If $G$ is a non-cyclic group of order $3^n,\ n>0$, I want to show it has at least $6$ normal subgroups.
My attempt: If $G$ is abelian and non-cyclic, by Fundamental Theorem of Finite Abelian Group, it contains $\mathbb{Z}_3 \times \mathbb{Z}_3$ as a subgroup. Since $\mathbb{Z}_3 \times \mathbb{Z}_3$ has exactly $6$ subgroups, they give the $6$ normal subgroups required.
If $G$ is nonabelian, we have $3$ normal subgroups easily: $\{e\}, G,$ and the center $Z$. Also, $|Z|\ne 3^{n-1}$, otherwise, $G/Z$ would be cyclic and $G$ would be abelian. By Sylow Theorems, we also have a normal subgroup of order $3^{n-1}$. How can I find $2$ more normal subgroups?
Best Answer
Note that $G/Z(G)$ cannot be cyclic. If $G/Z(G)$ is abelian, then it contains a subgroup isomorphic to $C_3\times C_3$ and all its subgroups are normal, and you can lift them back to $G$ to get six distinct normal subgroups (all of which contain $Z(G)$).
If $G/Z(G)$ is nonabelian, then you can apply an inductive argument, since it is a noncyclic group of order $3^k$ with $k\lt n$.