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 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
 A: 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 $$(xy)^{p^{r-1}}=x^{p^{r-1}}y^{p^{r-1}}=x^{p^{r-1}}\neq 1.$$
