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?

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# All groups of order 99 are abelian.

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