group of order $p^n$

Question

If $p$ is any prime number and $G$ is a group of order $p^2$, then $G$ is abelian, and either $G\cong \mathbb Z_{p^2}$ or $G\cong \mathbb Z_p \times \mathbb Z_p$.

How about order $p^n$ where $n$ is any integer? Is there similar conclusion, i.e. If $p$ is any prime number, $n$ is an integer, and $G$ is a group of order $p^n$, then $G$ is abelian; either $G\cong \mathbb Z_{p^n}$ or $G=H \times K$, where $H$ and $K$ are subgroups of $G$, $H$ is a group of order $p^i$, $K$ is a group of order $p^j$ and $i+j=n$?

Answer 1

The classification of $p$-groups is considered a very hard problem. “Most” groups are $p$-groups, in a precise sense that would take too long to explain (in fact, it is conjectured that if you take the number of isomorphism classes of order $2^k\leq n$, and divide by the number of isomorphism classes of groups of order at most $n$, the limit as $n\to\infty$ will be $1$.)

A $p$-group must have nontrivial center. If $G$ is a $p$-group, we let $Z_0(G)=\{e\}$, and let $Z_{i+1}(G)$ be the subgroup of $G$ such that $Z_{i+1}(G)/Z_i(G)$ is the center of $G/Z_i(G)$. The least value of $c$ such that $Z_c(G)=G$ is called the “class of $G$” (so abelian groups are class $1$, center-by-abelian are class $2$, etc).

The coclass of a group of order $p^n$ is $n-c$. The analysis of $p$-groups by co-class is relatively recent, as these things go. They also informed a series of important conjectures about $p$-groups that, in a way, brought order out of the chaos.

The Co-class Theorems that give some unifying structure. The strongest one is:

Theorem. There is a function $f(p,r)$ such that every finite $p$-group of coclass $r$ has a normal subgroup $K$ of class at most $2$ and index at most $f(p,r)$. If $p=2$, one may require $K$ to be abelian.

There is some work on the finer structure, but this gives you an idea of how far the state of the art is from something like what you were hoping for.