Normal Sylow p-Subgroups in Group of Order 12
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Let $G$ be of order $12$.
Then $G$ has either:
Suppose there are $4$ Sylow $3$-subgroups $P_1$, $P_2$, $P_3$ and $P_4$.
- $P_i \cap P_j = \set e$
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 $4$.
Hence the result.