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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?

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How about just looking for a group that is not Abelian; every cyclic group is Abelian. For n=6,24, use, e.g., the symmetric groups $S_3$ and $S_4$ – FBD Aug 12 '13 at 7:20
Just take direct products of cyclic groups whose orders not relatively prime. For instance, with 24, take $\mathbb{Z}_6\times \mathbb{Z}_4$ – Prahlad Vaidyanathan Aug 12 '13 at 7:22

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.

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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.

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See cyclic number for a criterion describing which integers allow only cyclic groups. – Jeppe Stig Nielsen Aug 12 '13 at 9:00

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