$\operatorname{ord}(a^k)$ is a divisor of $\operatorname{ord}(a)$

Question

I'm trying to prove this for a finite group $G$.

Along with the book's hint, I have the following proof:

Let $\operatorname{ord}(a)=n$. Then

$$(a^k)^n=a^{kn}=a^{nk}=(a^n)^k=e.$$

It follows that $kn$ is a multiple of $n$, so $kn=nb$ for an integer $b$. So, $k=b$.

But I'm not sure what $k=b$ gives me.

You know that $(a^k)^n=e$. Why don't you try to show that, for any element $g$ of any finite group $G$, if $g^m=e$, then the order of $g$ divides $m$? — Andrew, Feb 27 '16 at 01:26
Indeed, it's true (and supremely helpful in general) that $g^m=e$ if and only if the order of $g$ divides $m$. — Greg Martin, Feb 27 '16 at 01:43

score 1 · Accepted Answer · answered Feb 27 '16 at 01:49

1

Andrew's comment is good; a next step is to use the division algorithm and suppose toward a contradiction that ord$(a) = n = q $ ord$(a^k) + r$ (since you know ord$(a^k) \le $ord(a)).

answered Feb 27 '16 at 01:49

Sheridan Grant

250

$\operatorname{ord}(a^k)$ is a divisor of $\operatorname{ord}(a)$

1 Answers1

Linked