Is there a way to construct non-cyclic groups of any order?

Question

I know that it is possible to construct non-cyclic of small orders such as 4: $\mathbb{Z_2} \times \mathbb{Z_2}$ but how would I construct non-cyclic groups of orders 6 and 24 without needing to look them up?

Answer 1

This is not always possible. For example, every group of prime order is cyclic.

More generally, it is possible to prove that every group of order $n$ is cyclic if and only if $n$ has prime factorization

$$n = p_1p_2 \ldots p_t$$

where $p_i$ are distinct and $p_i \not\equiv 1 \mod{p_j}$ for each $i, j = 1, \ldots, t$.

So if a noncyclic group of order $n$ exists, one of the following (or both) happens.

• $n$ is divisible by $p^2$ for some prime $p$. In this case $\mathbb{Z}_p \times \mathbb{Z}_p$ is noncyclic, and thus $\mathbb{Z}_p \times \mathbb{Z}_p \times \mathbb{Z}_{n/p^2}$ is noncyclic of order $n$.

• $n$ is divisible by primes $p$ and $q$ such that $p \equiv 1 \mod{q}$. In this case there exists a nonabelian group $T$ of order $pq$, so $T \times \mathbb{Z}_{n/pq}$ is noncyclic of order $n$.

For references see here and here.

Answer 2

Hint: Consider semi-direct products. For example, with $6$, you know that $\text{Aut}(\mathbb{Z}_3)$ has order $2$, and so there is a non-trivial homomorphism $\varphi:\mathbb{Z}_2\to\text{Aut}(\mathbb{Z}_3)$. Similarly, $3\mid |\text{Aut}(\mathbb{Z}_2^3)|$. Note then that a semi-direct product is abelian, if and only if its actually a direct product.

PS: The answer to your question is no. See my answer here for a classification of the integers $n$ such that the only group of order $n$ is cyclic.

As an example, if $p$ an $q$ are primes, $p<q$ such that $q\not\equiv 1 \mod p$ then any group of order $pq$ is cyclic. For example, $15=3\cdot 5$ has only the cyclic group.