Normal Sylow p-Subgroups in Group of Order 12

Theorem
Let $G$ be of order $12$.

Then $G$ has either:
 * a normal Sylow $2$-subgroup

or:
 * a normal Sylow $3$-subgroup.

Proof
Note that a Sylow $2$-subgroup of $G$ is of order $4$.

From Sylow $3$-Subgroups in Group of Order 12, there are either $1$ or $4$ Sylow $3$-subgroups.

Suppose there is exactly $1$ Sylow $3$-subgroup $P$.

Then from Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.

Suppose there are $4$ Sylow $3$-subgroups $P_1$, $P_2$, $P_3$ and $P_4$.

Each intersection $P_i \cap P_j$ for $i, j \in \set {1, 2, 3, 4}, i \ne j$ is the trivial subgroup of $G$:


 * $P_i \cap P_j = \set e$

Thus $G$ contains:
 * The identity element $e$
 * $8$ elements of order $3$, of which $2$ each are in $P_1$, $P_2$, $P_3$ and $P_4$
 * $3$ more elements, which (along with $e$) must form the Sylow $2$-subgroup of order $4$.

This Sylow $2$-subgroup $Q$ must be unique.

Hence by Sylow $p$-Subgroup is Unique iff Normal, $Q$ is normal.

Hence the result.