Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ is a normal subgroup of $G$.
Then use Sylow’s theorem show that the number of $p$-Sylow subgroups is $1$ for $p=3, 11$.

Proof.

Since the order of $G$ is $33=3\cdot 11$, we consider $p$-Sylow subgroups of $G$ for $p=3, 11$.
Let $n_p$ be the number of $p$-Sylow subgroups of $G$. It suffices to show that $n_p=1$ for $p=3, 11$. In fact, by Sylow’s theorem any two $p$-Sylow subgroups are conjugate. If there is only one $p$-Sylow subgroup $P$, then $g^{-1}Pg=P$ for any $g\in G$, hence $P$ is normal.

By Sylow’s theorem, the number of $p$-Sylow subgroups of $G$ satisfies
\[n_p \equiv 1 \pmod p\]
and $n_3|11$ and $n_{11}|3$.

From these two constraints, we see that we must have $n_3=n_{11}=1$.
Therefore, for each $p$, there is a unique $p$-Sylow subgroup of $G$.

By the consideration above, we conclude that any $p$-Sylow subgroup of $G$ is a normal subgroup of $G$.

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