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Groups (OCR fp3)

The question is in the image attached. I'm referring to question 6 part b (i).
The expected answer is 9.

However, surely a group under mod 3 can not have more than 3 distinct elements, and we require every element in a group to be distinct? I am confused!

Thanks.

Reply 1

around

Original post by quint101

Our group consists of elements of the form ax+b, for a, b numbers mod 3. There are 3 possible choices for a (0, 1, 2) and 3 possible choices for b, giving 9 total elements in the group.

Go through and write out all the elements yourself if you don't trust.

Reply 2

quint101

Original post by around

Obviously I get where 9 comes from!

But my question was why can a group under mod3 have more than 3 elements? For example: x, x+1, x+2 and 2x+1 here are all elements of the group. However, at least two of the are the same modulo 3. So (as as all elements in a group are distinct), we can not have more than three maximum element? :s-smilie:

EDit: I am incorrectly assuming x is an interger. Then at least two from x, x+1, x+2 and 2x+1 do not have equal in mod 3. Could you please confirm this? If we were given all elements as integers, then the order of any group under mod3 cannot exceed 3 right?

Thanks.

Reply 3

electriic_ink

Original post by quint101

a group under mod 3 can not have more than 3 distinct elements.

This isn't true. It may be true if the group is generated by a single element but in this case it isn't. edit: Quick example: consider the set {(0,0), (0,1), (1,0), (1,1)} of vectors in R^2 under addition mod 2.

I am incorrectly assuming x is an interger.

x doesn't have to be an integer. x could be a real number, a function or even a matrix - it doesn't matter. All the polynomials x, x+1, x+2 and 2x+1 are different since they have different co-efficients mod 3.

(edited 12 years ago)

Reply 4

ghostwalker

Original post by quint101

An element of Q requires 2 parameters to define it, each of which is defined mod3, hence the 9 elements. You could represent them as number pairs (a,b), where each of a,b is one of {0,1,2}.

You're not interested in evaluating the polynomials for any particular value of x.
Indeed you're not even told what the domain of x is, or even that it has a domain since you're not considering them as a function from X to Y; solely as polynomials in their own right.

Edit: Too slow :sigh:

(edited 12 years ago)

Reply 5

nuodai

Original post by quint101

Just to reiterate what the two above me have said: polynomials don't always have to be seen as functions, and in some branches of mathematics it's often useful to distinguish between a "polynomial" and a "polynomial map", where the former is just a mathematical object that obeys certain rules, and the latter is a true function. A polynomial is basically just a vector, where instead of writing

(a,b,c)

you write

a+bx+cx^2

, for instance; except spaces of polynomials come with more structure because, whereas you can't "multiply" two vectors, you can multiply two polynomials. The

x

is an "indeterminate", and needn't be considered as taking any kind of value.

(edited 12 years ago)

Reply 6

quint101

Ok, thank you all. I understand this now.

I know here this is not the case, but just checking:

"If there exists a group under mod 3 where every element an an interger, then the groups order cannot be greater than 3"

Please confirm (or correct) the above.

Original post by nuodai

(a,b,c)

you write

a+bx+cx^2

, for instance; except spaces of polynomials come with more structure because, whereas you can't "multiply" two vectors, you can multiply two polynomials. The

x

is an "indeterminate", and needn't be considered as taking any kind of value.

Reply 7

nuodai

Original post by quint101

"If there exists a group under mod 3 where every element an an interger, then the groups order cannot be greater than 3"

The only way you could define what you mean by "group under addition modulo 3 in which every element is an integer" would be a "subgroup of

\mathbb{Z}_3

", so yes.

Reply 8

quint101

Original post by nuodai

The only way you could define what you mean by "group under addition modulo 3 in which every element is an integer" would be a "subgroup of

\mathbb{Z}_3

", so yes.

Yep, that exactly what I was thinking. A further question to confirm: if in our initial question we were given that x is an integer, the order of the group still would be 9 right (because we are considering a "polynomial" not a "polynomial map") ?

Thank you!

Reply 9

nuodai

Original post by quint101

Even if you were considering a polynomial map (in this case) they'd all be distinct, because the corresponding polynomial maps have different images. These are summarized in the following table:

Unparseable latex formula:

\begin{array}{c|c} f & (f(0),f(1),f(2)) \\ \hline[br]0 & (0,0,0) \\[br]1 & (1,1,1) \\[br]2 & (2,2,2) \\[br]x & (0,1,2) \\[br]x+1 & (1,2,0) \\[br]x+2 & (2,0,1) \\[br]2x & (0,2,1) \\[br]2x+1 & (1,0,2) \\[br]2x+2 & (2,1,0) \end{array}

So they're all different even when considered as functions in this case.

But really this isn't something even first-year undergraduates should be worrying about. Just interpret the

x

to be "anything" so that two polynomials are equal if and only if all their coefficients are equal.

Reply 10

Raiden10

Original post by quint101

No, x, x+1, x+2 and 2x+1 are not numbers or "elements of Z3" but functions on Z3. There are indeed only 3 elements of Z3. But there are a whopping 27 functions on Z3. For example the polynomial x^2 isn't even in the above table and that has 9 elements (all functions on sets like Z3 can be represented by polynomials - there are no transcendental functions). If you pick a pigeonhole-principle fight with a 27 or 9 element set, and you only have a 3 element set, you're gonna lose.