Why_must_H_be_normal?
Nathan Kamgang
motivates the article
Introduction to Modular Arithmetic and Equivalence Classes
What’s the remainder of 6 when it’s divided by 5? One!
We can write: $6 = 5 \cdot 1 + 1$
What’s the remainder of 10 when divided by 5? Zero!
We can write: $10 = 5 \cdot 2 + 0$
You’ve probably encountered modular arithmetic, which is a formal way of asking “what’s the remainder of number $a$ when it’s divided by $n$?”
Formal Definition: We write $a \equiv b \pmod{n}$ (read as “$a$ is congruent to $b$ modulo $n$”) when $n$ divides $(a - b)$.
This means $(a - b)$ is a multiple of $n$. For example, $6 \equiv 1 \pmod{5}$ because $6 - 1 = 5$ is divisible by 5.
The Challenge: Negative Numbers
What’s the remainder of -1 when it’s divided by 5?
Our human intuition of division breaks down when it comes to negative numbers. Hopefully the definition of modular arithmetic can help us!
We ask ourselves: what’s the smallest non-negative number $b$ (where $0 \leq b < 5$) such that $-1 - b$ is divisible by 5?
Try it!
[Space for answer]
One answer is 4, since $-1 - 4 = -5$ is a multiple of 5. Therefore $-1 \equiv 4 \pmod{5}$.
Building Equivalence Classes
If we go through the integers one after the other and ask “what’s the smallest non-negative number $b$ such that this integer minus $b$ is a multiple of 5?” we can place each integer in a group based on the answer. We get these 5 groups:
$$[0] = {\ldots, -10, -5, 0, 5, 10, 15, \ldots}$$
$$[1] = {\ldots, -9, -4, 1, 6, 11, 16, \ldots}$$
$$[2] = {\ldots, -8, -3, 2, 7, 12, 17, \ldots}$$
$$[3] = {\ldots, -7, -2, 3, 8, 13, 18, \ldots}$$
$$[4] = {\ldots, -6, -1, 4, 9, 14, 19, \ldots}$$
These sets we created are called equivalence classes or residue classes modulo 5. The partition has three important properties:
- Every integer belongs to exactly one set (no overlaps)
- Every integer can be found in one of these sets (complete coverage)
- Each set is infinite (extends in both directions)
Try to see why this must be true from the definition of modulus we gave earlier!
A Fun Observation
Take any element from set $[1]$ and any element from set $[2]$, then add them.
I’ll do it for some random elements:
- From $[1]$: pick $6$ | From $[2]$: pick $12$ | Sum: $6 + 12 = 18$
- From $[1]$: pick $-4$ | From $[2]$: pick $7$ | Sum: $(-4) + 7 = 3$
- From $[1]$: pick $11$ | From $[2]$: pick $-3$ | Sum: $11 + (-3) = 8$
Now take the remainder of each sum when divided by 5:
- $18 \equiv 3 \pmod{5}$, so $18 \in [3]$
- $3 \equiv 3 \pmod{5}$, so $3 \in [3]$
- $8 \equiv 3 \pmod{5}$, so $8 \in [3]$
What do you observe? The result is always in set $[3]$!
This suggests a rule: $$[1] + [2] = [3]$$
More generally, if $a \in [i]$ and $b \in [j]$, then $a + b \in [(i+j) \bmod 5]$.
You’re used to adding numbers, but through this rule, we’ve created an arithmetic for sets—an arithmetic of equivalence classes! How cool is that?
Why Does This Work?
Why is this arithmetic possible?
Choosing different elements from the same equivalence class has no effect on the result! This property is called well-definedness, and it’s what makes the arithmetic work.
Let’s verify: since $[1] + [2] = [3]$, we can use any representatives from these classes:
$$[1] = [6] = [11] = [-4] = \ldots$$ $$[2] = [7] = [12] = [-3] = \ldots$$ $$[3] = [8] = [13] = [-2] = \ldots$$
Checking with different representatives:
- $1 + 2 = 3 \equiv 3 \pmod{5}$ ✓
- $6 + 7 = 13 \equiv 3 \pmod{5}$ ✓
- $11 + 12 = 23 \equiv 3 \pmod{5}$ ✓
- $(-4) + (-3) = -7 \equiv 3 \pmod{5}$ ✓
No matter which representatives we choose, the result always lands in $[3]$! The classes can be represented by any of their elements without losing generality.
Telling why this arithmetic is important
what’s 6/5 what’s this divided by 5 what’s a negative integers divided by 5 now we invoke the def of modular arithmetic to give the answer
if we ask every single integers this question and assign them a group for their answer we form this structure(then i show them those classes)
observe what we remark about those classes they partition and every elements fits in
If i add any element of those class then it does not matter the result is the same remainder five.
So those classes can be represented a lot of different ways
Modular arithmetic partition the set of z and creates a new arithmetics
the arithmetic created by z is on equivalent classes
The new arithmetics doesn’t care about the specific element that represents the classes
how to reproduce this useful idea for the group the answer is the quotient
Modular arithmetic, formalized by Guass, Was so succesful in mathematics, that it was only a matter of time since people try to apply to groupu theory. How can one go about replicating the arithemetic we developped now for group element?
brief reminder of the property we want to replicate
We started by partition z using the modular operation Then we defined an arithemetic of equivalent classes
Do you know a tool that can partition a group? Lagrange’s theorem is a foundational result in group theory. The proof of the theorem rest on that idea that we can partition a group into the left cosets of its subngroup(it works for the right coset too). SO take a subgroup N of G. and consider the partitions created by that subgroups: an image of that partionning.
Half of our job to apply this set arithmetic is done since we already have partitions akin to the equivalences classes we had? Now we have to define an arithmetic on those partitions the arithemetic rule we created. Let’s remember that rule on equivalence classes we developped. We said that adding an elements from two classes and takingaskin gth remainder of the result will always be the same no matter the elements we chose? restate the rule
NOw translated for our grou pusing + notation Translated for our group in altex using * notation isn’t the most faithful translation of that idea when applied to our group elements instead?