Give an example of a group $G$ where $n|ord(G)$ but there is no $g\in G$ such that $ord(g)=n$

Question

I thought about Some kind of: $D_n\times \Bbb{Z}_p$ but it does not work. I don't know where to start looking for it, I mean I know where to do not look for it, cannot be cyclic ;), now I can see that the group of symmetries will not work as well. But I am guessing and checking. I think this is not the right approach

Answer 1

Try $S_{3}$ with $n=6$, where $S_{3}$ denotes the symmetric group on $3$ symbols

Answer 2

Take $\mathbb{Z}_2\times\mathbb{Z}_2$. It has no element of order $4$.

Answer 3

A somewhat trivial set of examples: $|G|$ divides $|G|$, so take any group which is not cyclic and it'll have no element of order $|G|$.

For a less trivial example, look at $G = \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. Every non-identity element has order $2$, but $|G| = 8$.

Answer 4

$S_4$ has order $24$ but no elements of order $6, 8, 12, 24$.

Answer 5

Perhaps the following is worth mentioning. If $n$ is prime, then Cauchy's theorem tells you that a finite group of order divisible by $n$ has an element of order $n$.

We will show

Suppose $n$ is not prime. Then there is a group $G$ of order divisible by $n$, with no elements of order $n$.

If $n = p^{k}$ is a power of the prime $p$, with $k > 1$, then the direct product of $k$ groups of order $p$ will have order $n$, but all non-trivial elements will have order $p < n$.

If $n$ is not a power of a prime, but it is of the form $p^{k} m$, where $p$ is a prime, $k >1$, and $\gcd(p, m) = 1$, then the argument just used applies to show that there is a group of order $n$ with no elements of order $n$. First build a group $H$ of order $p^{k}$ with no elements of order $p^{k}$, and then take $G$ to be the direct product of $H$ by the cyclic group of order $m$.

If $n = p_{1} \cdots p_{k}$ is a product of $k > 1$ distinct primes, with $p_{1} < \dots < p_{k}$, consider the symmetric group $H$ on $p_{k}$ letters. It does not have an element of order $p_{1} p_{k}$, because if $a$ where such an element, then $a^{p_{k}}$ and $a^{p_{1}}$ would be commuting elements of order resp. $p_{1}$ and $p_{k}$. But an element of order $p_{k}$ is a $p_{k}$-cycle, and only commutes with its powers. Now take the direct product $G$ of $H$ by a cyclic group of order $p_{2} \cdots p_{k-1}$.

Answer 6

Maybe needless to say: if $G$ is a finite group then the following are equivalent.

(a) For all $d \mid |G|$ there is a $g \in G$ with $ord(g)=d$.
(b) $G$ is cyclic.

The proof is almost trivial. Hence, any finite non-cyclic group provides an example to the question.