Prove that Order of Element Divides Order of Group

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|.

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

Subgroups of Cyclic Groups

Prove that a group of prime order is cyclic

Every Subgroup of a Cyclic Group is Cyclic: Proof

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