Reasoning Example for Quotient Group Definitions

Consider the map $φ : Z \to C^{\times}$ where $n \mapsto i^{n}$ . Let's verify $φ$ is a group Homomorphism (or Group Morphism). Take any $n_{1}, n_{2} \in Z$ . Notice:

\begin{aligned} φ (n_{1} + n_{2}) & = i^{n_{1} + n_{2}} \\ = i^{n_{1}} i^{n_{2}} \\ = φ (n_{1}) φ (n_{2}) \end{aligned}

If we consider the values of $n$ that map to $i^{n}$ :

Value $n$	$φ (n)$
...	...
-1	$- i$
0	1
1	$i$
2	$- 1$
3	$- i$
4	1
...	...
Notice it's very similar to Z over nZ - The Integers Modulo n for $n = 4$ . So the image of $φ$ is ${1, i, - 1, - i}$ . In general the formula is:

φ (n) = {\begin{cases} 1 & n \equiv 0 mod 4 \\ i & n \equiv 1 mod 4 \\ - 1 & n \equiv 2 mod 4 \\ - i & n \equiv 3 mod 4 \end{cases}

Notice here that we've:

partitioned $Z$ into four subsets:
- $C_{1} = φ^{- 1} (1) = {\dots, - 4, 0, 4, \dots}$ , where here $φ^{- 1}$ is the preimage (not the inverse)
- $C_{2} = φ^{- 1} (i) = {\dots, - 3, 1, 5, \dots}$
- $C_{3} = φ^{- 1} (- 1) = {\dots, - 2, 2, 6, \dots}$
- $C_{4} = φ^{- 1} (- i) = {\dots, - 1, 3, 7, \dots}$
... but only one of these partitions is a Kernel of $φ$ . Namely:
- $C_{1} = {n \in Z | φ (n) = 1} = Ker (φ)$ since $1$ is the identity in $C^{\times}$ . Thus $C_{1} \leq Z$ .
- The other 3 are not subgroups, but they look very similar to $C_{1}$ , only with some offset in the integers. You can say each of the $C_{2}, C_{3}, C_{4}$ is a "translation" of $C_{1}$ . For example:
  - $C_{2} = {\dots, - 3, - 1, 5, \dots} = 1 + C_{1}$
  - Similarly $C_{3} = 2 + C_{1}$
  - $C_{4} = 3 + C_{1}$ .

Here we have a clear bijection between ${C_{1}, C_{2}, C_{3}, C_{4}} \to {1, i, - 1, - i}$ . Here the right is a Group, and since we have a bijection then the left set is also a group, provided we do similar operations from $C^{\times}$ . Namely, the operation on ${C_{1}, \dots, C_{4}}$ is to "add representatives" to get to the resulting partition, for each $n \in C_{k}$ being our representative for $C_{k}$ .

Operation	Result
$C_{2} ⋆ C_{3}$	$9 + 10 = 19 \in C_{4}$ so $C_{2} ⋆ C_{3} = C_{4}$ . Doing $1 + (- 2)$ or any other example yields similar results.
$C_{3} ⋆ C_{3}$	$C_{1}$ since $6 + 6 = 12 \in C_{1}$ .
...	...
What's the pattern? Notice that:

Note

Iff you have $a, b \in C_{k}$ for some $k \in {1, 2, 3, 4}$ then:

φ (a) = φ (b)

which is true by the definition of $C_{k}$ . This shows that $φ$ is well-defined. But using our notation for groups then:

φ (a) φ (b)^{- 1} = 1 \in C^{\times}

iff:

φ (a - b) = 1 \in C^{\times}

which only happens when $a - b \in Ker (φ)$ .

We've defined an equivalence relation $\sim$ on $Z$ by:

a \sim b ⟺ φ (a) = φ (b) ⟺ a - b \in Ker (φ)

So we'll adjust our notation:

C_{1} = [0] = [4] = \dots or use C_{1} = \overset{―}{0} = \overset{―}{4} = \dots

Thus:

C_{2} = [1], C_{3} = [2], C_{4} = [3]

Then:

C_{2} ⋆ C_{3} = C_{4} \to [1] + [2] = [1 + 2] = [3]

So instead of using $⋆$ we can just use normal integer addition to fill our table.

$⋆$	$[0]$	$[1]$	$[2]$	$[3]$
$[0]$	$[0]$	$[1]$	$[2]$	$[3]$
$[1]$	$[1]$	$[2]$	$[3]$	$[0]$
$[2]$	$[2]$	$[3]$	$[0]$	$[1]$
$[3]$	$[3]$	$[0]$	$[1]$	$[2]$
So we claim that:

Claim

$({[0], [1], [2], [3]}, ⋆)$ is a Group.
The map $γ : Z \to {[0], [1], [2], [3]}$ where $n \mapsto [n]$ is a Homomorphism (or Group Morphism)
The map $α : {[0], [1], [2], [3]} \to {1, i, - 1, - i}$ is an Isomorphism

The General Case

For the more general scenario, say we have some $φ : G \to H$ is a Homomorphism (or Group Morphism). Then:

$Ker (φ) = φ^{- 1} (1_{H})$ (see Kernel) is a normal subgroup of $G$
For each $h \in H$ if $h \notin image (φ)$ then $φ^{- 1} (h) = \emptyset$ . Otherwise then $φ^{- 1} (h) = g \cdot Ker (φ) = {g k | k \in Ker (φ)}$ . Altnernatively, this set is $Ker (φ) \cdot g = {k g | k \in Ker (φ)}$ for any $g \in G$ with $g \in φ^{- 1} (h)$ . We call these the cosets of $Ker (φ)$ .

So in our $C_{i}$ example we had 4 cosets. Similar to above, there's an equivalence relation on $G$ where:

g_{1} \sim g_{2} ⟺ φ (g_{1}) = φ (g_{2}) ⟺ g_{1} g_{2}^{- 1} \in Ker (φ)

The set of distinct cosets (ie: the set of equivalence classes) is a group under the operation:

[g_{1}] \cdot [g_{2}] = [g_{1} g_{2}]

Then this group is isomorphic to the image of $φ$ .

We'll do a lot more of this among many different lemmas.