# 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 <=. Commented Aug 13, 2019 at 2:52
• @LordSharktheUnknown Thanks! Updated! Commented Aug 13, 2019 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. Commented Aug 13, 2019 at 8:49
• @José Carlos Santos Do you really think that it's a duplicate? Commented Aug 13, 2019 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. Commented Aug 14, 2019 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 $a$. Commented Aug 14, 2019 at 6:18