Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $|\ker(\phi)|=p^3$.

share|cite|improve this question
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. – Babak 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? – Babak S. Aug 24 '12 at 10:33
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
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.

share|cite|improve this answer
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
@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? – Babak 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

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.

share|cite|improve this answer
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 :) – Babak S. Aug 24 '12 at 13:37
I like this! +1 – amWhy Mar 13 '13 at 1:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.