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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?

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  • $\begingroup$ 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. $\endgroup$ Aug 29 '21 at 2:23
  • $\begingroup$ 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 $G$. $\endgroup$
    – Derek Holt
    Aug 29 '21 at 8:14
  • $\begingroup$ @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. $\endgroup$ Aug 29 '21 at 18:56
  • $\begingroup$ @ArturoMagidin Yes good point! $\endgroup$
    – Derek Holt
    Aug 29 '21 at 19:55
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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 $k\lt n$.

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  • $\begingroup$ Thank you. This is very helpful. $\endgroup$ Aug 29 '21 at 2:48

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