Note that $20449=11^2 \cdot 13^2$.
Let $G$ be a group of order $20449$.
We prove by Sylow’s theorem that there are a unique Sylow $11$-subgroup and a unique Sylow $13$-subgroup of $G$.
Hence $G$ is the direct product of these Sylow subgroups.

Since these Sylow subgroups are of order $11^2$ and $13^2$, respectively, they are abelian.
Since the direct product of abelian groups is abelian, the group $G$ is abelian.

Proof.

Observe that $20449=11^2 \cdot 13^2$.
Let $G$ be a group of order $20449$.

Let $n_{11}$ be the number of Sylow $11$-subgroups of $G$.
By Sylow’s theorem, $n_{11}$ satisfies
\begin{align*}
&n_{11}\equiv 1 \pmod{11} \text{ and }\\
&n_{11} \text{ divides } 13^2.
\end{align*}

The second condition yields that $n_{11}$ could be $1, 13, 13^2$.
Among these numbers, only $n_{11}=1$ satisfies the first condition.
So there is a unique Sylow $11$-subgroup $P_{11}$ of $G$, hence $P_{11}$ is a normal subgroup of $G$.

Similarly, let $n_{13}$ be the number of Sylow $13$-subgroups of $G$.
Sylow’s theorem yields that $n_{13}$ satisfies:
\begin{align*}
&n_{13}\equiv 1 \pmod{13} \text{ and }\\
&n_{13} \text{ divides } 11^2.
\end{align*}
From the second condition, we see that $n_{13}$ could be $1, 11, 13$.
Among these numbers, only $n_{13}=1$ satisfies the first condition.
So there is a unique Sylow $13$-subgroup $P_{13}$ of $G$, hence $P_{13}$ is a normal subgroup of $G$.

Note that the orders of $P_{11}$ and $P_{13}$ are $11^2$ and $13^2$, respectively.
The intersection of $P_{11}$ and $P_{13}$ is the trivial group.
Thus, we have
\begin{align*}
|G|=\frac{|P_{11}P_{13}|}{|P_{11}\cap P_{13}|}=|P_{11}P_{13}|.
\end{align*}
This yields that $G=P_{11}P_{13}$.

In summary, we have

Sylow subgroups $P_{11}$ and $P_{13}$ are normal in $G$.

$P_{11}\cap P_{13}=\{e\}$.

$G=P_{11}P_{13}$.

These implies that $G$ is the direct product of $P_{11}$ and $P_{13}$:
\[G=P_{11}\times P_{13}.\]

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