Is there any way to prove that a group of order 63 has a subgroup of order 21 just using Cauchy's Theorem (i.e., without Sylow)?
We know that there is an element $x$ of order $7$ and an element $y$ of $3$. Can we conclude that $xy$ has order 21, since $7$ and $3$ are coprime?
By Cauchy you have a subgroup $S$ of order 7. Its normalizer $N :=N_G(S)$ has order 7, 21 or 63. If 21, then you are done. If 63, then $N$ is normal, so take the preimage of a subgroup of order 3 of $G/N$, which exists by Cauchy. If 7, then $N$ has (counting itself) 9 conjugates, which intersect trivially, as $N$ has prime order. This gives $54=9\cdot 6$ elements of order 7. Having one element of order 1, this leaves $8=63-54-1$ elements unaccounted for, some of which have order 3 by Cauchy. $N$ acts by conjugation on the set $T$ of elements of $G$ of order 3. Elements of order 3 come in pairs (paired with their inverses) so $T$ has even order at most 8, i.e., 2, 4, 6 or 8. Hence $N$ has a fixed point $t$, which it centralizes (as the action is by conjugation), so $N$ and $t$ generate a subgroup of order 21 (or you observe that the centralizing element contradicts the normalizer having order 7, if you prefer to end the proof this way).