Prove all groups of order 99 are abelian:(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: All groups of order 99 are abelian.

**Physics Forums | Science Articles, Homework Help, Discussion**