# All groups of order 99 are abelian.

by JFo
Tags: abelian, groups, order
 P: 92 There is only 1 Sylow 3-subgroup and 1 Sylow 11-subgroup in a group of order 99. Denote these as $S_3$ and $S_{11}$. $|S_3| = 9$ and $|S_{11}| = 11$. $S_{11}$ is cyclic and and every nonzero element of $S_{11}$ is of order 11. This implies that $S_3 \cap S_{11} = \{e\}$. Therfore $|S_3S_{11}| = 99$ so $S_3S_{11} = G$. Also $G \simeq S_3 \times S_{11}$. $S_3$ is order of a prime squared and thus is abelian, and $S_{11}$ is abelian because it is cyclic. Therefore $S_3 \times S_{11}$ is abelian and hence G is abelian. QED Is this right? Wow, thanks much for your help Matt!