Abstract-Algebra-1-Cosets-and-Lagranges-Theorem, guaranteed and verified 100% Pass

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Cosets and Lagrange’s Theorem

Def. Let 𝐻 be a subgroup of 𝐺. The subset 𝑎𝐻 = {𝑎ℎ| ℎ ∈ 𝐻}, where 𝑎 ∈ 𝐺
is called the left coset of 𝐻 containing 𝑎.

The subset 𝐻𝑎 = {ℎ𝑎| ℎ ∈ 𝐻}, where 𝑎 ∈ 𝐺 is called the right coset of 𝐻
containing 𝑎.


Ex. Find the left cosets and right cosets of the subgroup 4ℤ of ℤ.


4ℤ = {… , −12, −8, −4, 0, 4, 8, 12, … }.
The left coset containing 𝑚 ∈ ℤ is set
𝑚 + 4ℤ = {… , 𝑚 − 12, 𝑚 − 8, 𝑚 − 4, 𝑚, 𝑚 + 4, 𝑚 + 8, 𝑚 + 12, … }
𝑚 = 0: 0 + 4ℤ = {… , −12, −8, −4, 0, 4, 8, 12, … }
𝑚 = 1: 1 + 4ℤ = {… ,1 − 12, 1 − 8, 1 − 4, 1, 1 + 4, 1 + 8, 1 + 12, … }
= {… , −11, −7, −3, 1 , 5, 9, 13, … }
𝑚 = 2: 2 + 4ℤ = {… , −10, −6, −2, 2, 6, 10, 14, … }
𝑚 = 3: 3 + 4ℤ = {… , −9, −5, −1, 3, 7, 11, 15, … }
𝑚 = 4: 4 + 4ℤ = {… , −12, −8, −4, 0, 4, 8, 12, … }
So the left cosets start repeating. Thus there are 4 distinct left cosets:
4ℤ, 1 + 4ℤ, 2 + 4ℤ, and 3 + 4ℤ.

Notice that cosets are either identical to another coset or disjoint (they have no
common elements) and the union of the disjoint cosets is exactly the group ℤ. This
is a feature of all cosets of a subgroup. This is called a partition of a group into the
cosets of a subgroup. The cosets 𝑎𝐻 and 𝑏𝐻 will either be the same

(e.g. 0 + 4ℤ and 4 + 4ℤ) or disjoint (e.g. (0 + 4ℤ) ∩ (1 + 4ℤ) = 𝜙) and the
union of all disjoint cosets is the original set 𝐺.

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Notice that ℤ, + is an abelian group. Thus the left coset 𝑚 + 4ℤ and the right
coset 4ℤ + 𝑚 are the same sets. So all of the left cosets are the same as the right
cosets. If 𝐺 is not abelian 𝑎𝐻 need not be the same as 𝐻𝑎 (although it might be).



Ex. Find the partition of ℤ6 into cosets of the subgroup 𝐻 = {0,3}.



Left (or right since ℤ6 is abelian) cosets of 𝐻 will be of the form:

𝑚 + 𝐻 where 𝑚 ∈ ℤ6 .
𝑚 = 0: 0 + 𝐻 = {0, 3} = 𝐻 (the set 𝐻 is always a coset associated
with the identity element in G)

𝑚 = 1: 1 + 𝐻 = {1, 4}
𝑚 = 2: 2 + 𝐻 = {2, 5}
𝑚 = 3: 3 + 𝐻 = {3, 0} = {0, 3} = 𝐻
𝑚 = 4: 4 + 𝐻 = {4, 1} = 1 + 𝐻
𝑚 = 5: 5 + 𝐻 = {4, 2} = 2 + 𝐻

So 0 + 𝐻, 1 + 𝐻, 2 + 𝐻 are the three distinct and disjoint cosets of 𝐻 and
notice that:

ℤ6 = {0 + 𝐻 } ∪ {1 + 𝐻 } ∪ {2 + 𝐻 } = {0,3} ∪ {1, 4} ∪ {2, 5}.


Notice that all of the cosets of 𝐻 have the same number of elements as 𝐻. This
is always the case.