The order of an element divides the order of a finite group. If G is a finite group and a ∈ G, then o(a) divides |G|. Let us now recall the definition of the order of an element in a group.
Let a ∈ G. The order of a in G is the smallest positive integer n such that an = e, the identity element in G. The order of a is denoted by o(a). In this article, we will show that n divides |G|.
Table of Contents
Proof
Let a ∈ G be an element of order n in a finite group G.
Consider the subgroup generated by a, and denote it by H.
So H = <a>.
Then by definition of cyclic groups, o(a) = |H|.
As H is a subgroup of a finite group G, so by Lagrange’s theorem of finite groups, we deduce that |H| divides |G|.
Noting o(a) = |H|, we obtain that o(a) divides |G|.
So the order of an element in a finite group divides the order of that group.
Solved Problems
Question 1: Is there any element of order 4 in Z6?
Answer:
The group Z6 has order 6.
As 4 does not divide 6, by the above fact we conclude that there does not exist any element of order 4 in Z6.
More Topics: An Introduction to Group Theory
Abelian Group | Quotient Group
Normal Subgroup | Simple Group
Prove that a group of prime order is cyclic
Every Subgroup of a Cyclic Group is Cyclic: Proof
FAQs
Answer: The order of an element in a group is the least positive integer n such that the n-th power of that element is the identity of the group. For example, the identity element has order 1.