n00b Group Theory question

PaulieBoy

Is this a group : ({n: n=10k for some k which is an element of Z} , addition as the operator)

This is my best attempt :

1) Closure : n = n , so is closed ? I am working on the assumption that n is the only element under consideration, so there is nothing to add ?

2) Identity : n + e = e = e + n . Not sure about this! I think e = 0, and would exist when k=0. So n can take the value 0.

3) Inverse : n + a^-1 = e . Inverse is -n, again I'm 'iffy' about this answer!

Not sure about any of the above, so any input would be great as it would help with this question and some others.
Thanks for your time.

Michael Collins

Is the group defined for a single k picked from the integers, or is k the set of all integers.

The way it is stated suggests the former. In which case the group (should it exist) just has a single element.

The latter suggests the group is a multiple of 10, so the group consists of {...-20, -10, 0, 10, 20...}.

Can you clarify which of these is being asked?

Fremen

PaulieBoy wrote: »

Is this a group : ({n: n=10k for some k which is an element of Z} , addition as the operator)

Start by listing five or six elements of the set.

1) Closure : n = n , so is closed ?

I think you've misunderstood the definition of the set. It should be read as "the set of all n such that n = 10k for some integer k". It's very important to get this notation for defining sets clear in your head for maths in general.

Now what's the definition of closure? What do you need to show here?

2) Identity : n + e = e = e + n . Not sure about this! I think e = 0, and would exist when k=0. So n can take the value 0.

3) Inverse : n + a^-1 = e . Inverse is -n, again I'm 'iffy' about this answer!

Once you understand what set it is you're working with, this will probably become clear.

sponsoredwalk

Here's a great way to think about a group:

What is a group? Algebraists teach that this is supposedly a set with two
operations that satisfy a load of easily-forgettable axioms. This definition
provokes a natural protest: why would any sensible person need such
pairs of operations? "Oh, curse this maths" - concludes the student
(who, possibly, becomes the Minister for Science in the future).

We get a totally different situation if we start off not with the group but
with the concept of a transformation (a one-to-one mapping of a set
onto itself) as it was historically. A collection of transformations of a set is
called a group if along with any two transformations it contains the result
of their consecutive application and an inverse transformation along with
every transformation.

This is all the definition there is. The so-called "axioms" are in fact just
(obvious) properties of groups of transformations. What axiomatisators
call "abstract groups" are just groups of transformations of various sets
considered up to isomorphisms (which are one-to-one mappings
preserving the operations). As Cayley proved, there are no "more
abstract" groups in the world. So why do the algebraists keep on
tormenting students with the abstract definition?
link

Okay, so if you think of +, addition, as a function acting on elements of
the set ℤ it becomes clear. Here the function + acting on elements of ℤ is
what is meant by group transformations in what I quoted above. All this
means is that after a sequence of applications of the function + on the
elements of ℤ, i.e. a sequence of transformations of elements of ℤ, if
you end up with

i) (a + b) + c = a + (b + c) (Associativity)

ii) a + e = e + a = a (Identity)

iii) a + b = b + a = e (Invertability)

then you have a group! Sometimes you need to worry about closure,
that a + b is in ℤ, but that depends on the set & function involved really.
Furthermore it's clear to think in terms of functions, i.e. a group of
transformations of elements of G, i.e. a collection of transformations of
elements of G, i.e. a sequence of applications of + to G etc... to know
what is in what set.

Now I'll assume that G = {n | n = 10k, k ∈ ℤ}, that G is the set of all
n such that n is equal to 10k where k is an element of ℤ.

Let a = 10p, b = 10q, where p & q are in ℤ. We know G has plenty of
elements by the fact that it's elements depend on ℤ. Since + is defined on
the set G we find that applying the function + to two elements a & b in
G takes the form

+ : G × G → G where (a,b) ↦ +(a,b) = (a + b).

Think of +(a,b) as f(x,y) & (a + b) as whatever f(x,y) is equal to (if that
helps).

Now (a + b) = (10p + 10q) = 10(p + q) & (p + q) is an element of ℤ so
set (p + q) = s then 10s is in G. Just as n = 10k in the definition of G we
set z = 10s & then (z ∈ G). To be pedantic,

+ : G × G → G has (a,b) ↦ +(a,b) = (a + b) = z

To be even more pedantic

+ : G × G → G has (a,b) ↦ +(a,b) = (a + b) = (10p + 10q) = 10(p + q) = 10s = z

Basically the idea is just that, upon applying the function + to two
elements of G you get one element z out of it. This illustrates closure.

So with all of this in mind try satisfying the rest of the "axioms".

PaulieBoy

Great help guys, thanks !
Yes, it's kinda obvious now that you say it! The set is { ...-10,0,10,20....}
I just could not see that.

So bearing that in mind :

1) Closure : I am still of the opinion that n is the only element ?
If so : n = n so is closed ? There is 'nothing' to add to n, no other element?
If not then n+any element within the set is an element of that set so it is closed either way. Any help clarifying this would be great!

2) Identity : n + 0 = e . 0 is in the set, so 0 is the identity element.

3) Inverse : For every n there is an inverse +(-n) which is an element of the set. So a^-1 = (-n)

So it is a group. I think!
Any help clarifying 1) would be great .
I understand that n takes on different values, but that there is only one n, with one value attached depending on k.
Hence why I think there is only 'one' n ....


Fantastic help all, cheers :-)

LeixlipRed

Closure here means that if you "operate" two elements of the set on each other you will get an element that is also in the set. i.e. if you add 20 to 30 you get 50 which is still inside the set. But can you prove in general that's it's closed?

PaulieBoy

My take on this is that it is a set with just one element,n , where that element is obtained from the the value of k.

Say: ( {all positive integers} with addition as the operator)
We know there are an infinite number of integers. For any two a and b : a+b = an integer. We have closure!

Say: ({1} with addition as the operator)
There is one element in the set , 1, with addition as the operator.
We don't have a and b anymore ? we just have a, so under addition a = a ?
Set is closed ?

Where am I going wrong ?

LeixlipRed

Well you're contradicting yourself. You say the set has one element, n. Yet above you said the set was {....,-10,0,10,20,...}.

Your confusion stems from this statement:

{n: n=10k for some k which is an element of Z}

which in plain english means a set where n is an element and n is of the form 10k where k is an integer. i.e. there are infinitely many k's hence infinitely many n's.

Now is this set closed (i.e if we add together every two elements in this set do we stay inside the set? Or looking at it the other way we don't want to be able to add two elements of the set together and end up with something that isn't in our set?).

So reword in english, all elements of our set are multiples of 10. If we add two multiples of 10 together is that number a multiple of ten...?

Michael Collins

LeixlipRed wrote: »

...Your confusion stems from this statement:

{n: n=10k for some k which is an element of Z}

which in plain english means a set where n is an element and n is of the form 10k where k is an integer. i.e. there are infinitely many k's hence infinitely many n's...

Could this also mean a set where n is an element and n is of the form 10k where k is a single, specific integer i.e. the set with one element? A bit weird I know but maybe this question is trying to get across a specific point? (Only one integer value ok k will make the above description a valid group.)

LeixlipRed

Ah, ok, I read it as being all n of that form. But it doesn't say that. I see you pointed that out earlier but I didn't notice. Maybe the OP can clarify.

PaulieBoy

It's written as {n : n = 10k for some k which is an element of Z}

My read of this is : n such that n = 10k for some k which is an element of Z, and that this means that this is a set with a single element , n , whose value is determined by the value of k.
The value of n , is 'drawn' from the set { .... -10, 0, 10, .....}

Am I correct in this ? If so then my original post is my attempt to solve this from that point of view...

If this is so : Closure is restricted to one element , n, and n is n, so it's closed!
Help!!!

LeixlipRed

Ahh, now I see. Well what if you add n to itself?

PaulieBoy

LeixlipRed wrote: »

Ahh, now I see. Well what if you add n to itself?

Right! Now we are at the very heart of my question :

If I have only one element under an operator - addition in this case - do I do as you suggest : n + n = 2n which is in the set , and thus is closed. Or

As there is only one element ,n , then it's n = n , which is in the set , and thus is closed.

If you only have one element can you add it to itself and end up with another element , likewise if you were using the multiplication operator then :

n x n = n^2 , or for division n/n = 1 , is this the correct approach to take ?
Therefore my approach n=n for all operators is wrong ?

Thanks for your time all on this .

Fremen

It's a really weird question. You sure it's not phrased as "for all" rather than "for some"?

PaulieBoy

Fremen wrote: »

It's a really weird question. You sure it's not phrased as "for all" rather than "for some"?

Most deffo "for some" not "for all" .
Modern Algebra An Introduction by John R. Durbin 2nd ed.
Section 5, Q 5.5.
And off course they only give some of the answers!!

Michael Collins

Well let's assume it means just the single element group so (aka the Trivial Group). Maybe have a go at proving the other "for all k" case yourself after.

You really have several different sets of one element here, one for each different value of k.

Given that the group has to be closed under addition, what is the only value of k that will work?

Take k = 1 for example, then we just have the set containing: {10}

Is it closed? Nope: [latex] 10 + 10 = 20 \notin \hbox{Group} [/latex]
So this is not a group as we've fallen at the first hurdle. One value of k gives you a closed set under addition, what is it?

PaulieBoy

Michael Collins wrote: »

One value of k gives you a closed set under addition, what is it?

If k = 0 then 10K = 0 and so the set is {0} , and 0+0 = 0.
Which is closed.
The identity would be 0.
The inverse would also be 0
So it is a group if k=0 ?

Thanks for the help :-)

sponsoredwalk

I find it very very hard not to read

{n : n = 10k for some k which is an element of Z}

as

G = {n | n = 10k, k ∈ ℤ},

where

G = {n₁, n₂, n₃, ...}

&

n₁ = k₁, n₂ = k₂, n₃ = k₃ etc...

I've seen this notation used plenty of times & never had it refer to a
single element set. Any time I come across such notation I'll post it.
I really doubt the question would be phrased so ambiguously to allow it
to be a group if k = 0 while not a group for any other k.

Also, to read it as {n | n = 10k, "for all k in ℤ" } would mean that every
element n would be equal to an infinite number of values, again makes
no sense. If n = 10k for every value of k in ℤ then n₁ = 1, n₁ = 2, n₁ = 3,
etc... & n₂ = 1, n₂ = 2, n₂ = 3 etc... while if every n = 10k for some unique
k in ℤ you don't have such nonsense.

If you can answer it both ways then there really is no problem, this
is not a mathematical problem it's a problem with the ambiguous phrasing
of a question.