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I just need some hints to prove this:

Let $|G|=p^3$ be a a non-abelian group. If every subgroup of $G$ is normal, then $p=2$ and $G=Q_8$.

I know the following facts about a non-abelian group $G$ of order $p^3$:

  1. $Z(G)=G'=Z_p$

  2. If $p$ is an odd number, then the function $\phi:G\longrightarrow Z(G)$ given by $ \phi(g)=g^p$ is a homomorphism and $|\ker(\phi)|=p^2$ or $|\ker(\phi)|=p^3$.

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How can a normal subgroup of order 2 be embedded in a group? –  Jack Schmidt Aug 24 '12 at 9:50
    
If the group be finite, 2 must divide the order of the group. –  B. S. Aug 24 '12 at 9:59
    
You can say more about the embedding. A normal subgroup of order 2 is always contained in the center of the group (finite or not). –  Jack Schmidt Aug 24 '12 at 10:02
    
@JackSchmidt: You are noting that if $G$ has a subgroup of order 2, say $N$, then according to the problem $N$ is normal in $G$ and therefore $N\subseteq Z(G)=Z_p$. This leads to a contradiction unless $p=2$. Am I right, Jack? –  B. S. Aug 24 '12 at 10:33
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Babak, if you can write a proof now, you ought to post it as an answer - and then after it has been up for a while, you can accept it. –  Gerry Myerson Aug 24 '12 at 12:16
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2 Answers

up vote 3 down vote accepted

Suppose $p$ is odd. Consider the kernel of $\phi$ in the second fact. It consists exactly of the elements of order dividing $p$, and so there are either $p^2$ or $p^3$ of these; always more than $p$. However, by the first fact, there are exactly $p$ central elements of order dividing $p$. In particular, for every odd prime $p$ and non-abelian $p$-group $G$, there is a non-central element of order $p$, and the subgroup it generates is not normal (since it is order $p$ and not central by assumption).

Suppose $p=2$. Then there are two very explicit cases, $D_8$ which doesn't work, and $Q_8$ which does.

Groups like this, in which every element of order $p$ are central, have been studied by JG Thompson and others. Maps like $\phi$ always exist, and serve to build the upper exponent-$p$ series of the group. In particular, if $Z(G)$ is cyclic, $p$ is odd, and every element of order $p$ is central, then $G$ itself is cyclic. If $Z(G)$ has rank 2, then the “socle series” of $G$ has factors of rank at most 2.

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Suppose $p$ is odd. Then there are two very explicit examples, $C_{p^2} \rtimes C_p$ and $\begin{pmatrix} 1 & \mathbb F_3 & \mathbb F_3 \\ 0 & 1 & \mathbb F_3 \\ 0 & 0 & 1\end{pmatrix}$. Neither works. –  PseudoNeo Aug 24 '12 at 13:26
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@PseudoNeu: yup, just trying to use the facts given. :-) –  Jack Schmidt Aug 24 '12 at 13:32
    
@JackSchmidt: Would this problem be valid if $G$ be just a non-abelian finite $p$- group? –  B. S. Aug 24 '12 at 13:34
    
The conclusion is nearly true: $Q_8 \times \mathbb{Z}_2 ^n$ is the only extra groups. The argument seems more complicated though, since we no longer control $Z(G)$. My argument heavily relies on "all order p elements are central", and there are non-abelian p-groups where all order p-elements are central, but not all subgroups are normal. These groups are bigger than $p^3$ of course. –  Jack Schmidt Aug 24 '12 at 16:17
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In light of Jack's neat comments:

Corollary (1): Let $G$ be a finite $p$-group and let $N$ be a normal subgroup of $G$. Then if $|N|=p$ then $N\leq Z(G)$.

This is a consequence of a well-known theorem noting if $G$, a finite $p$-group, has a non-trivial normal subgroup $N$ then $N\cap Z(G)\neq1$.

Corollary (2): Let $G$ be a non-abelian group of order $p^3$, where $p$ is an odd prime. Then $G$ has normal subgroups $N$ such that $$Z(G)<N<G$$ and $N\cong\mathbb Z_p\times\mathbb Z_p$.

According to the second corollary, a non-abelian group $G$, of order $p^3$, where $p$ is odd prime, has normal subgroups which are not central. This corollary excludes $Q_8$. Suppose that $1\neq x\in G$ is an element of order $p$ and $N=<x>$. Since we accept that for our group every subgroups are normal in $G$ so we would have a contradiction if $p$ be odd prime.

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Your corollary 2 is true, but in the comments above you said that all subgroups of order $p^2$ were $\mathbb{Z}_p \times \mathbb{Z}_p$. This fails for the first explicit example given by Pseudo-Neu (there is one normal subgroup of order $p^2$ that is cyclic). –  Jack Schmidt Aug 24 '12 at 13:33
    
(So your answer is fine :-) –  Jack Schmidt Aug 24 '12 at 13:33
    
@JackSchmidt: Thanks. You helped me step by step to find the answer :) –  B. S. Aug 24 '12 at 13:37
    
I like this! +1 –  amWhy Mar 13 '13 at 1:01
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