# All groups of order 99 are abelian.

by JFo
Tags: abelian, groups, order
 P: 92 Prove all groups of order 99 are abelian: I'm stuck right now on this proof, here's what I have so far. proof: Let G be a group such that |G| = 99, and let Z(G) be the center of G. Z(G) is a normal subgroup of G and |Z(G)| must be 1,3,9,11,33, or 99. Throughout I will make repeated use of the theorem which states if the factor group G/Z(G) is cyclic, then G is abelian. Case 1: Assume |Z(G)| = 99, then Z(G) = G, and G is abelian. Case 2: Assume |Z(G)| = 33, then |G/Z(G)| = 3, a prime, so G/Z(G) is cyclic, and thus G is abelian. Case 3: Assume |Z(G)| = 9, then |G/Z(G)| = 11, a prime, so G/Z(G) is cyclic and G is abelian. Case 4: Assume |Z(G)| = 3, then |G/Z(G)| = 33 which factors into (3)(11). There is a theorem which states that if a group is order of a product of two distinct primes p,q with p
 HW Helper Sci Advisor P: 9,395 How many Sylow-3 subgroups does a group of order 99 have? How many Sylow-11 subgroups?
 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!
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