Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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

2 Answers 2

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.

share|improve this answer

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.