If $a,b$ commute in $G$, have finite order, $⟨a⟩\cap⟨b⟩\neq ⟨1⟩$ and neither is contained in the other, describe $|ab|$. If $a,b$ commute in $G$, have finite order, $⟨a⟩\neq ⟨b⟩$, and $⟨a⟩\cap⟨b⟩\neq ⟨1⟩$, describe $|ab|$.
If $\langle a \rangle \cap \langle b \rangle = \langle 1 \rangle$ then $o(ab)=\hbox{lcm} (o(a), o(b))$. If $\langle a \rangle = \langle b \rangle$ there are plenty of counterexamples where it isn't true. If $\langle a \rangle \subset \langle b \rangle$ it is again easy to provide a counterexample (a cyclic group).
  In the other cases, however, it seems that $o(ab)=\hbox{lcm} (o(a), o(b))$ always.
However, I have a hunch that it is false; could you give a counterexample or a method to unravel this problem?
 A: No, you cannot conclude that the order is the least common multiple in this situation. 
For an easy example, consider $G=C_2\times C_3$, let $a=(1,1)$ and $b=(0,2)$. Then $\langle b\rangle \subseteq \langle a\rangle$ but they are not equal, $o(a) = 6$, $o(b)=3$, and $ab=(1,0)$ has order $2$.  (This is just $C_6$, with $a$ a generator and $b=a^2$). 

Added. The problem has been changed to exclude either group being contained in the other. This only requires an easy tweak: take $G=C_2\times C_3\times C_2$, $a=(1,1,0)$, $b=(0,2,1)$. Then $o(a)=6$, $o(b)=6$, $ab=(1,0,1)$ has order $2$, and neither subgroup is contained in the other.

In general, the best you can say is that:
$$\begin{align*}
\frac{\mathrm{lcm}(o(a),o(b))}{\gcd(o(a),o(b))} &\Bigm| \frac{\mathrm{lcm}(o(a),o(b))}{|\langle a\rangle\cap \langle b\rangle|}\\
\strut\\
\frac{\mathrm{lcm}(o(a),o(b))}{|\langle a\rangle\cap \langle b\rangle|}&
\Bigm| o(ab)\\
\strut\\
o(ab) &\Bigm| \mathrm{lcm}(o(a),o(b)).
\end{align*}$$

The key here is that you can always express an element of finite order as a product of pairwise commuting elements of prime power order, pairwise coprime. To see this, suppose $x$ has order $ab$ with $\gcd(a,b)=1$. Then there exist integers $r$ and $t$ such that $1 = ar+bt$. Note that $\gcd(r,b)=\gcd(r,t)=\gcd(a,t)=1$. We have that $x = (x^{ar})(x^{bt})$. Now, $x^a$ has order $b$, and $\gcd(r,b)=1$, so $o(x^{ar}) = o((x^a)^r) = b/\gcd(b,r) = b$. Symmetrically, $o(x^{bt}) = a$. So $x$ is a product of an element of order $a$ and an element of order $b$. Lather, rinse, and repeat (or do induction on the number of prime factors) to get that if $x$ has order $n$, and $n=p_1^{a_1}\cdots p_r^{a_r}$ is the prime factorization of $n$, then $x$ can be written as $x=x_1\cdots x_n$ with $o(x_i)=p_i^{a_i}$. 
So now think about your $a$ and $b$; primes that only occur in one of $o(a)$ and $o(b)$ will occur in $o(ab)$; and more generally a bit of work shows that if the highest power of $p$ that divides $o(a)$ is different from the highest power of $p$ that divides $o(b)$, then the highest power of $p$ that divides $o(ab)$ is the maximum of the two (giving you the "correct" exponent for the least common multiple). So the key lies in situations where the highest power of $p$ that shows up in $o(a)$ and in $o(b)$ is equal. Clearly, you can try to arrange things so that the $p$-part of $a$ and the $p$-part of $b$ cancel out (or just partially cancel out) but the rest of the parts are disjoint, so that you get $o(ab)$ strictly smaller than $\mathrm{lcm}(o(a),o(b))$, while also ensuring that neither of $\langle a\rangle$ and $\langle b\rangle$ contain each other. That is what I did: the $2$-parts of $a$ and $b$ don't interact, but the $3$-parts cancel each other. 
