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# group theory pairs of elements watch

1. 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.

2. (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.
3. (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
4. (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.
'See what you're saying now.

5. (Original post by ghostwalker)
Cartesian product doesn't come into this.

'See what you're saying now.

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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6. (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.
7. (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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8. (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.
9. (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.

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.
10. (Original post by cooldudeman)
...
Done editing.
11. (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.
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?
12. (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
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,

Now do I find the other combinations of <a,b> that also give all of Q_8?
Yes.
13. (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.
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???
14. (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}.

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
15. (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.

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

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

Well you've got 8 elements to choose from, and you've only really talked about 4 or 5 of them.
17. (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.
18. (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?
19. (Original post by cooldudeman)
so there are 12.
Yep.
20. (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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