# Prove: $\text{ord}(a^m) = \frac{\text{lcm}(m,n)}{m}$ [duplicate]

This is Pinter $$10.G.5$$

Let:

$$a \in G$$

$$\text{ord}(a) = n$$

Prove: $$\text{ord}(a^m) = \frac{\text{lcm}(m,n)}{m}$$

Use $$10.G.3$$ and $$10.G.4$$ to prove this.

Here is $$10.G.3$$:

Let $$l$$ be the least common multiple of $$m$$ and $$n$$. Let $$l/m = k$$. Explain why $$(a^m)^k = e$$.

Here is $$10.G.4$$:

Prove: If $$(a^m)^t = e$$, then $$n$$ is a factor of $$mt$$. (Thus, $$mt$$ is a common multiple of $$m$$ and $$n$$.)

Conclude that: $$l = mk \leq mt$$

OK, let's begin.

By $$10.G.3$$:

$$(a^m)^{\text{lcm}(m,n)/m} = e \tag{5}$$

If $$\text{lcm(m,n)}/m$$ is the lowest number such that (5) is true, then:

$$\text{ord}(a^m) = \text{lcm}(m,n)/m \tag{9}$$

Let's assume that there is a number

$$q < \text{lcm}(m,n)/m \tag{7}$$

such that:

$$(a^m)^q = e \tag{6}$$

By $$10.G.4$$ with (6):

$$n \ \big|\ mq$$

$$l \lt mq$$

$$\text{lcm}(m,n) \leq mq$$

Isolate q:

$$\text{lcm}(m,n)/m \leq q \tag{8}$$

So (9) must be true.

That's how I approached it. If anyone notices any issues, I'd be happy to know about them.

Even if it is considered correct, do you feel there is a better way?

• You should use \le rather than <=. – Angina Seng Aug 13 '19 at 2:52
• @LordSharktheUnknown Thanks! Updated! – dharmatech Aug 13 '19 at 2:54
• @JyrkiLahtonen Can you explain how this is a duplicate of the question you linked to? There is no requirement in this question that G be cyclic. Whereas the question you linked to is for cyclic groups. – dharmatech Aug 13 '19 at 8:49
• @José Carlos Santos Do you really think that it's a duplicate? – Michael Rozenberg Aug 13 '19 at 9:06
• @YuiToCheng Once again not a proper dupe target becauyse - among other things - the OP is working with LCMs not GCDs, and is working from specific lemmas. Please be more careful in choosing proper dupe targets. – Bill Dubuque Aug 14 '19 at 13:32

Yes, $$\,(a^{m})^{k} = 1\!\! \overset{(1)\!\!}\iff n\mid mk \iff m,n\mid mk\!\! \overset{(2)\!\!}\iff {\rm lcm}(m,n)\mid mk\iff {\large{\frac{{\rm lcm}(m,n)}m\,\mid}}\, k$$
$$\!(1)$$ and the (omitted) final conclusion is by Corollary' here, and $$(2)$$ is the LCM Universal Property
• @BillDubuque Do you really think this question has not been handled on our site? The OP's complaint about my choice of dupe was mostly because "their version is not only about cyclic groups". Yet they only work within the cyclic group generated by $a$. – Jyrki Lahtonen Aug 14 '19 at 6:18