10
$\begingroup$

For which $n$ is there a group of order $n^2$ without a subgroup of order $n$.

Such groups can not be nilpotent.

This question is related to Sudokus as composition tables of finite groups.

$\endgroup$
  • $\begingroup$ Must $n$ be even? $\endgroup$ – i. m. soloveichik Jul 30 '14 at 22:20
  • 2
    $\begingroup$ No, there is an example with $n=5^3 \times 3$ that is similar to the one with $n=24$, but with $|N| = 5^6$. $\endgroup$ – Derek Holt Jul 30 '14 at 22:32
12
$\begingroup$

I have a feeling that there is not much we can say in general, but here is an infinite family of examples.

Let $p$ be an odd prime such that $3$ divides $p+1$ (for example $p = 5$).

One example is $G = A_4 \times ((C_p)^2 \rtimes C_3)$, where the semidirect product is nontrivial. Then $G$ has order $(6p)^2$, but $G$ has no subgroup of order $6p$.

Proof: Let $H$ be a subgroup of order $6p$. Then $H$ has a subgroup of order $2$. Since any such subgroup is contained in the $A_4$ factor, it follows that $H \cap A_4$ has order divisible by $2$. Now $A_4$ has no subgroup of order $6$, so $H \cap A_4$ must have order $2$.

Now note that because $3$ does not divide $p-1$, any group of order $3p$ is cyclic. Thus $H$ contains an element of order $3p$. There is no such element in $(C_p)^2 \rtimes C_3$, so it must be of the form $xy$, where $x \in A_4$ has order $3$ and $y \in (C_p)^2 \rtimes C_3$ has order $p$. But then $x \in H \cap A_4$, a contradiction since $H \cap A_4$ has order $2$.


The above gives examples for $n = 30, 66, 102, 138, \ldots$

With GAP it is possible to check that $n = 28$ also gives examples, for example $\operatorname{SmallGroup}(784, 160)$ and $\operatorname{SmallGroup}(784, 162)$. I am not sure if there are any examples for smaller $n$.

$\endgroup$
  • 1
    $\begingroup$ With GAP I went through groups of order $n^2$, where $n \leq 30$ and $n \neq 24$, and the examples in this answer are the only ones among them (one for $n = 30$ and two for $n = 28$). There are $8681$ groups of order $24^2 = 576$, so going through them manually might take a very long time $\endgroup$ – Mikko Korhonen Jul 30 '14 at 19:19
  • 3
    $\begingroup$ Not so long with a bit of parallel processing, but it would have helped if I had started at the far end. SmallGroup(576,8661) has no subgroup of order $24$. That's the only example, and $24$ is the smallest $n$ with this property. $\endgroup$ – Derek Holt Jul 30 '14 at 19:53
  • $\begingroup$ @DerekHolt: Thanks for saving me the trouble! $\endgroup$ – Mikko Korhonen Jul 30 '14 at 20:06
  • $\begingroup$ I checked n=31...44; all such groups have a subgroup of index n. $\endgroup$ – i. m. soloveichik Jul 31 '14 at 15:54
8
$\begingroup$

Just to expand my comment, the smallest $n$ for which this occurs is $n=24$, and the single example of a group of order $24^2$ with no subgroup of order $24$ is ${\mathtt{SmallGroup}}(576,8661)$, a semidirect product $N \rtimes C_9$, where $N$ is elementary abelian of order $64$, and the $C_9$ acts irreducibly on $N$.

It is a Frobenius group of degree $64$ and can also be accessed as $\mathtt{PrimitiveGroup}(64,1)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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