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 $p\in\mathbb P$?
 A: 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: 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 $\frac{\mathbb{Z}}{p\mathbb{Z}}$. 
