Let $G$ to be group of order 99 and has a normal subgroup of order 9. prove that $G$ is abelian and find how many non isomorphic groups are there

Question

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.

I will use that if $G/Z(G)$ is cyclic, then $G$ is abelian.

I know that I have a normal subgroup $|N|=9 $

if I prove that $Z(G)=N$ then $|G /Z(G)| = 99/9 = 11 $

11 is prime so $G/Z(G)$ is cyclic, then $G$ is abelian

should I prove that $Z(G)=N$ or can I assume that? if yes, how to find the non isomorphic groups? and if I can't assume that $Z(G)=N$, how can I prove that ?

Answer 1

Hint: $G/C_G(N)$ is a subgroup of $\text Aut(N). N $ is abelain and $|\text Aut(N)| = 2\cdot3$ or $2^4 \cdot 3.$ Also $|G/C_G(N)|$ divides $11.$ This shows that $G = C_G(N).$ Hence $N \subseteq Z(G).$ The number of abelian group of order $99$ can be computed using Structure Theorem of Finitely Generated Abelian Groups.

Answer 2

Since you've already studied Sylow theorems I think it can be a fast proof. Let us denote by $\;n_p\;$ the number of Sylow $\;p$- subgroups, so in this case we must have

$$\begin{align}&n_{11}=1\;,\;\;\text{since this is the only natural that both divides}\;3^2=9\;\text{and is } =1\pmod{11}\\ &n_3=1\;,\;\;\text{since this is the only natural that both divides}\;11\;\text{ and is} =1\pmod 3\end{align}$$

thus, both Sylow subgroups of $\;G\;$ ( say $\;P_{11}\,,\,\,P_3\;$) are normal, abelian (why?) and also

$$P_{11}\cap P_3=\{1\}\;,\;\;G=P_{11}P_3$$

As you can see, we don't need to be given that $\;P_3\;$ is normal: it follows at once from the Sylow theorems.

From the above, it follows these Sylow subgroups generate their direct product and thus

$$G=P_{11}P_3\cong P_{11}\times P_3$$