Odd order n smaller than 27

Question

I have a group $G$ that is a group of matrices of the form

$$\left( \begin{array}{ccc} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{array} \right)$$

where $a,b,c \in \Bbb Z_3$.

I have shown it is of order $27$ and non-abelian. How do I show that there's only one non-abelian group of order $n$ less than $27$, and how do I find $n$?

Answer 1

First, enumerate the odd numbers beneath $27$. You'll notice something: everything is either a prime $p$, of the form $p^2$ for a prime $p$, or of the form $pq$ for primes $p<q$. You should prove and then use the following theorems.

• A group of order $p$ prime is cyclic, hence abelian.
• A group of order $p^2$, $p$ prime, is abelian. (There are also precisely two of them). To prove this, use that $p$-groups have nontrivial center; then think about $G/Z(G)$.
• If $p<q$ are prime and $p \mid q-1$, then there are precisely two groups of order $pq$, one of them non-abelian. If $p<q$ are prime and $p \nmid q-1$, then there is only one group of order $pq$, and it is abelian. To prove this, think about semidirect products.