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The order of a finite group G is always a multiple of the order of any element a ∈ G

My teacher gave some true or false questions to do and this was one of the question which the answer is true but I'm trying to understand why.

So I tried with the group $Z_{12}$ which is a finite group and the elements are $\{0,1,2,3,4,5,6,7,8,9,10,11\}$

So if im right, lets say I pick the element 3 then the order of 3 is 4 and 12 is a multiple of 4 and 4 is a multiple of 4. So is this what this idea saying? or do I have it wrong?

kero
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    You've correctly understood the statement. The statement is a special case of Lagrange's theorem, but I am not aware of any proof of it that is not essentially a proof of Lagrange's theorem, so you might as well go read about that. @Nameless Wrong. Also cyclic is a special case of abelian. – anon Mar 13 '15 at 01:07
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    @nameless Why? If $G$ is any finite group, and $a \in G$, the order of $a$ is the order of the subgroup generated by $a$. But the order of any subgroup must divide the order of the entire group. $G$ doesn't have to be cyclic. – layman Mar 13 '15 at 01:07

1 Answers1

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Another way to say this is that the order of every element divides $|G|$. Notice the order of $a$ is the order of $\langle a \rangle$ (the subgroup generated by $a$).

By Lagrange's theorem the order of this subgroup divides $|G|$

Asinomás
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