Noncyclic group of order $3^n$, $n>0$, must have $6$ normal subgroups

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?

• The Sylow theorems don't tell you there are normal subgroups of order $3^{n-1}$. They tell you there are subgroups of order $3^{n-1}$, and you can deduce they must be normal in a variety of ways (e.g., the index is the smallest prime dividing the order of the group; or normalizer is strictly larger because we have a $p$-group), but Sylow by itself does not prove normality. Aug 29 '21 at 2:23
• Here is a solution with no case subdivision: $\{ 1 \}$, $G$, $\Phi(G)$ and (since $G/\Phi(G)$ is elementary abelian of order at least $9$), at least three normal subgroups strictly between $\Phi(G)$ and $G$. Aug 29 '21 at 8:14
• @DerekHolt: You need the four subgroups strictly between $\Phi(G)$ and $G$, if you want to avoid having to deal with the case of $G$ itself elementary abelian separately. Aug 29 '21 at 18:56
• @ArturoMagidin Yes good point! Aug 29 '21 at 19:55

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 $$k\lt n$$.