Question:
Prove that a group $G$ of order 143 is abelian.
Solution:
$G$ contains subgroups $H$ and $K$ of orders 11 and 13, respectively.
The subgroups $G/H$ and $G/K$ have orders 13 and 11, respectively, and hence are cyclic,
and a fortiori, abelian.
This means $G' \subset H \cap K = \{e\}.$
So, $gh = ghg^{-1}h^{-1}hg = ehg = hg.$
I have used the
Lemma:
If $G'$ is the subgroup generated by the set
$$U = \{ ghg^{-1}h^{-1} \colon g,h \in G\},$$
then $G/G'$ is abelian and any normal subgroup $N$ which satisfies $G/N$
is abelian also satisfies $N \supseteq G'.$
$G'$ is called the commutator subgroup of $G.$
It is the smallest subgroup containing $U,$ and any subgroup containing $U$ contains $G'.$
Proof:
For $g,h$ in $G,$ we have $gh N= hg N = ghh^{-1}g^{-1}hg N$ implying that $h^{-1}g^{-1}hgN = N.$
So, $h^{-1}g^{-1} hg \in N.$
As $g,h$ were arbitrarily chosen in $G,$ we get $U \subset N.$
Problem:
This would also apply to all groups of order $pq$ where $p$ and $q$ are distinct primes. This is known to be false.
Best Answer
The subgroups $H$ and $K$ may not be normal, so you cannot take the quotients and conclude that they contain the commutator subgroup.
For instance, when $G=S_3$ it has subgroups of order $2$ and $3$, but your argument does not apply since none of the subgroups of order $2$ are normal.
So in your particular problem with $|G|=143$, you need to give some additional argument that the subgroups of order $11$ and $13$ must be normal for your proof to be complete.