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.

cooldudeman
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 28102015 23:07
Last edited by cooldudeman; 28102015 at 23:28. 
ghostwalker
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 29102015 06:54
(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.
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. 
cooldudeman
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 29102015 07:37
(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.
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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 29102015 07:42
(Original post by cooldudeman)
Sorry I meant cartrsian product.
Lol I meant recall info as in the part ehere it says recall on the question.
Will await your reworking. 
cooldudeman
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 30102015 00:52
(Original post by ghostwalker)
Cartesian product doesn't come into this.
'See what you're saying now.
Will await your reworking.
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DFranklin
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 30102015 01:00
(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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 30102015 01:21
(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.
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 30102015 02:50

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 30102015 08:34
(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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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 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
So,
Now you need to find two generators, which together generate (i.e. all possible combinations of those two elements gives) the whole of . Then repeat to find all possible pairs that generate the group.Last edited by ghostwalker; 30102015 at 09:52. 
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 30102015 09:53
(Original post by cooldudeman)
... 
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 31102015 14:36
(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.
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 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
So,
Now you need to find two generators, which together generate (i.e. all possible combinations of those two elements gives) the whole of . Then repeat to find all possible pairs that generate the group.
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?Last edited by cooldudeman; 31102015 at 14:46. 
ghostwalker
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 01112015 08:54
(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  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
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?
So,
Now do I find the other combinations of <a,b> that also give all of Q_8?Last edited by ghostwalker; 01112015 at 08:59. 
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 01112015 17:00
(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  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,
Yes.
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???Last edited by cooldudeman; 01112015 at 17:04. 
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 01112015 17:11
(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?
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???Last edited by ghostwalker; 01112015 at 17:13. 
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 01112015 18:10
I even tried <1,1>, it doesn't give all of it.
please help 
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 01112015 19:18
(Original post by cooldudeman)
I cant think of anymore...!
I even tried <1,1>, it doesn't give all of it.
please help 
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 01112015 19:26
(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.
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.Last edited by cooldudeman; 01112015 at 19:30. 
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 01112015 19:30
(Original post by cooldudeman)
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?
Does <i,i> work? 
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 01112015 19:35
(Original post by cooldudeman)
so there are 12. 
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 01112015 19:38
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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