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    The definition of <a,b.> I think I have from lectures is
    The elements of <X> are all elements of G that can be expressed using (dotproduct) and ^(-1) from elements of X. I think X is just a subgroup of a group G.

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    (Original post by cooldudeman)
    The definition of <a,b.> I think I have from lectures is
    The elements of <X> are all elements of G that can be expressed using (dotproduct) and ^(-1) from elements of X. I think X is just a subgroup of a group G.
    Agree with part (a).

    Part b. I suspect you're talking about the generating set for a (sub)group. See the first two paragraphs here. Or Google further.

    There is no "dotproduct", it's the group operation.

    The "recall info"? Come on; what's that supposed to mean?

    Those in mind, I'd reread and redo part (b). It's currently beyond redemption, IMO.
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    (Original post by ghostwalker)
    Agree with part (a).

    Part b. I suspect you're talking about the generating set for a (sub)group. See the first two paragraphs here. Or Google further.

    There is no "dotproduct", it's the group operation.

    The "recall info"? Come on; what's that supposed to mean?

    Those in mind, I'd reread and redo part (b). It's currently beyond redemption, IMO.
    Sorry I meant cartrsian product.

    Lol I meant recall info as in the part ehere it says recall on the question.

    Will have another go soon and let you know.

    Thanks
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    (Original post by cooldudeman)
    Sorry I meant cartrsian product.
    Cartesian product doesn't come into this.

    Lol I meant recall info as in the part ehere it says recall on the question.
    :cool: 'See what you're saying now.

    Will await your reworking.
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    (Original post by ghostwalker)
    Cartesian product doesn't come into this.



    :cool: 'See what you're saying now.

    Will await your reworking.
    Shouldn't say 34. I think I got 37 of them. I tried to think of all possibilities. Is this remotely right? If not then please help.

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    (Original post by cooldudeman)
    Shouldn't say 34. I think I got 37 of them. I tried to think of all possibilities. Is this remotely right? If not then please help.

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    I'm not sure I'm following what you're doing, but there are only 28 possible ways of choosing 2 (distinct) elements from a set of size 8, so I don't think your answer can possibly be right.
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    (Original post by DFranklin)
    I'm not sure I'm following what you're doing, but there are only 28 possible ways of choosing 2 (distinct) elements from a set of size 8, so I don't think your answer can possibly be right.
    Does it have to be distinct though?

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    (Original post by cooldudeman)
    Does it have to be distinct though?

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    Well yes, since otherwise your group would be generatable from a single element and the quaternion group isn't cyclic.
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    (Original post by cooldudeman)
    Shouldn't say 34. I think I got 37 of them. I tried to think of all possibilities. Is this remotely right? If not then please help.

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    Difficult to know how to proceed to be honest, and I have to go out in 15 minutes, with 40 minutes worth of other "stuff" to do before I do.

    \langle a,b\mid \text{ some relationships }\rangle =G

    Means that the group G can be generated from the two elements a,b, subject to some relationshpis on a,b.

    We'll ignore the relationship part as you're dealing with generators for a specific group whose structure is known.

    Note G is a group, not just a set; there is a group operation on its elements.

    So, we can reduce this to \langle a,b\rangle =G for now; the format used in the question.

    Notice also that these are equal, "<a,b>" is a group; it is not an element of G.
    a,b stand for two elements of the group, AND you need to replace them with actual elements of the group, e.g <i>, just a single generator, would generate the group i^0,i^1,i^-1,i^2, etc. Which works out to {1,i,-1,-i}
    So, \langle i \rangle=\{1,i,-i,-1\}

    Now you need to find two generators, which together generate (i.e. all possible combinations of those two elements gives) the whole of Q_8. Then repeat to find all possible pairs that generate the group.
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    (Original post by cooldudeman)
    ...
    Done editing.
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    (Original post by ghostwalker)
    Difficult to know how to proceed to be honest, and I have to go out in 15 minutes, with 40 minutes worth of other "stuff" to do before I do.

    \langle a,b\mid \text{ some relationships }\rangle =G

    Means that the group G can be generated from the two elements a,b, subject to some relationshpis on a,b.

    We'll ignore the relationship part as you're dealing with generators for a specific group whose structure is known.

    Note G is a group, not just a set; there is a group operation on its elements.

    So, we can reduce this to \langle a,b\rangle =G for now; the format used in the question.

    Notice also that these are equal, "<a,b>" is a group; it is not an element of G.
    a,b stand for two elements of the group, AND you need to replace them with actual elements of the group, e.g <i>, just a single generator, would generate the group i^0,i^1,i^-1,i^2, etc. Which works out to {1,i,-1,-i}
    So, \langle i \rangle=\{1,i,-i,-1\}

    Now you need to find two generators, which together generate (i.e. all possible combinations of those two elements gives) the whole of Q_8. Then repeat to find all possible pairs that generate the group.
    I'm a bit confused why you say i^-1 would give -i because is this i even the imaginary value? it doesn't say on the question, or are we supposed to assume it is. In the question it says kj=-i.

    Also if we have <i, j>, we can generate

    -1 fromi^2=j^2,
    1 from (i^2)(j^2)=(-1)(-1)
    k from ij,
    -k from ji,
    -i from kj=(ij)j,
    -j from ik=i(ij)

    Do we need to get i and j too? They are already in the <i,j> aren't they. But if we did have to 'get' them, I would do:
    i from jk=j(ij)
    j from ki=(ij)i

    This is the whole of Q_8 right?

    Now do I find the other combinations of <a,b> that also give all of Q_8?
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    (Original post by cooldudeman)
    I'm a bit confused why you say i^-1 would give -i because is this i even the imaginary value? it doesn't say on the question, or are we supposed to assume it is. In the question it says kj=-i.

    i^0,i^1,i^-1,i^2, etc. as in i^0,i^1,i^-1,i^2,i^-2,i^3,i^-3,i^4,i^-4,....
    Although it so happens i^-1 does equal -i in Q_8 - try deriving it, or check it.


    Also if we have <i, j>, we can generate

    -1 fromi^2=j^2,
    1 from (i^2)(j^2)=(-1)(-1)
    k from ij,
    -k from ji,
    -i from kj=(ij)j,
    -j from ik=i(ij)

    Do we need to get i and j too?

    They are already in the <i,j> aren't they. But if we did have to 'get' them, I would do:
    i from jk=j(ij)
    j from ki=(ij)i
    You already have i,j so you don't additionally need to derive them from i,j.
    Note; All combinations of i,j includes combinations of just i's, and just j's. E.g. i^2=-1, i^3=-i, etc.

    This is the whole of Q_8 right?
    Yep.

    So, \langle i,j\rangle = Q_8

    Now do I find the other combinations of <a,b> that also give all of Q_8?
    Yes.
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    (Original post by ghostwalker)
    i^0,i^1,i^-1,i^2, etc. as in i^0,i^1,i^-1,i^2,i^-2,i^3,i^-3,i^4,i^-4,....
    Although it so happens i^-1 does equal -i in Q_8 - try deriving it, or check it.




    You already have i,j so you don't additionally need to derive them from i,j.
    Note; All combinations of i,j includes combinations of just i's, and just j's. E.g. i^2=-1, i^3=-i, etc.



    Yep.

    So, \langle i,j\rangle = Q_8



    Yes.
    Thanks. I feel like the only pairs we can have are <i, j> and <i, k> and <j, k>. Am I right?

    Ones with -1 or +1 with i, j or k would not be able to generate X \ {i, j, k} where X is an element of {i, j, k}.

    For eg, if we have <-1, i>, it cant generate j or k right?

    But what would be a good reasoning to why this is???
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    (Original post by cooldudeman)
    Thanks. I feel like the only pairs we can have are <i, j> and <i, k> and <j, k>. Am I right?
    So far, so good. There are more.

    Ones with -1 or +1 with i, j or k would not be able to generate X \ {i, j, k} where X is an element of {i, j, k}.
    My brain hurts reading that.

    For eg, if we have <-1, i>, it cant generate j or k right?
    Correct.

    But what would be a good reasoning to why this is???
    Well <1,i>, for example, generates the subgroup {1,-1,i,-i} - closed under the group operation (obviously, otherwise it wouldn't be a subgroup) and doesn't contain j or k, nor -j,-k
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    (Original post by ghostwalker)
    So far, so good. There are more.

    I cant think of anymore...!

    I even tried <1,-1>, it doesn't give all of it.

    please help
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    (Original post by cooldudeman)
    I cant think of anymore...!

    I even tried <1,-1>, it doesn't give all of it.

    please help
    Well you've got 8 elements to choose from, and you've only really talked about 4 or 5 of them.
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    (Original post by ghostwalker)
    Well you've got 8 elements to choose from, and you've only really talked about 4 or 5 of them.
    oh crap, forgot about the negatives of ijk
    so the only potential ones are i, j, k, -i, -j, -k so no. of ways to choose 2 from 6 elements is 6C2=15

    So the only pairs that would generate Q_8 would be any combination of the elements above right?

    Very tedious to write them all out, don't you think...

    So it is like <x, y> with x,y in {i,j,k,-i,-j,-k} with x not equal to y, right?

    EDIT: Just realised that ones like <i,-i> wont work! So that gets rid of 3 pairs so there are 12.
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    (Original post by cooldudeman)
    oh crap, forgot about the negatives of ijk
    so the only potential ones are i, j, k, -i, -j, -k
    Yes.

    so no. of ways to choose 2 from 6 elements is 6C2=15

    So the only pairs that would generate Q_8 would be any combination of the elements above right?
    No

    Does <i,-i> work?
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    (Original post by cooldudeman)
    so there are 12.
    Yep.
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    (Original post by ghostwalker)
    Yes.



    No

    Does <i,-i> work?
    yeah i actually edited my post. you just missed it.

    So the ones would be:

    i, j
    -i, j
    i, -j
    -i, -j

    i, k
    -i, k
    i, -k
    -i, -k

    j, k
    -j, k
    j, -k
    -j, -k

    right?
 
 
 
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