Question

solve

Answer 1

@aegis_penguin come in here

Answer 2

ok this is me

Answer 3

so the way this place works, is you can post a question and get help from everyone, but we can also just come in here and talk to each other

Answer 4

the reason it is good is because we can do things like this

Answer 5

Cool

Answer 6

\(\int_0^3\sum_6^9dx\).....

Answer 7

How do you do that? Use symbols and all?

Answer 8

you might want to start to familiarize your self with latex

Answer 9

you can do it by hand

example \(\int_0^{10}f(x)dx\) is done with the code `\(\int_0^{10}f(x)dx\)`

Answer 10

or you can click down on the bottom where is says \(\large \sum\) Equation

Answer 11

but its best just to learn the latex code, it is pretty easy.

Answer 12

That's easier.

Answer 13

hehe, so what are you guys working on in group theory?

Answer 14

also, be careful place gets pretty addictive.....

Answer 15

I have answered 4000 questions....

Answer 16

We;re currently working on subgroups like the cyclic group

Answer 17

Really darn that's a lot.

Answer 18

you got any questions?

Answer 19

What is a cyclic group and the integer modulus notation meaning.

Answer 20

Abelian groups are cool.

Answer 21

One example would be \(\mathbb{Z}_5\), we start with 0 then add 1 to it, and we have 1, add 1 to that and get 2......until we get to 4 then when we add one to that, we start over

so it goes 0,1,2,3,4,0,1,2,3,40,1,2,3,4

note that 3+3 = 6 = 1
because if you start from the left of that list and count to the right after 6 moves you are on 1

Answer 22

so \(\mathbb{Z}_5=\{0,1,2,3,4\}\)
and the operation is addition modulo 5

Answer 23

What I don't get is how the modulus has to do with the remainder?

Answer 24

note that if we wanted to do a huge number like 3*37, we get 3*37 = 111 and dividing that by 5 we get a remainder of 1. so 3*37 = 1 modulo 5

Answer 25

in other words

5 divides 3*37-1

Answer 26

Also what is the Cartesian product of integer modlus something another integer modulus something?

Answer 27

one thing at a time

Answer 28

:)

Answer 29

so what is 3*18 mod 7?

Answer 30

its the remainder when 3*18 is divided by 7

Answer 31

I mean in general how do you find the Cartesian product of two sets with modulus on the product of the two?

Answer 32

this might be confusing because I switch from addition to multiplication, we will stay with addition for now

Answer 33

ok well first you need to understand how to do modulo addition before you can understand that

Answer 34

do you get it?

Answer 35

Okay

Answer 36

4+9 = 1 mod 6

Answer 37

Yes that makes sense.

Answer 38

7+6 mod 3?

Answer 39

the remainder when 7+6 is divided by 3

Answer 40

0

Answer 41

7+6 = 13
13/3 = 4 R1

Answer 42

so its 1

Answer 43

the only time it will be 0 is if the number divides the other number

example

2+6 = 0 mod 4

but

2+6 = 3 mod 5

Answer 44

I thought it was zero since /integer^ 3 = {0,1,2} and counting this 13 times I get 0. So I divide the number by the modulus and the remainder is the answer?

Answer 45

0,1,2,0,1,2,0,1,2,0,1,2,0,1,2...

if you count like that, dont count the elements, count the spaces between elements

Answer 46

0 to 1 is one move
1 to 2 is two moves
.
.
.
0 to 1 is ten moves

Answer 47

but after you understand that, dont use the counting method.

imagine if I asked what is 234353252+3423423423423 mod 7?

Answer 48

that might take a while

Answer 49

A very long time.

Answer 50

7+10 = 2 mod 5

because the 17/5 = 3 R2

Answer 51

I dont know why I put "the"

Answer 52

4+9 = ? mod 4

Answer 53

13 mod 4?

Answer 54

what is that?

Answer 55

I don't know.

Answer 56

what is the remainder when you divide 13 by 4?

Answer 57

1

Answer 58

correct

Answer 59

13 mod 5?

Answer 60

3

Answer 61

correct

Answer 62

ok now you know how to do modulo addition

Answer 63

Proving somethings in the book is hard.

Answer 64

now lets look at the direct product of two cyclic groups \(\mathbb{Z}_5, \) and \(\mathbb{Z}_4\)

so the operation is the first one is addition modulo 5, and the operation for the second one is addition modulo 4

if we look at the direct product \(\mathbb{Z}_5\times \mathbb{Z}_4\) the elements will look like \((x,y)\) where \(x\in \mathbb{Z}_5\) and \(y\in \mathbb{Z}_4\)

Answer 65

Okay so far it makes sense.

Answer 66

But why are they cyclic groups?

Answer 67

so when we add, we add component wise in the respected operation

\((3,2)+(3,3) = (3+3 mod 5, 2+3mod4)=(1,1)\)

Answer 68

we dont have to use cyclic groups, its just good for study

Answer 69

we use \(\mathbb{R}\times \mathbb{R}\) all the times, we call it \(\mathbb{R}^2\) and \(\mathbb{R}\) is not cyclic

Answer 70

all the time*

Answer 71

I don't understand how to add components like how you got (3+3mod5,2+3mod4)?

Answer 72

ok lets do \((1,2)+(4,3)\) in \(\mathbb{Z}_5\times\mathbb{Z}_4\)

we do 1+4 = ? mod 5
and 2+3 = ? mod 4

Answer 73

?? that was a question

Answer 74

Okay that makes sense, how come they didn't teach me this at group theory.

Answer 75

I dont think you guys should be spending that much time with it, but you asked....

Answer 76

\((1,2)+(4,3)=(0,1)\) right?

Answer 77

Can you show all the steps on how to do the calculations?

Answer 78

1+4 = 5 = 0 mod 5 because it has no remainder when divided by 5
2+3 = 5 = 1 because 5 leaves a remainder of 1 when divided by 4

Answer 79

how late will you be up? my food is ready and my wife is calling for me, but I can be back in a hour or so....

Answer 80

Probably up to 12;30, but this site is really cool.

Answer 81

:)
ill be back in a hour

Answer 82

ask anything in the mean time

Answer 83

Thanks Zach =)

Answer 84

What about (2,3) + (0,1) in Z5XZ4?

Is it (1,0)?

Since 2+0 = 2 and dividing 2 by 5 gives a remainder of 1
and 3+1= 4 divided by 4 gives a remainder of 0

So (1,0) ?

Answer 85

2+0 = 2 and 2 = 2 mod 5
remember \(\mathbb{Z}_5=\{0,1,2,3,4\}\) so if our element is in there, its just itself

3=3mod4
7=7mod8
but
7=2mod5 because 7 is not in Z_5

Answer 86

@aegis_penguin still here?

Answer 87

Oh sorry I'm back.

Answer 88

I'm really sleepy. When are you on?

Answer 89

I been doing math since the morning to evening and had classes till 9 pm.

Answer 90

I think my attention span reached its limit. I have some homework problems and I'll make sure to write some of them here. Are you able to go PSU at 11pm to 1pm during any of the following days: Monday, Tuesday, Wednesday, Thursday?

Answer 91

wed or thur

Answer 92

What are centralizers and normalizers in a group? That's the last thing we covered in class.

Answer 93

Also what makes a subgroup?

Answer 94

A subgroup is anytime that you have a group that is contained in another group. It is just like a subset of a set. The thing with subgroups is that we dont have to check all the axoims of a group, because if the set is a subset of a group, then for sure it is associative because the group is associative.

Answer 95

centralizers and normalizers you can Google the definition. That is all I would do to give you an exact definition. You will not really get to see what those two things are about until later in algebra.