Question

has anyone dealt with group theory?

Answer 1

a more specific question might be useful.

Answer 2

how do you use the one step subgroup test to prove it is a subgroup

Answer 3

they have prooofs all over the internet about how it is derived but how is it applied

Answer 4

i never came across something called a onestep subgroup test in my classes

Answer 5

we had at best a 2 step, or 2 properties to compare

Answer 6

okay how do you apply that one?

Answer 7

i have some of the stupidest questions

Answer 8

inverses and closure i believe were the properties for a subgroup in my class

Answer 9

for instance (z/mz) under multiplication = (a(bar) element (z/mz) such that gcd(a(bar),m)=1
show it is a group

Answer 10

how does you material define this one step test? is it really just one step? does closed with respect to conjugates come into play?

Answer 11

yeah that is the same as us

Answer 12

Z/mZ seems to be notation for Z mod m

Answer 13

it is

Answer 14

well to show a group is to show 4 properties; closure, inverses, identity, ugh, and something else

Answer 15

associativitity :)

Answer 16

one step: let G be a group and h nonempty set of G. If ab^-1 is in h whenever a,b are in h then H is a subgroup of G

Answer 17

or is the question to show that the the set of elements that are gcd(a,m)=1 is a subgroup of Z mod m

Answer 18

show it is a group on its own

Answer 19

in doing so you will clearly have it be a subgroup of z mod m

Answer 20

and multiplication is defined as regular old grade achool multiplication right?

Answer 21

*school

Answer 22

yeah

Answer 23

2 bar *3 bar =6 bar

Answer 24

and by bar you are refering to a mod residue class

Answer 25

but alll the elemenets in G need to be relatively prime with m

Answer 26

yep

Answer 27

but how do i show that?

Answer 28

well, lets try an example: Z mod 9

0,1,2,3,4,5,6,7,8

the identity is 1, needs to be in there.

gcd(1,m) is 1 for all m right?

Answer 29

yep and 2,4,5,7,8

Answer 30

is 0? i dont think so.

Answer 31

gcd(9,0) = 9

an inverse is such that aa' = id

Answer 32

bc 0/9 =/=1

Answer 33

the group G = {1,2,4,5,7,8} mod 9

Answer 34

gcd(0,m)=m

Answer 35

the inverse are there:; 1 and 8, 2 and 7, 4 and 9

Answer 36

its simple to show for some specific mod m, but for some general mod we would have to start making valid assumptions

Answer 37

its under mutlitiplication

Answer 38

pfft, yeah ... mutliplication

Answer 39

2*5 4*7 8*8

Answer 40

those are the inverses

Answer 41

pretty tough question i think.

Answer 42

1 2 4 5 7 8
1 1 2 4 5 7 8
2 2 4 8
4 4 8 7
5 5 1 2
7 7 5 1
8 8 7

yeah, how to generalize it when ive forgotten most of the thrms is tough

Answer 43

yeah i beleive it.

Answer 44

i might give up and call it a night.

Answer 45

i will reward the effort

Answer 46

closure:

let a,b in H (calling H the subgroup of G)

since gcd(a,m)=1 and gcd(b,m)=1, then gcd(ab,m)=1 shows closure

can we use this or do we have to develop it?

Answer 47

1 is the identity in H since gcd(1,m)=1 shows identity

Answer 48

if a(bar) doesnt divide n and b(bar) doesnt divide n then ab (bar) wont divide n so therefore it is closed

Answer 49

how do we show inverse.

Answer 50

thinking .... we need to show that for any a in G, given that gcd(a,m)=1 then gcd(a',m)=1

Answer 51

this is where the forgotten thrms would come in useful :)

Answer 52

true

Answer 53

well, we know that if gcd(a,m)=1 and gcd(b,m)=1 then gcd(ab,m)=1

gcd(aa',m) = gcd(1,m) = 1

Answer 54

What is the question? I like group theory :)

Answer 55

trying to show the for some Z mod m group, the elements that are gcd(a,m)=1 form a subgroup

Answer 56

yep

Answer 57

if we already know that gcd(a,m)=1 and gcd(b,m)=1 implies that gcd(ab,m)=1 then we have both closure and a relatively same showing of identity

Answer 58

i dont know how to show inverse

Answer 59

bc associattivity is straight forward bc multiplication is

Answer 60

gcd(aa',m)=1 shows inverse since gcd(1,m)=1

Answer 61

Right, so you just need to show inverses now. If: $$\gcd(a,n)=1$$and: $$ab\equiv 1\pmod n,$$then: $$\gcd(b,n)=1.$$

Answer 62

aa' may not equal 1 though. It might be a number that is congruent to 1 modulo n.

Answer 63

yeah but you havent shown that every a has an inverse b

Answer 64

aa' = id in G, and id in multiplication is 1 .... assuming we all knnow the context of course

Answer 65

every a has an inverse since Z mod m is a cyclic group

Answer 66

we havent proved its a group yet let alone a cyclic group. thats part c

Answer 67

as of yet

Answer 68

do we have to show that Z mod m is a cyclic group? or just a group? i was under the assumption that we were trying to show that the elements of Z mod m that meet the definition gcd(a,m)=1 form a subgroup

Answer 69

first a group

Answer 70

if forming a subgroup is to be proven, then its assumed that Z mod m is known already to be a group

Answer 71

and then later a cyclic subgroup of order 4 in z/40z

Answer 72

but not under the conditions that they are realitively prime bc we do not know that every element has an inverse yet

Answer 73

we do if we know that Z mod m is a group from which we are pulling our elements for the subgroup from.

Answer 74

otherwise we seem to be having the cart before the horse to me

Answer 75

I agree.

Answer 76

Otherwise we should just start proving that Z mod m is a group under addition.

Answer 77

*multiplication, but yeah :)

Answer 78

its not a group under multiplication. 0 has no inverse.

Answer 79

i can take an random set off the integers that doesnt mean that i will be a group necisarrily

Answer 80

oh yeah .... the years play a toll i swear they do lol

Answer 81

its specified that the group only contains elements that have a gcd(a,m)=1 gcd(0,m) is not 1 there fore it is not in the set

Answer 82

a subgroup is not defined for some random non group set

Answer 83

a subset is not a subgroup

Answer 84

yeah we have a subset of numbers we need to prove that it is actually a group

Answer 85

Ah, so it seems like the way this problem stated, they might want you to not think about the larger group your set is a part of, but just to show its a group on its own? Thats weird >.<

Answer 86

yeah that is correct.

Answer 87

its late, so ill let yall iron it all out from here :)

Answer 88

So you have to go off the group axioms then. I feel like that just makes the problem harder than it has to be, but when in Rome =/

Answer 89

Thanks amistre64 :)

Answer 90

is this true a*m+1=id and because a*m+1 is not divisible by m then it is in the group showing therefor that all elements have an invserse.

Answer 91

thasnk amistre64

Answer 92

The reason 1 is in your set is because gcd(1,m)=1. To show that its the identity, you have to show that 1*x=x*1 for any x such that gcd(x,m)=1

Answer 93

er, 1*x=x*1=x

Answer 94

no inverse i have the idenity.

Answer 95

Ah, ok. Then for inverses you need to show for every a in your set, there exists b such that gcd(b,m)=1 and ab=1 mod m

Answer 96

yes

Answer 97

Do you know this theorem? If gcd(a,b)=d, then there exist integers x and y such that: $$ax+by=d$$

Answer 98

yep