Abstract-Algebra-1-Factor-Quotient-Groups, guaranteed and verified 100% Pass

1


Factor/Quotient Groups

Def. Let 𝐻 be a normal subgroup of a group 𝐺 (i.e. 𝑔𝐻 = 𝐻𝑔, for any 𝑔 ∈ 𝐺).

We define 𝐺/𝐻 (called “𝐺 𝑚𝑜𝑑 𝐻”) to be the set of distinct cosets of 𝐻 in 𝐺.

𝑮/𝑯 = {𝒂𝟏 𝑯, 𝒂𝟐 𝑯, … }


We define the product of two elements (i.e. cosets) of 𝐺/𝐻 by:

(𝑥𝐻 )(𝑦𝐻 ) = {(𝑥ℎ1 )(𝑦ℎ2 )⃒ℎ1 , ℎ2 ∈ 𝐻} = 𝑥𝑦𝐻



𝐺/𝐻 is a group with this multiplication and is called a factor group or
quotient group.



First, let’s show if 𝐻 is a normal subgroup of 𝐺 then

(𝑥𝐻 )(𝑦𝐻 ) = {(𝑥ℎ1 )(𝑦ℎ2 )⃒ℎ1 , ℎ2 ∈ 𝐻} is equal to 𝑥𝑦𝐻.


(𝑥ℎ1 )(𝑦ℎ2 ) = 𝑥 (ℎ1 𝑦)ℎ2 , because multiplication is associative.
Since 𝐻 is normal i.e. 𝑦𝐻 = 𝐻𝑦 for all 𝑦 ∈ 𝐺, there is an ℎ3 ∈ 𝐻

such that ℎ1 𝑦 = 𝑦ℎ3 .


So (𝑥ℎ1 )(𝑦ℎ2 ) = 𝑥(ℎ1 𝑦)ℎ2

= 𝑥 (𝑦ℎ3 )ℎ2
= 𝑥𝑦(ℎ3 ℎ2 ) ∈ 𝑥𝑦𝐻.

, 2


Let’s show 𝐺/𝐻 is a group.
0) We just saw that it’s closed under multiplication.
1) The multiplication is associative because the group multiplication in 𝐺 is
associative.
(𝑎𝐻 )(𝑏𝐻𝑐𝐻 ) = (𝑎𝐻 )(𝑏𝑐𝐻 ) = (𝑎𝑏𝑐 )𝐻
(𝑎𝐻𝑏𝐻 )(𝑐𝐻 ) = (𝑎𝑏𝐻 )(𝑐𝐻 ) = (𝑎𝑏𝑐 )𝐻.
2) The identity element is the coset 𝑒𝐻 = 𝐻.
3) Given 𝑎𝐻, 𝑎−1 𝐻 is the inverse element (coset) in 𝐺/𝐻 since
(𝑎𝐻 )(𝑎−1 𝐻) = (𝑎𝑎−1 )𝐻 = 𝑒𝐻 = 𝐻.


Ex. Let 𝐺 = ℤ and 𝐻 = 4ℤ = {… , −8, −4, 0, 4, 8 … }. Identify the

elements of 𝐺/𝐻 = ℤ/4ℤ .


Since 𝐺 is abelian, 𝐻 is a normal subgroup of 𝐺.

The factor group ℤ/4ℤ is the set of cosets of 𝐻 = 4ℤ in 𝐺 = ℤ.

That is, the elements of 𝐺/𝐻 are:

0 + 4ℤ = {… , −8, −4, 0, 4, 8 … }
1 + 4ℤ = {… , −7, −3, 1, 5, 9 … }
2 + 4ℤ = {… , −6, −2, 2, 6, 10 … }
3 + 4ℤ = {… , −5, −1, 3, 7, 11 … }.
If we want to “multiply” two elements, say 2 + 4ℤ and 3 + 4ℤ,
we do it by:
(2 + 4ℤ)(3 + 4ℤ) = (2 + 3) + 4ℤ = 5 + 4ℤ = 1 + 4ℤ.