# How can $1=\frac{\operatorname{ord} a}{\gcd(\operatorname{ord}a,k)}\mid\frac{k}{\gcd(\operatorname{ord}a,k)}$?

Let $$G$$ be a group and $$a\in G$$ an element with finite order. For any $$k\in\mathbb{Z}$$,

$$\operatorname{ord}a^k=\frac{\operatorname{ord} a}{\gcd(\operatorname{ord}a,k)}$$

ord-order ,gcd-greatest common divisor

Proof:

Consider $$k,m\in\mathbb{Z}$$ such that:

$$(a^k)^m=e\implies a^{km}=e\implies \operatorname{ord} a \mid km\implies \frac{\operatorname{ord}a}{\gcd(\operatorname{ord}a,k)}\mid\frac{k}{\gcd(\operatorname{ord}a,k)}m$$

Therefore:

$$1=\frac{\operatorname{ord}a}{\gcd(\operatorname{ord}a,k)}\mid\frac{k}{\gcd(\operatorname{ord}a,k)}\implies\frac{\operatorname{ord} a}{\gcd(\operatorname{ord}a,k)}|m$$

So we have:

$$\operatorname{ord}a^k=\frac{\operatorname{ord} a}{\gcd(\operatorname{ord}a,k)}$$

Question:

I am struggling with this material I am reading, in the sense I do not understand why $$1=\frac{\operatorname{ord} a}{\gcd(\operatorname{ord}a,k)} \mid \frac{k}{\gcd(\operatorname{ord}a,k)}$$ is true. I have been looking in books but I found nothing of this. How can $$1=\frac{\operatorname{ord} a}{\gcd(\operatorname{ord}a,k)}\mid\frac{k}{\gcd(\operatorname{ord}a,k)}$$? The least common divisor between two numbers is not necessarily equal to the first or second. $$\frac{15}{3}$$ and $$\frac{20}{3} \neq 1$$ and $$3=\gcd(15,20)\neq 1$$.

Let $|a|$ represents the order of $a$ and let the g.c.d. of $|a|$ and $k$ be $g$.
$(a^k)^\frac{|a|}{g}=(a^{|a|})^\frac{k}{g}=e$ and so $|a^k|$ divides $\frac{|a|}{g}$.
Conversely, $(a^k)^{|a^k|}=e$ implies $a^{k|a^k|}=e$ and so $|a|$ divides $k|a^k|$. Then $\frac{|a|}{g}$ divides $\frac{k|a^k|}{g}$ and, since $\frac{|a|}{g}$ and $\frac{k}{g}$ are coprime, $\frac{|a|}{g}$ must divide $|a^k|$.
Therefore $\frac{|a|}{g}=|a^k|$, as you required.
• Thanks for your answer! I have a doubt. How do you know $\frac{|a|}{g}$ and $\frac{k}{g}$ are coprime? – Pedro Gomes Nov 21 '17 at 18:40
• Since the g.c.d. of $|a|$ and $k$ is $g$. – S. Dolan Nov 21 '17 at 19:03