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All groups of order 99 are abelian. 
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#1
May1906, 03:15 AM

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<q, then G is cyclic if q is not congruent to 1 modulo p. Since 11 is not congruent to 1 mod 3 G/Z(G) is cyclic, and so G is abelian. OK, here's where I get stuck! Case 5: Assume Z(G) = 11, then G/Z(G) = 9 = 3^2. Since G/Z(G) is order of a prime squared, G/Z(G) is abelian. Thus by the theorem of finitely generated abelian groups, G/Z(G) is either isomorphic to Z_9 or Z_3 x Z_3. If its isomorphic to Z_9, then G/Z(G) is cyclic and were done. But if its isomorphic to Z_3 x Z_3 then I don't know how to proceed. Case 6: Assume Z(G) = 1, the G/Z(G) = 99. I'm not sure how to proceed from here. Any suggestions? 


#2
May1906, 04:09 AM

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P: 9,396

How many Sylow3 subgroups does a group of order 99 have? How many Sylow11 subgroups?



#3
May1906, 05:20 AM

P: 92

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


#4
May1906, 05:35 AM

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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<q<r are primes is solvable.



#5
May1906, 05:44 AM

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 psubgroups. 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? 


#6
May1906, 06:24 AM

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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? 


#7
May1906, 06:36 AM

P: 92

Oh, your right. I got two theorems mixed up. Thanks again for your help



#8
May2006, 06:38 PM

HW Helper
P: 2,567

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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