14.4. Order of a Group Element

So now, let us define order of a group element. So, imagine you are given a finite group and for convenience, I will be using the multiplicative notation. However, whatever we define here holds for any group. And now consider an arbitrary group element g. We define a function from the set of natural numbers to the group and my function is the following. The domain will be {0,1,2,…,∞} and the co-domain is the group. The way I go from the domain to co-domain is, if I want to map the element , I go to g^n. Now, it is easy to see that since my group G is a finite group, it will have finite number of elements whereas, my domain is infinitely large then by pigeonhole principle, I know that there exists at least 2 non-zero values m and n such that m > n and both m as well as n get mapped to the same group element, namely g^m = g^n.

In this chunk, we start with the definition of the order of a group element. The order refers to how many times we need to multiply the element g by itself before we get back to the identity element of the group. The text also mentions we’re using multiplicative notation for convenience. Essentially, when we keep raising the element g to higher powers (g^1, g^2, and so on), we notice that due to the finite nature of the group, we will eventually hit repeating values because there are only a limited number of distinct elements in the group. This repetition indicates that we have found a certain exponent where g raised to that exponent equals the identity element, which is the fundamental aspect of determining the order of the group element.

Think of a clock as an analogy. If you think of each hour on the clock as an element of a group, moving from one hour to the next can represent raising the group element to a power. For example, if it's 12 o'clock (the identity hour) and you keep adding one hour, you'll eventually circle back to 12 again—the order of the clock would be 12 hours.

Now, since g^m is also a group element, if we multiply both sides of the equation with g^(-n), then we get g^(m-n) = 1. Note that 1 is the identity element in multiplicative notation. Since m > n, m - n is positive. This in turn implies that there is at least one positive integer k, such that for the element g which I have arbitrarily chosen here, g^k = 1.

Here, we are using the conclusion reached in the previous chunk. By manipulating our previous findings algebraically, we conclude that there is a positive integer k (specifically, m - n) such that when we raise g to that power, we get back the identity. This is crucial because the order of an element is defined as the smallest positive integer that satisfies this condition. If we keep choosing powers of g (like g^1, g^2, etc.), we will continue finding new elements until we eventually get back to the identity, giving us the order.

Imagine you have a bicycle with gears. When you pedal to go forward, you're effectively multiplying the power of your push. As you shift gears, you might feel like you're going at a consistent speed until you reach a point where you can no longer gain speed at one particular gear; at that point, you might need to shift back to the easiest gear to continue moving forward efficiently. The smallest gear setting that can get you back to the start (like the identity of the group) is akin to finding the order.

Of course, there might be multiple values of k for which g^k will be 1, it depends upon how many (m,n) pairs are there. But at least 1 positive integer k is definitely there such that g^k is the identity element. Among all those positive integers k such that, g^k is equal to the identity element, the smallest positive integer is called as the order of the element g.

In this chunk, it is discussed that while there may be several integers k that satisfy the equation g^k = 1, we focus on the smallest among them, which is known as the order. This distinction is critical as it provides a unique identifier (the smallest step that returns us to the identity) for the element in question. If the group were infinite, it could be that the element never returns to the identity, in which case we would say the order is infinite.

Think of a hamster wheel. If the hamster runs indefinitely but never stops, we might consider its running as similar to having no order since it stays in motion and doesn’t return to a resting state (the identity). On the other hand, if it runs for a set duration and then stops, we can measure its 'length of run'—the smallest time it takes to return to its rest position—this would represent the order.

So now, let us discuss some interesting properties of the order of a group element. So again, I will stick to the multiplicative notation. So, imagine you are given an element g and it is given to you that its order is n; that means, I know that g^n = 1 then my claim is the following. If you have g^k also giving you the identity element then that is possible if and only if, k is a multiple of n.

This chunk introduces a property of orders concerning their multiples. If we have established that the order of g is n, then any exponent k that makes g to the power k equal the identity must be a multiple of n. This property ensures that all 'returns to the identity' occur at intervals defined by the order. The proof would typically show that if k is any integer, it can be expressed as a multiple of n, confirming that if g^k = 1, then k must be congruent to 0 (because those iterations lead back to the identity in sync with the order).

Consider a train schedule. If the train returns to its original station every hour (making its order 1), and comes back to station X precisely three times within three hours, that means it perfectly fits into a cycle where 1 is the base of this repetitive return. However, if another train operates on a different schedule (say every 15 minutes, i.e., 4 times in an hour), it still belongs to a system where the fundamental return period is 1. Each stop is a multiple of that return period.

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