definition of skewed dihedral group
Topics
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of group.
Target Context
- The reader will have a definition of skewed dihedral group.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( n\): \(\in \mathbb{N} \setminus \{0\}\)
\( g\): \(\in \{\text{ the objects }\}\)
\( h\): \(\in \{\text{ the objects }\}\)
\(*SD_{2 n}\): \(= \{[1], [g], ..., [g^{2 n - 1}], [h], [h g], ..., [h g^{2 n - 1}]\}\), where as aliases \([1] = [g^0], [g] = [g^1], [h] = [h g^0], [h g] = [h g^1]\), \(\in \{\text{ the groups }\}\) with the group operation specified below
\( f\): \(: \mathbb{Z} \to \{0, ..., 2 n - 1\}, j \mapsto m \text{ where } j = 2 n l + m \text{ where } l, m \in \mathbb{Z} \text{ and } 0 \le m \lt 2 n\)
//
Conditions:
\([g^j] [g^k] = [g^{f (j + k)}]\)
\(\land\)
\([g^j] [h g^k] = [h g^{f (-j + k)}]\)
\(\land\)
\([h g^j] [g^k] = [h g^{f (j + k)}]\)
\(\land\)
\([h g^j] [h g^k] = [g^{f (n - j + k)}]\)
//
2: Note
"skewed dihedral group" is not any standard term; in fact, the author has not seen any name for this concept (\(SD_4\) seems to be called "quaternion" though), and so, the author had to concoct one. The term is because while this concept resembles 'dihedral group', while dihedral group dictates that \([h]^2 = [1]\), this concept dictates that \([h]^2 = [g]^n\).
The purpose of using such brackets, \([ ]\), is to distinguish the single element \([h g^k]\) from the multiplication \([h] [g^k]\), for example: the logic here is that 1st, define the set, \(\{[1], [g], ..., [g^{2 n - 1}], [h], [h g], ..., [h g^{2 n - 1}]\}\), as the set of the distinct objects and 2nd, define the group operation on the set: \([h g^k]\) is just a symbol to identify the element, not the multiplication, which is not defined at the 1st step yet. Of course, such a symbol is chosen because \([h g^k]\) happens to equal \([h] [g^k]\) after all.
The purpose of the aliases is just to simplify the specification of the group operation: without the aliases, we would need to write an additionally specification for \([g^j] [h]\), for example.
Succinctly speaking, \(f\) takes the remainder (non-negative) of the argument divided by \(2 n\), which (the remainder) is uniquely determined.
Another definition of \(SD_4\) declared the existences of \(g, h\), \(g^{2 n} = 1\), \(h^2 = g^n\), and \(g^{-1} h = h g\), and let the group operation induced as to satisfy associativity, which might be possible, but we would need to confirm that the group indeed consisted of \(\{1, g, ..., g^{2 n - 1}, h, h g, ..., h g^{2 n - 1}\}\) and the group operation was indeed consistent: a multiplication could be induced in some multiple ways, and would all of them give the same result? For example, \((g^{-1} h) g = (h g) g = h (g g) = h g^2\) and \((g^{-1} h) g = g^{-1} (h g) = g^{-1} (g^{-1} h) = (g^{-1} g^{-1}) h = g^{-2} h = g^2 h\), but \(h g^2 = g^2 h\)? and we would need to prove it generally.
Instead, our definition explicitly declares the set of \(SD_{2 n}\) as \(\{[1], [g], ..., [g^{2 n - 1}], [h], [h g], ..., [h g^{2 n - 1}]\}\) (so, we do not need to check that the elements are exactly them) and explicitly specifies the group operation (so, we do not need to check the consistency), and check associativity with respect to the already-determined operation, which seems more straightforward for us: defining the set and then specifying the group operation is a standard way of defining a group.
Let us confirm that it is well-defined (it is indeed a group).
\(SD_{2 n}\) is closed under the operation, because the definition explicitly specified each result in \(SD_{2 n}\).
\([1] = [g^0]\) is indeed \(1\): \([g^0] [g^k] = [g^{f (k)}] = [g^k]\); \([g^j] [g^0] = [g^{f (j)}] = [g^j]\); \([g^0] [h g^k] = [h g^{f (k)}] = [h g^k]\); \([h g^j] [g^0] = [h g^{f (j)}] = [h g^j]\).
Let us see that each element has the inverse.
As the preparation, let us see that \(f (f (j) + k) = f (j + k)\), \(f (- f (j) + k) = f (- j + k)\), and \(f (j) = f (j + 2 n)\).
\(j = 2 n l + f (j)\). \(j + k = 2 n l + f (j) + k\). So, the remainder of \(j + k\) and the remainder of \(f (j) + k\) are the same, because the difference of \(j + k\) from \(f (j) + k\) is a multiple of \(2 n\), which means that \(f (f (j) + k) = f (j + k)\).
\(j = 2 n l + f (j)\). \(- j = 2 n (- l) - f (j)\). \(- j + k = 2 n (- l) - f (j) + k\). So, the remainder of \(- j + k\) and the remainder of \(- f (j) + k\) are the same, because the difference of \(- j + k\) from \(- f (j) + k\) is a multiple of \(2 n\), which means that \(f (- j + k) = f (- f (j) + k)\).
As the difference of \(j + 2 n\) from \(j\) is a multiple of \(2 n\), \(f (j) = f (j + 2 n)\).
\([1]^{-1} = [1]\).
For \(0 \lt j \lt 2 n\), \([g^j]^{-1} = [g^{2 n - j}]\), because \([g^j] [g^{2 n - j}] = [g^{2 n - j}] [g^j] = [g^{f (j + 2 n - j)}] = [g^{f (2 n)}] = [g^0] = [1] = 1\), because \(2 n = 2 n 1 + 0\).
For \(0 \le j \lt 2 n\), \([h g^j]^{-1} = [h g^{f (n + j)}]\), because \([h g^j] [h g^{f (n + j)}] = [g^{f (n - j + f (n + j))}] = [g^{f (n - j + n + j)}] = [g^{f (2 n)}] = [g^0] = [1] = 1\); \([h g^{f (n + j)}] [h g^j] = [g^{f (n - f (n + j) + j)}] = [g^{f (n - (n + j) + j)}] = [g^{f (0)}] = [g^0] = [1] = 1\).
Let us check associativity.
Let \(j, k, l \in \mathbb{N}\) be any \(0 \le j, k, l \lt 2 n\) hereafter.
\(([g^j] [g^k]) [g^l] = [g^{f (j + k)}] [g^l] = [g^{f (f (j + k) + l)}] = [g^{f (j + k + l)}]\) while \([g^j] ([g^k] [g^l]) = [g^j] [g^{f (k + l)}] = [g^{f (j + f (k + l))}] = [g^{f (j + k + l)}]\).
\(([h g^j] [g^k]) [g^l] = [h g^{f (j + k)}] [g^l] = [h g^{f (f (j + k) + l)}] = [h g^{f (j + k + l)}]\) while \([h g^j] ([g^k] [g^l]) = [h g^j] [g^{f (k + l)}] = [h g^{f (j + f (k + l))}] = [h g^{f (j + k + l)}]\).
\(([g^j] [h g^k]) [g^l] = [h g^{f (- j + k)}] [g^l] = [h g^{f (f (- j + k) + l)}] = [h g^{f (- j + k + l)}]\) while \([g^j] ([h g^k] [g^l]) = [g^j] [h g^{f (k + l)}] = [h g^{f (- j + f (k + l))}] = [h g^{f (- j + k + l)}]\).
\(([g^j] [g^k]) [h g^l] = [g^{f (j + k)}] [h g^l] = [h g^{f (- f (j + k) + l)}] = [h g^{f (- (j + k) + l)}]\) while \([g^j] ([g^k] [h g^l]) = [g^j] [h g^{f (- k + l)}] = [h g^{f (- j + f (- k + l))}] = [h g^{f (- j - k + l)}] = [h g^{f (- (j + k) + l)}]\).
\(([h g^j] [h g^k]) [g^l] = [g^{f (n - j + k)}] [g^l] = [g^{f (f (n - j + k) + l)}] = [g^{f (n - j + k + l)}]\) while \([h g^j] ([h g^k] [g^l]) = [h g^j] [h g^{f (k + l)}] = [g^{f (n - j + f (k + l))}] = [g^{f (n - j + k + l)}]\).
\(([g^j] [h g^k]) [h g^l] = [h g^{f (- j + k)}] [h g^l] = [g^{f (n - f (- j + k) + l)}] = [g^{f (n - (- j + k) + l)}]\) while \([g^j] ([h g^k] [h g^l]) = [g^j] [g^{f (n - k + l)}] = [g^{f (j + f (n - k + l))}] = [g^{f (j + n - k + l)}] = [g^{f (n - (- j + k) + l)}]\).
\(([h g^j] [h g^k]) [h g^l] = [g^{f (n - j + k)}] [h g^l] = [h g^{f (- f (n - j + k) + l)}] = [h g^{f (- (n - j + k) + l)}] = [h g^{f (- n - (- j + k) + l)}] = [h g^{f (- n - (- j + k) + l + 2 n)}] = [h g^{f (n - (- j + k) + l)}]\) while \([h g^j] ([h g^k] [h g^l]) = [h g^j] [g^{f (n - k + l)}] = [h g^{f (j + f (n - k + l))}] = [h g^{f (j + n - k + l)}] = [h g^{f (n - (- j + k) + l)}]\).
So, associativity holds.
So, \(SD_{2 n}\) is indeed a group.
Although \([h g^j]\) is defined to be a symbol, it indeed equals the multiplication of \([h]\) and \([g^j]\), as a result (of course, as we have anticipated the result, we have chosen that specific symbol). So, although we have so far bothered to use \([]\), notations like \(h g^j\) have in fact no practical ambiguity.
\([g]^{2 n} = 1\), \([h]^2 = [g]^n\), \([g]^{-1} [h] = [h] [g]\), and \([h]^2 = ([g] [h])^2\) hold as a result, because \([g]^{2 n} = [g]^{2 n - 1} [g^1] = [g^{2 n - 1}] [g^1] = [g^{f (2 n - 1 + 1)}] = [g^{f (2 n)}] = [g^0] = 1\), \([h]^2 = [h g^0] [h g^0] = [g^{f (n - 0 + 0)}] = [g^{f (n)}] = [g^n] = [g]^n\), \([g]^{-1} [h] = [g^{2 n - 1}] [h g^0] = [h g^{f (- (2 n - 1))}] = [h g^{f (- 2 n + 1)}] = [h g^{f (- 2 n + 1 + 2 n)}] = [h g^{f (1)}] = [h g^1] = [h] [g^1] = [h] [g]\): \([g] [g^{2 n - 1}] = [g^{2 n - 1}] [g] = [g^{f (1 + 2 n - 1)}] = [g^{f (2 n)}] = [g^0] = 1\), and \(([g] [h])^2 = ([g] [h]) ([g] [h]) = ([g] ([h] [g])) [h] = ([g] [h g]) [h] = [h g^{f (-1 + 1)}] [h] = [h g^{f (0)}] [h] = [h g^0] [h] = [h] [h]\).
Supposing that we already have a set, \(\{[1], [g], ..., [g^{2 n - 1}], [h], [h g], ..., [h g^{2 n - 1}]\}\), and a group operation on it, let us see a sufficient way to confirm that it is a skewed dihedral group: of course we can check the whole operation, but that could be tedious.
As we suppose that we have the group operation, it is guaranteed to be closed and associative, because otherwise, it would not be any group operation. Checking the associativity itself may seem tedious, but an easy case is that each element of the set is defined to be a map and each multiplication is defined to be the composition of maps, which guarantees the associativity because maps compositions are associative.
Now, this is a sufficient set of conditions to be checked: 1) \(\{[1], [g], ..., [g^{2 n - 1}]\}\) is cyclic; 2) \([h]^2 = [g^n]\); 3) \([h] [g^j] = [h g^j]\); 4) \([g] [h] = [h g^{2 n - 1}]\).
Let us see that it is indeed sufficient.
\([g^j] [g^k] = [g]^j [g]^k = [g]^{j + k} = [g]^{f (j + k)} = [g^{f (j + k)}]\).
\([g^j] [h g^k] = [g]^j [h] [g^k] = [g]^{j - 1} ([g] [h]) [g^k] = [g]^{j - 1} ([h g^{2 n - 1}]) [g^k] = [g]^{j - 2} [g] [h] [g^{2 n - 1}] [g^k] = [g]^{j - 2} ([g] [h]) [g^{2 n - 1}] [g^k] = [g]^{j - 2} ([h g^{2 n - 1}]) [g^{2 n - 1}] [g^k] = ... = [h g^{2 n - 1}] [g^{2 n - 1}] ... [g^{2 n - 1}] [g^k]\), where there are \(j - 1\) \([g^{2 n - 1}]\) s, \(= [h] [g^{2 n - 1}]^j [g^k] = [h] ([g]^{2 n - 1})^j [g]^k = [h] [g]^{(2 n - 1) j + k} = [h] [g^{f ((2 n - 1) j + k)}] = [h] [g^{f (- j + k + 2 n j)}] = [h] [g^{f (- j + k)}] = [h g^{f (- j + k)}]\).
\([h g^j] [g^k] = [h] [g^j] [g^k] = [h] ([g]^j [g]^k) = [h] ([g]^{j + k}) = [h] [g^{f (j + k)}] = [h g^{f (j + k)}]\).
\([h g^j] [h g^k] = [h] [g^j] [h g^k] = [h] ([g^j] [h g^k]) = [h] [h g^{f (- j + k)}] = [h] [h] [g^{f (- j + k)}] = ([h] [h]) [g^{f (- j + k)}] = [g^n] [g^{f (- j + k)}] = [g^{f (n + f (- j + k))}] = [g^{f (n - j + k)}]\).
So, the operation inevitably satisfies the condition to be \(SD_{2 n}\).
