# Abstract Algebra Done The Way It Should Have Been: Cosets, Normal SubGroups, and Factor Groups

In my previous post, I said when a subset of the domain of a Group forms a SubGroup of that Group. In this blog post, I continue showing how Abstract Algebra can be done in a more usable and logically correct way.

So let's start by defining the term Left Coset of a subset of the domain of a Group by an element of the Group.

Let

𝒢 = <G, e, op, inv>be aGroupand letHbe a subset ofG. Letgbe anelementofGthen theofLeft CosetHbygin𝒢is the set

{x ∊ G | (∃ h ∊ H)[x = op(g, h)]}.

Now, **𝒢 **if is a Group, and **H ⊆ U(𝒢)**, and **g ∊ G** I will denote the Left Coset of **H** by **g** in **𝒢 **with **LeftCoset(g, H, 𝒢)**.

I will not explicitly define the term **Right Coset** of a subset of the domain of a Group by an element of the Group because it is practically almost the same as the left variant…

Now

𝒢if is a Group, andH ⊆ U(𝒢)we can define the set of all Left Cosets ofHby an element ofU(𝒢)in the Group𝒢like

LeftCosets(H, 𝒢) = {S ∊ 𝒫(U(𝒢)) | (∃ g ∊ U(𝒢))[S = LeftCoset(g, H, 𝒢)]}.

It is very *normal* to ask ourselves what conditions should **H** satisfy in order for **LeftCosets(H, 𝒢) **to form a Group with respect to **𝒢**?

Requiring **H **to form a subgroup of **𝒢 **is definitely a good idea because this will ensure that the identity element of **𝒢 **is in** H **and **H** will also be closed under the operations of **𝒢**.

Now let **H **forms a subgroup of **𝒢** and let’s imagine that **LeftCosets(H, 𝒢) **forms a Group. And lets this Group be **< LeftCosets(H, 𝒢), E, Op, Inv>** and let **𝒢 = <G, e, op, inv>**. It is probably a good idea to have

**E = LeftCoset(e, H, 𝒢) = {op(e, h) | h in H} = {h | h in H} = H****(∀a ∊ G)(∀b ∊ G)[Op(LeftCoset(a, H, 𝒢), LeftCoset(b, H, 𝒢)) = LeftCoset(op(a, b), H, 𝒢)]****(∀a ∊ G)[Inv(LeftCoset(a, H, 𝒢)) = LeftCoset(inv(a), H, 𝒢))]**

Lets focus on the second condition for a moment. First lets examine why we would want it.

Let **𝒢 = <G, e, op, inv>** and let **𝒢 **be a Group** **and let **H **forms a subgroup of **𝒢 **let **a ∊ G **and b **∊ G** then since **e∊ G** we have **op(a, b) = op(op(a, e), op(b, e))** and **op(a, b) = op(op(a, b), e) **and of course **op(a, b) ∊ LeftCoset(op(a, b), H, 𝒢) **whcih means that **op(a, b )** is product of elements from **LeftCoset(a, H, 𝒢)** and **LeftCoset(b, H, 𝒢)**.

Lets see if condition 2 itself requires a condition on **H**. Let **x ∊ LeftCoset(a, H, 𝒢) **and **y ∊ LeftCoset(b, H, 𝒢) **then there exists **u** in **H** and **t** in **H** such that **x = op(a, u)** and **y = op(b, t)**. Ok let then **u** and **t** be elements of **H** such that **x = op(a, u)** and **y = op(b, t)**. Then we have **op(x, y) = op(op(a, u), op(b, t)) = op(a, op(op(u, b), t)))**.

And here is the thing if **op(u, b)** was an element of **RightCoset(b, H, 𝒢)** we would have that **op(x, y)** is an element of** LeftCoset(op(a, b), H, 𝒢)**.

Indeed lets **op(u, b) **be in **RightCoset(b, H, 𝒢)** then there exists **h **in** H** such that **op(u, b) = op(b, h)**. So let **h** be an of **RightCoset(b, H, 𝒢)** such that **op(u, b) = op(b, h)**. We now have **op(x, y) = op(a, op(op(b, h), t))) = op(a, op(b, op(h, t))) = op(op(a, b), op(h, t))**. But since **H **forms a subgroup of **𝒢 **we have **op(h, t)** in **H **and so **op(op(a, b), op(h, t)) **is in** LeftCoset(op(a, b), H, 𝒢)**.

So here is what we have just proved:

**(∀g ∊ G)[LeftCoset(g, H, 𝒢) ⊆ RightCoset(g, H, 𝒢)] **⇒ **(∀a ∊ G)(∀b ∊ G)(∀x ∊ LeftCoset(a, H, 𝒢))(∀y ∊ LeftCoset(b, H, 𝒢))[op(x, y) ∊ LeftCoset(op(a, b), H, 𝒢)]**.

The converse is easy to see and check.

Here is a small exercise for the curious reader:

Let **𝒢 = <G, e, op, inv>** and let **𝒢 **be a Group** **and let **H **forms a subgroup of **𝒢 **and let's define a relation **Eq(H, 𝒢) **over the set** G** such that

**(∀a ∊ G)(∀b ∊ G)[<a, b> ∊ Eq(H, 𝒢) **⇔ **(∃ h∊ H)[b = op(h, op(a, inv(h))]]. **Show that **Eq(H, 𝒢) **is an equivalence relation**.**

We saw that if **𝒢** is group and **H **forms a subgroup of **𝒢** and **(∀g ∊ U(𝒢))[LeftCoset(g, H, 𝒢) ⊆ RightCoset(g, H, 𝒢)]** at least one good thing happens. And here is the thing if we require that **(∀g ∊ U(𝒢))[LeftCoset(g, H, 𝒢) = RightCoset(g, H, 𝒢)]** is true a lot of good things happen. We are now ready for our next definition.

Let

𝒢be group andHforms a subgroup of𝒢.We say thatHforms asubgroup ofnormal𝒢if(∀g ∊ U(𝒢))[LeftCoset(g, H, 𝒢) = RightCoset(g, H, 𝒢)].

Here is the second exercise for the curious reader (one of the good things):

Let **𝒢 = <G, e, op, inv>** and let **𝒢 **be a Group** **and let **H **forms a normal subgroup. Prove that

**(∀g ∊ U(𝒢))[ [g]_Eq(H, 𝒢) = LeftCoset(g, H, 𝒢) = RightCoset(g, H, 𝒢) ]**.

Where **[g]_Eq(H, 𝒢)** is of course the equivalence class of** g **under **Eq(H, 𝒢)**.

We are almost ready for the definition of Factor Group / Quotient Group. We miss only two pieces to solve the Puzzle. And the longer of those two pieces will of course be left as an exercise for the curious reader.

Let **𝒢 = <G, e, op, inv>** and let **𝒢 **be a Group** **and let **H **forms a normal subgroup of **𝒢**. Lets define **Inv** to be the set **{<X, inv[X]> | X ∊ G/Eq(H, 𝒢)}** where **G/Eq(H, 𝒢 ) **is of course the Quotient set of **G** by **Eq(H, 𝒢 ) (**the set of all equivalence classes of **G** by **Eq(H, 𝒢 )**) and **inv[X]** is the image of the set **X** under the function **inv** (the set **{y ∊ G | (∃ x∊ X)[y = inv(x)]}** ). And let's define **Op** to be the set

**{<T, op[T]> | T ∊ (G/Eq(H, 𝒢))²}.** I will now prove that **Inv** is a unary operation and leave it to the curious reader to prove that **Op** is a binary operation**.**

First, it is obvious that **Inv** is a function with a domain **G/Eq(H, 𝒢)** from the way it is defined. So I will show that **Range(Inv)** is a subset of **G/Eq(H, 𝒢)**.

Let **X** be an element of **G/Eq(H, 𝒢)** then let **x** be an element of **G **such that** X = [x]_Eq(H, 𝒢)**. I will now show that **Inv(X) = inv[X] = [inv(x)]_Eq(H, 𝒢)**.

Let **y** be an element of **inv[X]**. Then let **g** be an element of **X** such that** y = inv(g)**. We have that **X = [x]_Eq(H, 𝒢)** and **g ∊ X** and therefore there exists an element **h** of **H** such that **g = op(h, op(x, inv(h)))**. Let *** = op **and **$ = inv**. I will now switch to infix notation for * and prefix for **$** because the expressions will be alot more readable and easy to understand. Let then **h** be an element of **H** such that **g = h * (x * $h)**. We now have **y = $g = $(h * (x * $h)) = $(x * $h) * $h = ($$h * $x) * $h = (h * $x) * $h = h * ($x * $h) = h * (inv(x) *$h) = op(h, op(inv(x), inv(h))) ∊ [inv(x)]_Eq(H, 𝒢)**. Now when we generalize this result we get **Inv(X) = inv[X] ⊆ [inv(x)]_Eq(H, 𝒢)**.

Let now **y** be an element of **[inv(x)]_Eq(H, 𝒢)**. Therefore there exists an element **h** of **H** such that **y = op(h, op(inv(x), inv(h)))**. So let **h** be an element of **H** such that **y = op(h, op(inv(x), inv(h))**. We have **y = inv(inv(y))** which leads to **y = inv(op(h, op(inv(inv(x)), inv(h)))) = inv(op(h, op(x, inv(h))))**. Lets **g = op(h, op(x, inv(h)))**. By the definition of **Eq(H, 𝒢)** we get that **g **and **x **are in **Eq(H, 𝒢 )**. Therefore **g** in an element of **X**. But remember we have **y = inv(g) **from which follows that **y** is an element of **inv[X]. **Now when we generalize this result we get **[inv(x)]_Eq(H, 𝒢) ⊆ Inv(X) = inv[X]**.

So I have just shown that really **Inv(X) = inv[X] = [inv(x)]_Eq(H, 𝒢)**. After generalization, we get **Range(Inv)** **⊆** **G/Eq(H, 𝒢 )** which is what I wanted to show in the first place. Therefore **Inv** is indeed a unary operation over the set **G/Eq(H, 𝒢)**.

Now if **𝒢 = <G, e, op, inv>** is a Group and **H **forms a normal subgroup of **𝒢 **I will denote with **⟪op, H⟫** the set **{<T, op[T]> | T ∊ (G/Eq(H, 𝒢))²}** and with** ⟪inv, H⟫** the set **{<X, inv[X]> | X ∊ G/Eq(H, 𝒢)}**. **⟪op, H⟫ **should be read as the natural lifting of the Group operation of **𝒢 **with respect to **H. **And similarly **⟪inv, H⟫ **should be read as the natural lifting of the inverse function of **𝒢 **with respect to **H.**

Here is the last exercise for this blog post:

If **𝒢 = <G, e, op, inv>** is a Group and **H **forms a normal subgroup of **𝒢 **show that **<G/Eq(H, 𝒢), H, ⟪op, H⟫, ⟪inv, H⟫>** is a Group (this Group is called the ** Quotient** Group of

**𝒢**by

**H**or the

**Group of**

*Factor***𝒢**by

**H**). I will denote this Group simply with

**𝒢 : H**.

When **G is a finite **set we have |**G/Eq(H, 𝒢)| = |G| / |H|** because **[e]_Eq(H, 𝒢) = H **hence the name ** Quotient** Group.

I will wrap up this blog post by saying that the usual way of defining Factor Groups is ** dependent** on the Axiom of Choice while the way I have shown is not and this is actually a very big win!