# Abelian groups of order $14,27,30,$ and $21$.

Which of the following statements is false?

1. Any abelian group of order $27$ is cyclic.

2. Any abelian group of order $14$ is cyclic.

3. Any abelian group of order $21$ is cyclic.

4. Any abelian group of order $30$ is cyclic.

For $2$, it has elements of order $2$ and $7$ and hence an element of order $14$. For $3$ it has elements of order $3$ and $7$ and hence $21$. For $4$, it has elements of order $2,3,$ and $5$ and hence an element of order $30$. So, $2,3,$ and $4$ are all cyclic groups. So I guess $1$ is false. Please help.

• Look at $(\Bbb Z/\Bbb 3Z)^3$. Dec 14, 2012 at 5:28
• @Wow 'it has elements of order 2 and 7 and hence an element of order 14' how Dec 1, 2018 at 15:55
• @BrianM.Scott 'it has elements of order 2 and 7 and hence an element of order 14'. Could you please explain it. I'm sorry its a question from 6 years ago but it will be helpful for me.Thanks Dec 1, 2018 at 15:58

The first statement may be false since we can consider the additive abelian group $$\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}.$$ This has 27 elements, but certainly isn't cyclic.
Let $p$ be prime and $V$ be an $n$-dimensional vector space over the field $\mathbb Z/p$. Then $V$ may be regarded as an abelian group whose operation is vector addition.
The order of $V$ is $p^n$ by taking all possible linear combinations of the $n$ basis elements.
For any $x$ in $V$, $\sum_{i=0}^px=px=0$ since $p=0$ in the base field. Thus, when $n>1$ there is no generator of $V$.
This proves there is a noncyclic group of order $p^n$ whenever $p$ is prime and $n>1$.