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 $4$.

Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]

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Hint.
We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.
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Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
Hint.
Use Sylow's theorem. To review Sylow's theorem, check […]

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Prove that every Sylow $11$-subgroup of $G$ is contained in the center $Z(G)$.
Hint.
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Prove that every finite group of order $72$ is not a simple group.
Definition.
A group $G$ is said to be simple if the only normal subgroups of $G$ are the trivial group $\{e\}$ or $G$ itself.
Hint.
Let $G$ be a group of order $72$.
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Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
It follows from […]

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Let $G$ be a group of order $|G|=pqr$, where $p,q,r$ are prime numbers such that $p<q<r$.
Show that $G$ has a normal subgroup of order either $p,q$ or $r$.
Hint.
Show that using Sylow's theorem that $G$ has a normal Sylow subgroup of order either $p,q$, or $r$.
Review […]

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Determine whether a group $G$ of the following order is simple or not.
(a) $|G|=100$.
(b) $|G|=200$.
Hint.
Use Sylow's theorem and determine the number of $5$-Sylow subgroup of the group $G$.
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