Group integers modulo n

[LCSphysicist]

TL;DR Summary

I am having trouble to understand the mechanism behind the group integer modulo 4.

Maybe my problem is misunderstand the concept of " a modulo n ". I would appreciate any help to get this concept and understand the grou´p

 

[fresh_42]

Summary:: I am having trouble to understand the mechanism behind the group integer modulo 4.

View attachment 273930
Maybe my problem is misunderstand the concept of " a modulo n ". I would appreciate any help to get this concept and understand the grou´p

Modulo $n$ means to consider the remainders by a division of an integer by $n$. In your case, you have the remainders $0,1,2,3$. But $5$ is also a possible remainder: $5=0\cdot 4+ 5$ or $1= (-1)\cdot 4+5$. And so are the others. Try to define the equivalence relation: When are two integers $n$ and $m$ in the same equivalence class?

 

[Mark44]

Something else that might be confusing is when they write "ac = e". Since the group operation is addition, ac means a + c, or 1 + 3 = 0 modulo 4.

 

[fresh_42]

Something else that might be confusing is when they write "ac = e#. Since the group operation is addition, ac means a + c, or 1 + 3 = 0 modulo 4.

Oops, I missed that. It is not only confusing, it is plain wrong, because we also have a multiplication on the classes without $e$. Additive groups shouldn't be written as multiplicative groups. It complicates the concept of a ring later on.

 

[Mark44]

Oops, I missed that. It is not only confusing, it is plain wrong, because we also have a multiplication on the classes without $e$. Additive groups shouldn't be written as multiplicative groups. It complicates the concept of a ring later on.

It certainly would have been clearer to use the symbol '+' for this additive group, instead of juxtaposing a and b. IOW, it would have been better to write "a + b = e".

 

[fresh_42]

I learned the convention that groups are written by multiplication if they are arbitrary groups and groups are all which are considered. If they are written by addition, then the convention said that those are abelian groups - another reason to use addition here. However, especially the $\mathbb{Z}_n$ are the standard examples of rings, hence we need both at the same time on the equivalence classes above. Maybe it is old fashioned to insist on such conventions, but they make life a lot easier. E.g. I do not like if matrices are written as $A=(a_{ji})$ instead of $A=(a_{ij})$. And yes, $\varepsilon$ can certainly be negative, but why?

 

[Stephen Tashi]

Summary:: I am having trouble to understand the mechanism behind the group integer modulo 4.

The concept of "addition modulo 4" can be taught without talking about groups. To introduce it in an elementary school class, we ask students to pretend that the only allowed integers are 0,1,2,3. When an addition is performed such as 3 + 2 , we require that result in ordinary arithmetic be replaced by dividing the sum by 4 and using the integer remainder from that division as the answer. So, in modulo 4 arithmetic 3+2 = 1 since, in ordinary arithmetic 3 + 2 = 5 = 2(4) + 1.

For calculations like 1 - 3, we observe that in ordinary arithmetic
1 - 3 = -2 and -2 = (-1)(4) + 2, so in modulo 4 arithmetic 1 - 3 = 2.

The algorithm requires that when we express an integer $z$ in the form $z = k(4) + r$ that we must do this in a way that makes the remainder $r$ non-negative. Sometimes this requires that we choose $k$ to be a negative number.

When we study group theory, we can take a more abstract approach. One way to define "the additive group of integers modulo 4" is to let each element of the group be an infinite set of integers.

For example, in the text you quoted, $a$ is the set of all integers whose division by 4 gives a remainder of 1. For example:

1 = 0(4) + 1

5 = 2(4) + 1

-3 = -1(4) + 1

9 = 2(4) + 1

$a = \{1,5,-3,9,...\}$

The text you quoted conveniently lists the number 1 as the initial element in the set $a$. You can interpret the contents of $e,c,d$ in a similar manner.

To define the operation $\circ$ of the group, we define addition of the elements $a \circ b$ to be the addition of the sets they represent, in the sense that we form a new set consisting of all possible sums of numbers created by adding one element from $a$ to one element of $b$.

For example:

$a \circ b = \{1,5,-3,9,...\} \circ \{2,6,-2,10,..\}$

$= \{ 1+2,1+6,1-2,1+10,...,5+2,5+6,5-2,5+10,...\}$

$= \{3,7,-1,11,..,7,11,3,15,...\} = c$

The calculation give $a\circ b = c$. To understand that the final result is $c$ you have to imagine doing an infinite number of additions, and you have to keep in mind that in set notation, we do not list the same element twice. So the set I denoted by "$\{3,7,1,11,...,7,11,3,15,...\}$" should actually be denoted using only a single "3" , a single "7", and a single "11",

An more advanced way to define the additive group of integers modulo 4 is to use the group theory concept of a "quotient group". I don't know if you've studied that concept yet.

 

[fresh_42]

An more advanced way to define the additive group of integers modulo 4 is to use the group theory concept of a "quotient group". I don't know if you've studied that concept yet.

That is not "more advanced", it is only a different notation! The equivalence classes are precisely the elements of the quotient, and the entire "quotient building" is nothing else as introducing an equivalence relation.

 

[r731]

Relabel e,a,b,c to [0],[1],[2],[3] respectively. It should make things easier to understand. Note that in the group [0] = [4] = [8] and similarly [1] = [-3] = [5] = [9] and so on.

 

[Vanadium 50]

I am not sure relabeling makes things easier to understand. I think it makes it easier for people to get the right answer for a tiny subset of problems without understanding.

Consider the group {2,4,6,8} under the operation multiplication mod 10. The identity is 6. This throws people for loops, but an important thing about studying groups is recognizing an identity by its properties, not by what it "looks like."

Sure, I can "relabel" {2,4,6,8} as {1,2,0,3} under addition mod 4, so now the identity looks like one, but I think this does the student a disservice by not letting them discover that these two groups have the same Cayley table and that there are only two possibilities that are groups: C4 and V4.