I've seen this discussed before but never understood it until now. So what I like to do is convert everything in the quaternion group to 0,1,2,3,4,5,6,7 and write out the Caley table for it. You'll see some interesting patterns. If you read out the elements of the row for -1 left to right, and the elements of the column for -1 top to bottom, you'll see they are the same set in the same order {-1,1,-i,i,-j,j,k,-k} or {1,0,3,2,5,4,7,6} in my notation. This corresponds to -1 commuting with all the elements. Of course 1 commutes with all the elements as well. So some elements commute and others do not. So far, I have only seen groups where either everything commutes (additive groups of integers modulo n) or most everything does not commute (Symmetric Group 3 of order 3!=3*2*1=6) If you repeat the same pattern for the other rows and columns (compare the i-th row with the i-th column), you'll see that some pairs of elements are transposed (2,3 becomes 3,2), and others are not. Gonna work out the centralizers and center and conjugates and so forth as an exercise. Definitely worth doing on the other 4 groups of order 8.
Nice video, smooth and compact rundown. However at the end you mention the dihedral group containing 8 elements but what you put down is D8 - which has 16 elements, can you explain this?
It's a notational thing D8 can have 8 or 16 elements depending on the author of the book,the context etc .. unfortunately it's one of those things where multiple notations are used
Hey , could you please help me how to find the normaliser of these elements? Should my reason be - since the group is non abelian (which implies that some element do not commute) and hence the normaliser is zero or how? Getting real confuse here . It'd mean alot to me if I get some help here.
There is mistake If i.i=1 ( self invertible ) Then, we know, i.j = k => i.i.j= i.k => 1.j = -j (as i.k = -j) => j= -j Contradiction because they are different. Only 1, -1 are self invertible In quaternion group multn is its own ops and '-' denote negation of the original element, not inverse
It baffles me that out of 18k views only < 300 liked and there are < 30 comments . I'll attribute this to multiple views of same viewers... I guess. I just think a like is nice... guys n gals c'mon
You need to show that for all a,b,c in the quaternion set, they satisfy (ab)c = a(bc). In principle, you can do this: There are 8 choices for a, 8 choices for b, and 8 choices for c, so there are 8^3 = 512 possible combinations. Make sure you also have a multiplication table. Now make a table. Put the possible values of a, b, and c. Then for each calculate (ab)c and a(bc). The table will look like this: | a | b | c | (ab)c | a(bc) | | 1 | 1 | 1 | 1·1 = 1 | 1·1=1 | ... | i | j | k | kk=-1 | i i=-1 | ... I only showed the case (a,b,c) = (1,1,1) and (a,b,c) = (i,j,k). There are 510 other cases to go. Maybe automate this using a computer program? The quaternion satisfies the associative property if and only if the two columns (ab)c and a(bc) are identical. There are also shorter way to prove associativity without going through 512 cases. See math.stackexchange.com/questions/401506/quaternion-group-associativity
well {1} is just the group containing the identity, so the inverse of 1 is 1, so it is it's own inverse. In general, if G is a group, and e is the identity, then {e} is a subgroup always, and the inverse of e is just e because e*e = e
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I've seen this discussed before but never understood it until now. So what I like to do is convert everything in the quaternion group to 0,1,2,3,4,5,6,7 and write out the Caley table for it. You'll see some interesting patterns. If you read out the elements of the row for -1 left to right, and the elements of the column for -1 top to bottom, you'll see they are the same set in the same order {-1,1,-i,i,-j,j,k,-k} or {1,0,3,2,5,4,7,6} in my notation. This corresponds to -1 commuting with all the elements. Of course 1 commutes with all the elements as well. So some elements commute and others do not. So far, I have only seen groups where either everything commutes (additive groups of integers modulo n) or most everything does not commute (Symmetric Group 3 of order 3!=3*2*1=6)
If you repeat the same pattern for the other rows and columns (compare the i-th row with the i-th column), you'll see that some pairs of elements are transposed (2,3 becomes 3,2), and others are not. Gonna work out the centralizers and center and conjugates and so forth as an exercise. Definitely worth doing on the other 4 groups of order 8.
How easily you explain the concepts. Again, a big big thankyou sir 🙏
thank you !.....really so helpful video for beginners .
glad it helped!!
Thanks. It's very helpfull
You are welcome!
Thank you! It was very helpful!
Nice video, smooth and compact rundown. However at the end you mention the dihedral group containing 8 elements but what you put down is D8 - which has 16 elements, can you explain this?
It's a notational thing D8 can have 8 or 16 elements depending on the author of the book,the context etc
.. unfortunately it's one of those things where multiple notations are used
You're one smart guy!!!
+Angela DiMauro LOL!!!!!!
Wow...
Hey , could you please help me how to find the normaliser of these elements? Should my reason be - since the group is non abelian (which implies that some element do not commute) and hence the normaliser is zero or how? Getting real confuse here . It'd mean alot to me if I get some help here.
Wow I didn’t get it at all I’ll watch it again
There is mistake
If i.i=1 ( self invertible )
Then, we know,
i.j = k
=> i.i.j= i.k
=> 1.j = -j (as i.k = -j)
=> j= -j
Contradiction because they are different.
Only 1, -1 are self invertible
In quaternion group multn is its own ops and '-' denote negation of the original element, not inverse
It baffles me that out of 18k views only < 300 liked and there are < 30 comments .
I'll attribute this to multiple views of same viewers... I guess.
I just think a like is nice... guys n gals c'mon
yeah that happens a lot with educational videos, I have tons with very low views:)
thx man:)
Are Quaternions related to solutions related to matrices?
because i has an intrinsic value in algebra sqrt(-1), what are j and k's
thanks sorc
glad it helped;)
Very helpful. Thanks!
The commutator is i^2
Its very helpful
Beautiful video!! could you recommend some bibliography about Quaternions please?
Helpful video. How do you get i(-i)=-1(i^2)?
It is actually i(-i)= - i.i = - i^2 = - .-1= +1 =1 as i^2= -1 and - . - = + ..... The dot represent multiplcation
How to prove that this group satisfy associative property. (i know that groups have those property by definition but I want to know how to prove it)
You need to show that for all a,b,c in the quaternion set, they satisfy (ab)c = a(bc).
In principle, you can do this:
There are 8 choices for a, 8 choices for b, and 8 choices for c, so there are 8^3 = 512 possible combinations. Make sure you also have a multiplication table.
Now make a table. Put the possible values of a, b, and c. Then for each calculate (ab)c and a(bc). The table will look like this:
| a | b | c | (ab)c | a(bc) |
| 1 | 1 | 1 | 1·1 = 1 | 1·1=1 |
...
| i | j | k | kk=-1 | i i=-1 |
...
I only showed the case (a,b,c) = (1,1,1) and (a,b,c) = (i,j,k). There are 510 other cases to go. Maybe automate this using a computer program?
The quaternion satisfies the associative property if and only if the two columns (ab)c and a(bc) are identical.
There are also shorter way to prove associativity without going through 512 cases. See math.stackexchange.com/questions/401506/quaternion-group-associativity
{1} what is the inverse of this subgroup?
well {1} is just the group containing the identity, so the inverse of 1 is 1, so it is it's own inverse. In general, if G is a group, and e is the identity, then {e} is a subgroup always, and the inverse of e is just e because e*e = e
you didn't ask, but random fact, in general whenever you have a group G, you always have TWO subgroups, {e} and G are always subgroups of G:)
@@TheMathSorcerer Appreciate it wizard-san!
Sir give more examples on it & examples on matrixes.f finally i got
Sir find all the maximal and minimal subgroup o symmetric group of order 4 s4
What about j operating with itself ...and also k