Short proof that if a and b of a group G commute then an element of order lcm(o(a),o(b) exists in G

Question

I've been looking for short proofs for a few days now, but could not find one.

Today I was finally able to write my own proof and I'd be happy if someone could verify if it's true.

General assumption:

We are talking in a group with a multiplicative operation.

First we use the following theorem(1) as a given:

If $ab=ba$ and $gcd(o(a),o(b))=1$ => $o(ab)=o(a)o(b)$

Statement: If $ab=ba$ => $\exists{c} $ : o(c)=lcm(o(a),o(b))

Proof:

Let: d=$gcd(o(a),o(b))$

Then:

$lcm(o(a),o(b))=\frac{o(a)o(b)}{d}$

$gcd(\frac{o(a)}{d},o(b))=1$

Lets now look at: $a^d$ => $o(a^d)=\frac{o(a)}{d}$

Finally we arrive at: $a^{d}b$.

Since $gcd(o(a^d),o(b))=gcd(\frac{o(a)}{d},o(b))=1$

=> By theorem(1)

$o(a^db)$ must be equal to $o(a^d)o(b)$.

$o(a^d)o(b)=\frac{o(a)}{d}o(b)=lcm(o(a),o(b))$

Conclusion: $o(a^{gcd(a,b)}b)=lcm(o(a),o(b))$

Long proofs I found:

https://yutsumura.com/the-existence-of-an-element-in-an-abelian-group-of-order-the-least-common-multiple-of-two-elements/

Proofs I could not understand:

Order of element equal to least common multiple

Note that $\gcd(\frac{o(a)}{d}, \frac{o(b)}{d}) = 1$ but it is not generally true that $\gcd(\frac{o(a)}{d}, o(b)) = 1$. (E.g. let $o(a) = 18, o(b) = 12$) — Brian Moehring, Aug 22 '20 at 21:32
Thanks for the context, effort, detail, and research you put into writing this question. +1 — amWhy, Aug 22 '20 at 21:33
How could it be false that $gcd(\frac{o(a)}{d},o(b))=1$? I mean by dividing $o(a)$ by $d$, we remove all common factors of $o(a)$ and $o(b)$. — Everstudent, Aug 22 '20 at 21:35
Unfortunately, as Brian Moehring stated, your proof hinges on the fact that, for two integers $n$ and $m$, $\gcd(n/\gcd(n,m),m)=1$, which is not true in general. For example, let $n=12$ and $m=18$, $\gcd(n,m)=6$, and $\gcd(12/6,18)=2$. This holes the proof below the waterline. — David A. Craven, Aug 22 '20 at 21:37
Hm, I see if $d=gcd(18,12)=6$. Then $\frac{o(a)}{d}=3$ and o(b)=12. But $gcd(3,12)=3$ and not $1$... — Everstudent, Aug 22 '20 at 21:38
What you need to do is look at all prime divisors of both $o(a)$ and $o(b)$, and then select all of those where $o(a)$ has the largest power of the prime, and separate then from all those where $o(b)$ has the largest power. (Edit: it turns out that is how the proof you reference does it.) — David A. Craven, Aug 22 '20 at 21:38
That's what the proof from yutsumura.com is about. I understood it, but for more context on why I want a simpler proof: The problem is from Pinter's Abstract Algebra. The exercise says to use the theorem(1) from my question and to pick an element of the form $a^{i}b$. So I was thinking there has to be a simpler proof.
If someone is interested - the location of the problem is: Chapter 10 of Pinter's Abstract Algebra - problem E5. — Everstudent, Aug 22 '20 at 21:41
It can be done this way, but you essentially have to do the same thing as I said above. — David A. Craven, Aug 22 '20 at 21:54
Duplicate of Order of elements is lcm-closed in abelian groups — Bill Dubuque, Feb 04 '23 at 16:12

score 1 · Accepted Answer · answered Aug 22 '20 at 22:15

The proof is essentially the one mentioned in your question. It's much easier to take a proper power of $b$, but it can be done with $b$. It's much more fiddly.

Let $\pi=\{p_1,\dots,p_r\}$ be the primes that divide $o(b)$ at least as much as they divide $o(a)$, and let $m$ be the $\pi$-part of $o(a)$, i.e., the number such that $m\mid o(a)$ and $o(a)/m$ is not divisible by any prime in $\pi$.

I claim that $g=a^m\cdot b$ has the correct order. For this we need to show that no power of $g$ is the identity before $n=\mathrm{lcm}(o(a),o(b))$. I think the easiest way to do this is to show that some power of $g$ has order each prime power dividing $n$. Let $p$ be a prime dividing $n$, and write $n=p^rs$ where $p\nmid s$. I claim $g^s$ has order $p^r$.

First let $p$ lie in $\pi$. Certainly $a^m$, which has order not divisible by $p$, satisfies $(a^m)^s=1$, so $g^s=b^s$. But this is easily seen to have order $p^r$.

Now let $p$ lie in the complement of $\pi$. This time both $a^s$ and $b^s$ need not be trivial, but crucially, the order of $b^s$ strictly divides $p^r$, by the way we chose $\pi$. However, the order of $a^s$ is exactly $p^r$.

So if we know that the product of an element $x$ of order $p^r$ and an element $y$ of order $p^t$ for $t<r$ (that commute) is $p^r$, then we are done. But clearly this is the case, because $xy$ has order some power of $p$, and $$(xy)^{p^{r-1}}=x^{p^{r-1}}y^{p^{r-1}}=x^{p^{r-1}}\neq 1.$$

The crucial thing is that if you raise an element of order $n$ to the power $n/d$, you obtain an element of order $d$. I am using this over and over again. — David A. Craven, Aug 22 '20 at 22:38

Short proof that if a and b of a group G commute then an element of order lcm(o(a),o(b) exists in G

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