Groups (OCR fp3)

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  1. quint101's Avatar
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    Groups (OCR fp3)
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ID:	142767The 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.
  2. around's Avatar
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    Re: Groups (OCR fp3)
    (Original post by quint101)
    Click image for larger version. 

Name:	Picture2.png 
Views:	83 
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ID:	142767The 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.
    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.
  3. quint101's Avatar
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    Re: Groups (OCR fp3)
    (Original post by around)
    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.
    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?

    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.
  4. electriic_ink's Avatar
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    Re: Groups (OCR fp3)
    (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.
    Last edited by electriic_ink; 19-04-2012 at 09:50.
  5. ghostwalker's Avatar
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    Re: Groups (OCR fp3)
    (Original post by quint101)
    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?
    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:
    Last edited by ghostwalker; 20-04-2012 at 06:59. Reason: Spelling.
  6. nuodai's Avatar
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    Re: Groups (OCR fp3)
    (Original post by quint101)
    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?

    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.
    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.
    Last edited by nuodai; 19-04-2012 at 10:21.
  7. quint101's Avatar
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    Re: Groups (OCR fp3)
    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)
    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.
  8. nuodai's Avatar
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    Re: Groups (OCR fp3)
    (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.
  9. quint101's Avatar
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    Re: Groups (OCR fp3)
    (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!
  10. nuodai's Avatar
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    Re: Groups (OCR fp3)
    (Original post by quint101)
    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!
    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:

    \begin{array}{c|c} f & (f(0),f(1),f(2)) \\ \hline

0 & (0,0,0) \\

1 & (1,1,1) \\

2 & (2,2,2) \\

x & (0,1,2) \\

x+1 & (1,2,0) \\

x+2 & (2,0,1) \\

2x & (0,2,1) \\

2x+1 & (1,0,2) \\

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.
  11. Raiden10's Avatar
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    Re: Groups (OCR fp3)
    (Original post by quint101)
    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?

    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?
    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.
    Last edited by Raiden10; 20-04-2012 at 00:31.
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