Element Order within a Group
The order of an element \(g\) in a group \((G, *)\) is defined as the smallest positive integer \(r\) for which \(g^r = e\), where \(e\) stands for the group's identity element.
In layman's terms, the order of \(g\) represents the number of times the group's operation * needs to be applied to \(g\) for it to return to the identity element. For example:
$$ \underbrace{g*g*\ldots*g}_{r \ times} = e $$
This notion is key to understanding the cyclic nature of groups.
The order of an element can be either finite or infinite.
- Finite Order
An element g has a finite order when it repeats a certain number of times, n, before it reaches the identity element of the group. - Infinite Order
An element g has an infinite order if it repeats endlessly, never encountering the group's identity element.
Note: Whether an element has a finite order depends greatly on the group's characteristics and the element itself. In finite groups, every element has a finite order that falls between 1 and the group's order (its size). Conversely, in infinite groups, it's possible for some elements to have an infinite order, indicating that no finite sequence of operations can return the element to the identity state.
A Hands-on Example
The Group of Integers Under Addition ( \( \mathbb{Z}, + \) ):
Take the element \(1\) in the additive group \( \mathbb{Z}, + \).
Given the identity element is \(0\), repeatedly adding \(1\) to itself will never result in \(0\), which means the element \(1\) has an infinite order.
Note: The same goes for the element \(2\), as any multiple of 2 remains distinct from the identity element. For instance, 2+2=4, 2+2+2=6, and so on. The unique member of the additive group (\(Z,+\)) with a finite order is the zero element since it's also the identity element, making its order one.
The Group of Modular Arithmetic Classes modulo \(n\) ( \( \mathbb{Z}_n, + \) ):
Considering \( \mathbb{Z}_4 \), the group formed by the set of all remainder classes modulo 4 under addition \( (\mathbb{Z}_4 , + ) \).
The set \( \mathbb{Z}_4 \) comprises four elements:
$$ \mathbb{Z}_4 = \{ 0 , 1, 2, 3 \} $$
The element \(2\) has an order of \(2\) because adding \(2\) to itself gives \(0\) modulo \(4\), aligning it with the identity element of the group.
$$ 2 + 2 = 4 \equiv 0 \mod 4 $$
Conversely, the element \(1\) possesses an order of \(4\), as it requires four self-additions to revert to the group's identity element, zero.
$$ 1+1+1+1=4 \equiv 0 \mod 4 $$
Note: Similarly, the identity element \(0\) has an order of \(1\), reaffirming its role as the identity element of the group.
Insights
Here are a few additional thoughts on the order of an element within a group.
- For an element g in group G with infinite order, when you take any two distinct integers, s and t, it turns out the elements gs and gt are different. This means the cyclic subgroup <g> that g generates is also infinite.
Example. Within the additive group (Z,+), let's consider the element g=1. Choosing two distinct integers, s=2 and t=3, it becomes evident that their powers differ: 12 equals 1+1=2, while 13 sums up to 1+1+1=3. It's worth noting that in group theory, powering an element means applying the group's operation, addition in this case, repeatedly to the same element.
- If an element g has a finite order of n, it gives rise to a finite cyclic subgroup <g> with n members. $$ <g> = \{ g^0, g^1, ..., g^{n-1} \} $$ In this case, g0 serves as the group's identity element, so g0=e. Here, for any two distinct integers s and t, the elements gs and gt are identical if and only if s≡t (mod n).
Example. In the group (Z5,+) of integers with addition modulo 5, take g=1 for instance. The powers g3=3 and g8=3 arrive at the same result because both s=3 and t=8, when divided by 5, have a remainder of 3. This demonstrates that the cyclic subgroup <1>={0,1,2,3,4} is finite.
And the story goes on.
