# Groups with order $p^3$ ($p$ prime) have two non commutative isomorphism classes

I read in an exercise, that a group with $p^3$ ($p$ prime) elements have $2$ non commutative isomorphism classes. Unfortunately there was just this statement without any explanation. We just solved it for the commutative case.

Hope someone could explain, why there are just two classes.

• Have you tried anything yourself or used a search engine? – blue Aug 9 '14 at 23:20
• @blue: My only idea was, that every such group have to be $|Z(G)|=p$ and that if two groups are isomorphic, then also the centre. But that helped not much. We didn't have any result in the lecture to classify non commutativ groups. – DerJFK Aug 9 '14 at 23:26
• I suggest you read math.uconn.edu/~kconrad/blurbs/grouptheory/groupsp3.pdf and then write up an answer based on what you find, and post it as an answer. – Gerry Myerson Aug 10 '14 at 0:19
• @Gerry Myerson:Thank you, I will try :) – DerJFK Aug 10 '14 at 6:39

The document groupsp3.pdf shows a detailed answer of all isomorphy classes of groups with order $p^3$ and $p$ prime.

There are $5$ classes: 3 abelian and 2 non-abelian

The abelian ones are easy to see: $\mathbb{Z}/p^3\mathbb{Z},\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$ and $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p^2\mathbb{Z}$

For the non-abelian we have to differentiate between $p=2$ and $p\not=2$.

Case $p=2$: The only non-abelian groups are $Q$ the Quaternion group and $D_4$ the Dihedral group.

Case $p\not=2$: It turns out that the following (up to isomorphy) groups are the only non-abelian groups with order $p^3$.

(1.) Heis$(\mathbb{Z}/p\mathbb{Z}) = \left\{ \left. \begin{pmatrix} 1 & a & b \\ 0 & 1 &c \\ 0 & 0 & 1 \end{pmatrix} \right| a,b,c \in \mathbb{Z} / p \mathbb{Z} \right\}$

(2.) $G_p =\left\{ \left. \begin{pmatrix} 1+mp & a \\ 0 & 1 \end{pmatrix} \right| a,m \in \mathbb{Z} / p^2 \mathbb{Z} \right\}$

If $p=2$ the groups (1.) and (2.) are isomorphic.

Now, we see above all $5$ isomorphy classes for an arbitrary prime $p$ and group with $|G|=p^3$.