝑦
Then
𝑥𝑦 −1 = (ℎ𝑦)𝑦 −1
[∵ 𝑦 ∈ 𝐺 ⇒ 𝑦 −1 ∈ 𝐺]
= ℎ(𝑦𝑦 −1 )
𝑏𝑦 𝑎𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑣𝑖𝑡𝑦
= ℎ𝑒
[∵ 𝑦𝑦 −1 = 𝑒]
=ℎ∈𝐻
𝑒 𝑖𝑠 𝑖𝑑𝑒𝑛𝑡𝑖𝑡𝑦
−1
Thus 𝑥𝑦 ∈ 𝐻
(ii) The Condition is Sufficient:
Let 𝑥, 𝑦 ∈ 𝐺 such that 𝑥𝑦 −1 ∈ 𝐻 where H is a subgroup of G. to prove that
𝐻𝑥 = 𝐻𝑦 we know that two right cosets of H in G are either disjoint or identical,
Thus to prove that 𝐻𝑥 ∩ 𝐻𝑦 ≠ ∅ i.e. there exist an element 𝑎 ∈ 𝐺 such that 𝑎 ∈
𝐻𝑥and 𝑎 ∈ 𝐻𝑦.
then
𝑎 ∈ 𝐻𝑥 ⇒ ∃ ℎ1 ∈ 𝐻 such that 𝑎 = ℎ1 𝑥
(𝑖)
also
𝑥𝑦 −1 ∈ 𝐻𝑥 ⇒ ∃ ℎ ∈ 𝐻 such that 𝑥𝑦 −1 = ℎ
⇒ (𝑥𝑦 −1 )𝑦 = ℎ𝑦
𝑥 (𝑦𝑦 −1 ) = ℎ𝑦
𝑥𝑒 = ℎ𝑦
𝑥 = ℎ𝑦
Hence by using (𝑖) and (𝑖𝑖)
𝑎 = ℎ1 𝑥 = ℎ1 (ℎ𝑦)
= (ℎℎ1 )𝑦 ∈ 𝐻
And 𝑎 ∈ 𝐻𝑥 ⇒ 𝑎 ∈ 𝐻𝑦
Hence 𝐻𝑥 ∩ 𝐻𝑦 ≠ ∅ ⇒ 𝐻𝑥 = 𝐻𝑦.
[∵ 𝑦 ∈ 𝐺]
𝑏𝑦 𝑎𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑣𝑖𝑡𝑦 𝑖𝑛 𝐺
[∵ 𝑦𝑦 −1 = 𝑒]
(𝑖𝑖)
[∵ ℎℎ1 ∈ 𝐻]
Properties of Cosets:
1. Any two left (or right ) cosets of a subgroup in a group are either identical or
disjoint.
2. For all 𝑎 ∈ 𝐺 there is a bijection between H and aH. Thus, every left coset of H
in G has the same cardinality (order/number of elements) as H. The same
statements are true for the right cosets of H in G.
3. For any two right cosets (or left cosets) are either equal or have no element in
common and also that union of all the right cosets (or left cosets) of H in G
will equal G.
4. For all 𝑎 ∈ 𝐺 ⇒ 𝑎𝐻 = 𝐻𝑎 iff 𝑎𝐻𝑎−1 = 𝐻.
5. Every element of G belongs to exactly one left coset of H in G. Similarly,
every element of G belongs to exactly one right coset of H in G. Thus, G is
equal to the disjoint union of the distinct left cosets of H in G. Also, G is equal
to the disjoint union of the distinct right cosets of H in G.
6. Let H be a subgroup of group G. Then the set of all left (or right) cosets of H in
G defines a partition of G
Theorem: Let G be a group and H be a subgroup of G, then G is the union of all
the right (or left) cosets of H in G (above mentioned 5th property).
i.e.
𝑮 = 𝑯 ∪ 𝑯𝒂 ∪ 𝑯𝒃 ∪ 𝑯𝒄 ∪ …
[∀ 𝒂, 𝒃, 𝒄 … ∈ 𝑮]
= 𝑯 ∪ 𝒂𝑯 ∪ 𝒃𝑯 ∪ 𝒄𝑯 ∪ …
or
= ∪ 𝑯𝒙 = ∪ 𝒙𝑯
Proof: Let e be the identity of G ⇒ e is also the identity of H.
Let 𝑥 ∈ 𝐺, then 𝑥𝑒 = 𝑥, where 𝑒 ∈ 𝐻
⇒ 𝑥 ∈ 𝐻𝑥
⇒ 𝑥 ∈ ∪ 𝐻𝑥
thus
𝐺 ⊂ ∪ 𝐻𝑥
Clearly, 𝐻𝑥 is a right coset of H in G i.e. Hx is a subset of G, thus 𝐻𝑥
𝐺
⇒ ∪ 𝐻𝑥 ⊂ 𝐺
Thus, from (𝑖) and (𝑖𝑖) we obtain:
𝐺 = ∪ 𝐻𝑥
Similar proof is for 𝐺 = ∪ 𝑥𝐻.
[∀ 𝒙 ∈ 𝑮]
[∀ 𝑥 ∈ 𝐺]
(𝑖)
⊂𝐺∀𝑥 ∈
(𝑖𝑖)
[∀ 𝑥 ∈ 𝐺]
Lagrange’s Theorem:
Statement: If G is a finite group of order m and H is a subgroup of G of
𝒎
order n, then 𝒏|𝒎, and is the number of disjoint cosets of H in G.
𝒏
Proof: Since 𝑜(𝐺 ) = 𝑚 and 𝑜(𝐻 ) = 𝑛, we have 𝑚 ≥ 𝑛. To prove n divides m, we
will show that there exist a positive integer k such that: 𝑚 = 𝑛𝑘
Let 𝑎 ∈ 𝐺, then 𝑎𝐻 is a left coset of H in G. Since 𝑜(𝐺 ) = 𝑚 ⇒ there are m
number of cosets in G but all these cosets are not distinct. Hence from the class
{𝑎𝐻 ∶ 𝑎 ∈ 𝐺 }, we choose the distinct left cosets. Since order of G is finite, there will
be a finite number of distinct left cosets of H in G.
Let
𝑎𝐻 = {𝑎1 𝐻, 𝑎2 𝐻, 𝑎3 𝐻, … , 𝑎𝑘 𝐻}
be the set of all distinct left cosets of H in G.
Then by property 5 any distinct cosets of H are disjoint, hence we can write;
𝐺 = 𝑎1 𝐻 ∪ 𝑎2 𝐻 ∪ 𝑎3 𝐻 ∪ … ∪ 𝑎𝑘 𝐻
by property 2, the cardinality of each cosets of H is the same as the cardinality of H,
hence;
𝑜(𝐺 ) = 𝑜(𝑎1 𝐻 ) + 𝑜(𝑎2 𝐻 ) + 𝑜(𝑎3 𝐻 ) + ⋯ + 𝑜(𝑎𝑘 𝐻 )
𝑘 𝑡𝑖𝑚𝑒𝑠
= 𝑜(𝐻) + 𝑜(𝐻) + 𝑜(𝐻) + ⋯ + 𝑜(𝐻)
𝑜 (𝐺 ) = 𝑘 ∙ 𝑜 (𝐻 )
𝑚 =𝑘∙𝑛
⇒ 𝑛|𝑚
𝑘=
and
𝑚
𝑛
is the number of distinct cosets of H in G.
This completes the proof.
This theorem implies that the subgroups cannot be of any size (order). There
is a strong restriction on the possible subgroups of a group i.e. the order of a
subgroup must divide the order of a group.
Example 5: Imagine a group G with 323 elements. 323 = 17 × 19this means that the
only positive divisors of 323 are 323, 19, 17 and 1. So any subgroup of G must have
order 323, 19, 17 or 1. The order of improper subgroup G is 323 and the trivial
subgroup has order 1. This means if G has any other subgroups then their orders are
17 and 19.
We saw that 17 and 19 divide the order of G, but this does not mean that there
are subgroups of G of order 17 and 19. Lagrange’s Theorem is an ‘IF, THEN’
theorem, which means that the converse may not be true. Just because a number
divides the order of a group, there is no guarantee that a subgroup of divisor’s size
will exist.
Index of a Group:
Definition: If G is a group and H is a finite subgroup of G, then the number
of distinct right (or left) cosets of H in G is known as the index of H in G.
Denoted by:
[𝐺 ∶ 𝐻]
If G is a finite group then the index of every subgroup of G is finite. Thus, for
every finite group G if the subgroup H has k distinct right (or left) cosets of itself in
G, then by Lagrange’s Theorem
𝑜(𝐺)
𝑜(𝐻)
=𝑘
𝑜(𝐺)
𝑜(𝐻)
i-e Both the order and index of a subgroup of a finite group divide the order of the
group.
⇒ [𝐺 ∶ 𝐻 ] = 𝑘 =
Example 6: Consider 𝐺 = (ℤ, +) the additive group of integers. If 𝐻 = {3ℤ} then we
know, 𝐻, 𝐻 + 1, 𝐻 + 2 are the only three distinct right cosets of H in G.
Thus [𝐺 ∶ 𝐻 ] = 3
Note: In general if m is a positive integer then the subgroup 𝐻 = {𝑚ℤ} of (ℤ, +) has a
finite index m.
Example 7: Find the distinct left (or right) cosets of E = {0,  2,  4,....} = {2n : n  z} in
the group (Z, +)
Solution. The only left cosets of E in Z are 0+E and 1+E
Here 0 + E = {0 + 2n : n  z} = E and 1 + E = {1 + 2n : n  z} = {  1,  3,  5....}
and E  {1 + E} = Z , E  {1 + E} = 
Therefore, the index of E in Z is 2