Recall that if there is a unique Sylow $p$-subgroup in a group $GH$, then it is a normal subgroup in $G$.

Proof.

Since $12=2^2\cdot 3$, a Sylow $2$-subgroup of $G$ has order $4$ and a Sylow $3$-subgroup of $G$ has order $3$.
Let $n_p$ be the number of Sylow $p$-subgroups in $G$, where $p=2, 3$.
Recall that if $n_p=1$, then the unique Sylow $p$-subgroup is normal in $G$.

By Sylow’s theorem, we know that $n_2\mid 3$, hence $n_p=1, 3$.
Also by Sylow’s theorem, $n_3 \equiv 1 \pmod{3}$ and $n_3\mid 4$.
It follows that $n_3=1, 4$.

If $n_3=1$, then the unique Sylow $3$-subgroup is a normal subgroup of order $3$.
Suppose that $n_3=4$. Then there are four Sylow $3$-subgroup in $G$.
The order of each Sylow $3$-subgroup is $3$, and the intersection of two distinct Sylow $3$-subgroups intersect trivially (the intersection consists of the identity element) since every nonidentity element has order $3$.
Hence two elements of order $3$ in each Sylow $3$-subgroup are not included in other Sylow $3$-subgroup.

Thus, there are totally $4\cdot 2=8$ elements of order $3$ in $G$.
Since $|G|=12$, there are $12-8=4$ elements of order not equal to $3$.

Since any Sylow $2$-subgroup contains four elements, these elements fill up the remaining elements.
So there is just one Sylow $2$-subgroup, and hence it is a normal subgroup of order $4$.

In either case, the group $G$ has a normal subgroup of order $3$ or $4$.

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Definition.
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Hint.
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Hint.
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Hint.
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