Groups and rings of order $p^2$.

Question

Up to isomorphism

1. there are exactly two abelian groups of order $p^2$.

2. there are exactly two groups of order $p^2$.

3. there are exactly two commutative rings of order $p^2$.

4. there is exactly one integral domain of order $p^2$.

   From "fundamental theorem of finite abelian groups", (1) is true, but I have no idea about others. Please help.

Thanks in advance.

Answer 1

Quick and completing comments:

$$\begin{align*}&(1)\;\;\Bbb Z_{p^2}\;,\;\;\Bbb Z_p\times \Bbb Z_p\\&(2)\;\;\text{As before}\\&(3)\;\;\text{As before}\;+\;\Bbb F_{p^2}\\&(4)\;\;\text{Any finite integral domain is a field so only}\;\;\Bbb F_{p^2}\end{align*}$$

Answer 2

Let me give you a sketch for the proves for 2.) and 4.), since 3.) should be clear by the comment of Ferra.

2.) Let $G$ be a nonabelian $p$-group of the order $p^2$. Since $p$-groups have nontrivial center, we get $Z(G)=C_p$. Generally if $G/Z(G)$ is cyclic (like in this case, by Lagrange) then $G$ must be abelian. Thus groups of the order $p^2$ are abelian and we get $C_{p^2}$ and $C_p \times C_p$ as the only possibilities.

4.) Let $R$ be a finite integral domain and $f: R\to R : x\mapsto ax$ be the map of left multiplication by an element $a\in R, a\neq 0$. We have to show that $a$ has an inverse element. For this, show that $f$ is injective (use the property of an integral domain). Since $R$ is finite, $f$ is also surjective, thus bijective. This means every element (besides $0$) has an inverse, which makes $R$ to a field.