A little while ago I noticed a very interesting group, I am sure it is well known in math circles but I find it intriguing for a number of reasons and I have recently learned more about Abstract Algebra which ties in some more interesting ideas. The group relates to imaginary numbers, it is the set {1,-1,i,-i} which is closed under multiplication. It has an identity which is 1. Each element has an inverse, -1 is its own inverse: -1 · -1 = 1 and -i · i = 1. The following cayley table shows all this:
• | 1 | i | -1 | -i |
1 | 1 | i | -1 | -i |
i | i | -1 | -i | 1 |
-1 | -1 | -i | 1 | i |
-i | -i | 1 | i | -1 |
If you are new to these ideas I recommend looking at my previous post on abstract algebra, especially since we will be looking at some isomorphisms again. In order to talk about the ideas in the post we need to get some group theory concepts out of the way first, don't worry they aren't hard in fact they're fairly intuitive. Also in this post I am restricting the discussion to finite groups, groups with a finite number of elements. In fact the first idea is the concept of group order.
The order of a finite group is simply the number of elements of the set the group is defined on and is denoted as |G|.
The group above has an order of 4.
The next idea is that of exponentiation, which can be thought of as a subset of "classical" exponentiation on the field of real numbers, see identities here. Exponentiation on group elements extends to integer exponents only and refers to repeated application of the binary operation to the element itself. So:
a^{n} = a·a·a·a...a ( a "times" itself n times)
a^{-n} = a^{-1}·a^{-1}·a^{-1}·a^{-1}...a^{-1} (a^{-1} "times" itself n times or the inverse of a^{n})
a^{0} = e (where e is the identity element)
The laws of exponents are:
a^{m} a^{n} = a^{m+n}
(a^{m})^{n} = a^{mn}
a^{-n} = (a^{-1})^{n} = (a^{n})^{-1}
Just as a group has an order an element of a group also has an order and it is related to exponentiation and it is defined as:
If there exists a nonzero integer m such that a^{m}=e, then the order of the element is defined to be the least positive integer n such that a^{n}=e. Also if there does not exist any nonzero integer m such that a^{m}=e, we say that a has order infinity.
So the elements in our group (⋅, {1,-1,i,-i}) above have the following orders:
1^{1} = 1 (this is the identity and has order 1) [1^{1} = 1, ...]
-1^{2} = 1 (has order 2) [-1^{1} = -1, -1^{2} = 1, ...]
i^{4} = 1 (has order 4) [i^{1} = i, i^{2} = -1, i^{3} = -i, i^{4} = 1, ...]
-i^{4} = 1 (has order 4) [-i^{1} = -i, -i^{2} = -1, -i^{3} = i, -i^{4} = 1, ...]
Order is also sometimes referred to as period and you should note that each of the above will cycle through each power listed in the square brackets so for both the order 4 elements {i, -i}: a^{1n}=a^{1}, a^{2n}=a^{2}, a^{3n}=a^{3}, a^{4n}=a^{4} for all integers n>0 and the values will cycle through the list which would be the case for any element of finite order.
As it turns out elements like i and -i which cycle through all of the elements of a group exactly once in the period have an order that is equal to the order of the group. These define what is known as a cyclic group. So our group is cyclic and any element which cycles through all elements once is the generator so i and -i are generators for the group.
Another interesting result from group theory is that any cyclic group of order n is isomorphic to the group (+,Z_{n}) which is the integers (mod n) under addition so our group above is just a renaming of Z_{4} see the similarity to the cayley table below:
+ | 0 | 1 | 2 | 3 |
0 | 0 | 1 | 2 | 3 |
1 | 1 | 2 | 3 | 0 |
2 | 2 | 3 | 0 | 1 |
3 | 3 | 0 | 1 | 2 |
Now I want to take a look at another important group, it's called the dihedral group, this one is known as D_{4} and has the cayley table:
R_{0} | R_{1} | R_{2} | R_{3} | V | H | D_{1} | D_{2} | |
R_{0} | R_{0} | R_{1} | R_{2} | R_{3} | V | H | D_{1} | D_{2} |
R_{1} | R_{1} | R_{2} | R_{3} | R_{0} | D_{1} | D_{2} | H | V |
R_{2} | R_{2} | R_{3} | R_{0} | R_{1} | H | V | D_{2} | D_{1} |
R_{3} | R_{3} | R_{0} | R_{1} | R_{2} | D_{2} | D_{1} | V | H |
V | V | D_{2} | H | D_{1} | R_{0} | R_{2} | R_{3} | R_{1} |
H | H | D_{1} | V | D_{2} | R_{2} | R_{0} | R_{1} | R_{3} |
D_{1} | D_{1} | V | D_{2} | H | R_{1} | R_{3} | R_{0} | R_{2} |
D_{2} | D_{2} | H | D_{1} | V | R_{3} | R_{1} | R_{2} | R_{0} |
The dihedral group is the group of symmetries of a regular polygon, including both rotations and reflections. In this case we will consider the rotations and reflections of a square denoted as D_{4} represented visually as four couterclockwise rotations of a square:
And the following four reflections, vertical, horizontal, and two diagonal reflections:
With each image is included the permutation, actually dihedral groups are permutation groups and D_{4} is a set of 8 permutations, hence the order of the group is 8. It is a subgroup of the group of all permutations of four items which is known as the Symmetric Group which has order 4!=24, it's denoted as S_{4}. The interesting thing here is the rotations form a subgroup that is isomorphic to our group from above and Z_{4}. In a sense this isn't really surprising since it is a cyclic group of order four meaning it cycles through 4 items just as a square cycles through four rotations.
As you can see from the color coding, the elements {R_{0}, R_{1}, R_{2}, R_{3}} form a subgroup, a subgroup is just a subset of the elements that is still a group. These subgroup elements forming the upper left hand corer of the table are color coded to show the isomorphism between this subgroup and Z_{4} and also our group (⋅, {1,-1,i,-i}). Also R_{0} is the identity which makes sense since it is a zero rotation, which means that you didn't rotate it, or you did by 360° aka one tau.
Our Z_{4} isomorphic group (⋅, {1,-1,i,-i}) also has a subgroup. Our original arrangement of the cayley table was ordered to show the Z_{4} isomorphism. To show our subgroup we need to rearrange the cayley table:
• | 1 | -1 | i | -i |
1 | 1 | -1 | i | -i |
-1 | -1 | 1 | -i | i |
i | i | -i | -1 | 1 |
-i | -i | i | 1 | -1 |
Now it should be fairly easy to the subgroup (⋅, {1,-1}) again forming the upper left corner in the table above, or shown by itself:
• | 1 | -1 |
1 | 1 | -1 |
-1 | -1 | 1 |
Now if you think about it our subgroup which is just another group is a group, 1 is the identity and -1 is its own inverse, actually it turns out that all subgroups of cyclic groups are also cyclic groups and since all cyclic groups are isomorphic to Z_{n} groups, this group is actually one that we have seen before it's just Z_{2} addition aka exclusive-or, etc.
Also there are some remarkable connections to number theory in regards to the order of subgroups and the order of elements, for example the order of our subgroup is 2 which evenly divides 4 the order of the group, the order of a subgroup will always evenly divide the order of the group. Also the order of each element evenly divides the order of the group. I suspect that these ideas have profound implications.
Going back to our original group there are a couple of interesting things about imaginary numbers, first are the following equations:
If we consider the sequence of raising i to successive powers, multiplying by I each time, if we start with 1, we get the sequence (1, i, -1, -i), these are actually coordinates in the complex plane and can be written as (0i+1, 1i+0, 0i-1, -1i+0). This cycle (increasing powers of i) is the following (counter clockwise) rotation in the complex plane:
Conversely if we divide, or repeatedly multiply by -i, we get the following sequence (1,-i,-1,i) which correspond to the complex numbers (0i+1, -1i+0, 0i-1, 1i+0) and is the opposite (clockwise) rotation:
Another interesting seemingly related concept is the idea of turning number or winding number of a curve, the red circle with clockwise rotation corresponds to curve with a winding number of -1 and the blue curve with positive increasing powers of i has a winding number of 1.
As we saw our group is cyclic and isomorphic to Z_{4} and isomorphic to the rotation subgroup of the dihedral group D_{4} and both of the generators of our group correspond to rotation in the complex plane. Counterclockwise rotation corresponds to the period of i and clockwise rotation corresponds to the period of -i.
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ReplyDeletePractice is truly importand in variable based math. You can't hope to take in everything from online courses or turorials. You have to experience the ideas and the standards of variable based math over and over so as to learn them. root finder
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