On Quotient Groups
Special Mathematics Lecture
Magnus B. Lyngby (612206006)
October 28, 2022

Contents
1 Introduction

12 Proof of Proposition 1.2.9
2.1 Proof that a necessary and sufficient condition for a normal subgroup is equality ofleft, right cosets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Proof that the quotient group is well-defined . . . . . . . . . . . . . . . . . . . . . .
2.3 Showing the size of quotient groups for finite groups . . . . . . . . . . . . . . . . . .

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Introduction|G|
for finite groups.
In this report I will prove Proposition 1.2.9. [1], as well as show that GG0 = |G0|
The proof will be broken up into three sections, one for each sub-statement of Proposition 1.2.9.

2Proof of Proposition 1.2.9

2.1

Proof that a necessary and sufficient condition for a normal subgroupis equality of left, right cosets

We want to prove:
G0 is a normal subgroup of G ⇐⇒ G0 [c] = [c]G0 ∀c ∈ G(1)By using the definition of a normal subgroup:
G0 = cG0 c−1 ∀c ∈ G Multiply both sides on the right by c(2)⇐⇒ G0 c = cG0 ∀c ∈ G Use the definition of left and right cosets(3)⇐⇒ [c]G0 = G0 [c] ∀c ∈ G(4)

By following the logic backwards we see that we have proved the statement.

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2.2

Proof that the quotient group is well-definedFirst I must prove that the product [a]G0 [b]G0 := [ab]G0 is well-defined for a normal subgroup. Theproduct is well-defined if ∀x ∈ [a]G0 and ∀y ∈ [b]G0 we have xy ∈ [ab]G0 . In words, the representative
for the equivalence class does not matter. Assuming we arbitrarily pick x, y belonging to theirrespective classes we know:[x]G0 = [a]G0(5)[x]G0 = [a]G0(6)This together with (3) lets us prove that the product is well-defined:[xy]G0 = (G0 x)y = [x]G0 y = [a]G0 y = G0 ay = a(G0 y)= a[y]G0 = a[b]G0 = aG0 b = G0 ab = [ab]G0(7)Which gives the desired result. Next we must show that the product defined on the equivalenceclasses forms a group. This is done by checking the three conditions in Def. 1.1.1.. Let a, b, c ∈ G:([a]G0 [b]G0 )[c]G0 = [ab]G0 [c]G0 = [abc]G0(8)[a]G0 ([b]G0 [c]G0 ) = [a]G0 [bc]G0 = [abc]G0(9)Thus the product is associative. Next observe that G0 is an equivalence class, simply represented
as [e]G0 , and it acts as the identity:[a]G0 [e]G0 = [ae]G0 = [a]G0(10)[e]G0 [a]G0 = [ea]G0 = [a]G0(11)Lastly I will show that [a−1 ]G0 acts as an inverse to any equivalence class represented by a ∈ G:[a]G0 [a−1 ]G0 = [aa−1 ]G0 = [e]G0 = G0−1[a−1]G0 [a]G0 = [aa]G0 = [e]G0 = G0(12)(13)

All this together shows that the quotient group is well-defined, and is a group.

2.3

Showing the size of quotient groups for finite groups

I want to show

GG0|G|= |G. We know the equivalence classes define a partition of G, per definition,0|

and we know |G0 | is the number of elements in the identity equivalence class. I want to show thatevery other equivalence class also has |G0 | elements, since then the number of equivalence classes,
G
G0

|G|
would equal |G, which is exactly what I want to show.
0|

Recall that we can transform from [e]G0 to [a]G0 by a, and we can transform the other way by
a−1 . Since the transformation is invertible, it must be bijective, which means every element in [e]G0
is mapped exactly once to every element in [a]G0 . Since a was arbitrarily chosen, this means every
equivalence class has size |G0 |, which proves the statement.

References[1]

Austen Vic and Richard Serge. Groups and their representations. 2022.

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