Even Order Groups: Counting Elements of Order 2

[SqrachMasda]

Prove: a group of even order must have an even number of elements of order 2

 

[Hurkyl]

What group do they generate?

(You sure on that? What about Z_2?)

 

[wais]

Firstly, let us define what we mean by "order" in a group. The order of an element in a group is the smallest positive integer n such that the element raised to the power of n is equal to the identity element of the group. In other words, the order of an element is the number of times we need to combine the element with itself to get the identity element.

Now, let us consider a group G of even order. This means that the number of elements in the group is even, let's say 2n. We want to prove that there must be an even number of elements of order 2 in this group.

Let's start by assuming that there is an odd number of elements of order 2 in G. This means that out of the 2n elements in the group, there are an odd number of elements that have order 2. Let's say there are 2k+1 elements of order 2, where k is a positive integer.

Since the identity element has order 1, this means that there are 2n-(2k+1) elements in the group that have order other than 1 or 2. Now, let us consider the element x that has order other than 1 or 2. This means that the order of x is greater than 2, let's say m. Therefore, we can write x^m = e, where e is the identity element.

Now, let's consider the elements x, x^2, x^3, ..., x^m. Since the order of x is m, this means that these elements are all distinct, as if they were not, then we would have a smaller number k such that x^k = e, contradicting the fact that the order of x is m.

But notice that x^2, x^4, x^6, ..., x^m are all elements of order 2, as (x^2)^2 = x^4, (x^4)^2 = x^8, and so on. This means that out of the 2n-(2k+1) elements that have order other than 1 or 2, there are at least m/2 of them that have order 2.

But we know that m is greater than 2, so m/2 is an integer. Therefore, we have at least one more element of order 2 in the