# Classification of all finite elementary $p$-groups.

Let $G$ be a finite group. For a prime number $p$, let us call $G$ an elementary $p$-group iff $\exp G=p$. I know that all elementary $2$-groups are abelian, and I also know the construction of non-abelian elementary $p$-groups of order $p^3$ for every odd prime number $p$. My question is that, can we list all the elementary $p$-groups for each $p\in\mathbb P$?

• I have never heard that definition of elementary $p$-group before. There is another more specialized definition, used by Brauer, meaning a direct product of any $p$-group with a cyclic group of order coprime rto $p$. Anyway, the answer to the question is no we can't. I don't ttink it has been done for order $p^8$. – Derek Holt Jun 4 '15 at 8:04
• @DerekHolt It surprised me since I thought such a classification is not so hard to reach. Anyway thank you for pointing out this. – user244919 Jun 4 '15 at 20:48

Consider the group $U(n,\mathbb{Z}_p)$ consisting of $n\times n$ upper triangular matrices over the filed $\mathbb{Z}_p$ of order $p$, in which diagonal entries are $1$. For simplicity, consider $n\leq p$, which forces that $U(n,\mathbb{Z}_p)$ is a $p$-group of exponent $p$. I think the determination of subgroups of this group is still open, so the answer to your question could be "NO".

• A nice construction. BTW, do you know if every non-abelian $p$-group is of order at most $p^{\frac{p(p-1)}2}$ and can always be embedded in $U(p,\mathbb{Z}_p)$? – user244919 Jun 5 '15 at 9:16
• sorry, what is your question here? – Groups Jun 5 '15 at 9:23
• Sorry, I mean every non-abelian elementary $p$-group... – user244919 Jun 5 '15 at 9:24
• I'm sorry, what I intensionally mean is not non-abelian elementary $p$-group, but indecomposable elementary $p$-group... – user244919 Jun 5 '15 at 10:59
• No. Instead of $\mathbb{Z}_p$, if we consider any finite field of prime power order, say $q=p^k$, then we get many elementary $p$-groups, which provide examples in the negative answer of your question. (Please let me know if this statement is not clear to you). – Groups Jun 5 '15 at 14:13

It may not be a complete answer but those group can be constructed inductively using semi-direct product (hence they cannot be too complicated). Take $G$ to be an elementary $p$-group. Take any maximal group $M$ in $G$ then we know that $M$ cannot but be normal and $[G:M]=p$. From this it follows that $G/M$ is a group of order $p$ hence it is cyclic of order $p$. Furthermore it is clear that $M$ must be an elementary $p$-group as well. Hence we have decomposed $G$ in an extension :

$$1\rightarrow M\rightarrow G\rightarrow \frac{\mathbb{Z}}{p\mathbb{Z}}\rightarrow 1$$

Now I claim that the extension is splitted, take $1$ to be a generator of $\frac{\mathbb{Z}}{p\mathbb{Z}}$ and $g_0\in G$ such that $g_0M=1$ then I claim that the function :

$$s:\frac{\mathbb{Z}}{p\mathbb{Z}}\rightarrow G$$

$$k\mapsto g_0^k$$

This is well defined because $g_0$ is of order $p$. I claim that this function is a group morphism which splits the extension above. Hence :

$$G\text{ is isomorphic to }M\rtimes s(\frac{\mathbb{Z}}{p\mathbb{Z}})$$

Hence any elementary $p$-group of order $p^n$ is a semi direct product of an elementary $p$-group of order $p^{n-1}$ and $\frac{\mathbb{Z}}{p\mathbb{Z}}$.

• Any element of $G\setminus M$ generates a complement of $M$, as its order is $p$ by assumption. – j.p. Jun 4 '15 at 7:07
• I wouldn't expect the groups to be uncomplicated just because they are semidirect products. Every extension embeds into a wreath product, which is a (special) semidirect product. – j.p. Jun 4 '15 at 7:12
• This decomposition is nice, but this semidirect poduct construction need not provide an elementary $p$-group, since it may well contain higher order element. – user244919 Jun 4 '15 at 7:19
• @j.p. for your first remark, I agree with you, we may not talk about extension and section... For your second, I did not say that it is now a piece of cake to classify them, just a remark... – Clément Guérin Jun 4 '15 at 7:20
• @user244919, I absolutely did not say that any semi-direct product of this form will give elementary $p$-group, I just said that they are all of this form. – Clément Guérin Jun 4 '15 at 7:20