# All groups of order 99 are abelian.

by JFo
Tags: abelian, groups, order
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 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
 Sci Advisor HW Helper P: 9,396 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!
 Sci Advisor HW Helper P: 9,396 All groups of order 99 are abelian. When ever you're given some group of order a small product of 2 primes (or possibly three primes) it is almost always going to be the case that looking at Sylow subgroups will help. Eg show that every group of order pqr where p
 P: 92 Thanks, thats useful advice. Now that I think of it though, the proof above doesn't rely on the fact that S_3 and S_11 are Sylow p-subgroups. I could have just stated that there exist subgroups of orders 9 and 11 in G (since 9 and 11 are powers of a prime dividing |G|), and the same results would follow. Is there another proof you had in mind when you gave that hint?
 Sci Advisor HW Helper P: 9,396 No, you can't just conclude that. You need to know that both subgroups are normal, and you do that because you can count the number of conjugates of them using Sylow's theorems. Consider a group of order 6. It has a subgroup of order 2 and a subgroup of order 3 (whcih is normal) but it is not necessarily isomorphic to C_3xC_2. And how are you going to state that there is a subgroup of order 9 in G if you aren't going to use the fact that it is a sylow subgroup?
 P: 92 Oh, your right. I got two theorems mixed up. Thanks again for your help
 HW Helper P: 2,566 Not really relevant, but I always hated the wording of the theorem about G/Z(G) being cyclic implying G is abelian. If G is abelian, Z(G)=G and G/Z(G) = 1. So a much better way to state this theorem is that G/Z(G) may not be a cyclic group of order greater than one.

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