3
$\begingroup$

I am a beginner in group theory and I'm looking for finite groups that satisfy some properties.

The only example I've found so far is:

$$G_{q,c} = \{ f: \mathbb{F}_{q} \to \mathbb{F}_{q}, z \mapsto az+b | b \in \mathbb{F}_{q}, a \in (\mathbb{F}_{q}^{\times})^c \}$$ For $c$ such that $\frac{q-1}{c}$ is odd.

The properties are:

  1. $G$ has odd order and is not abelian. (The most important requirement)
  2. Let $p$ be the smallest prime dividor of $|G|$. The $p$-Sylow groups of $G$ must be cyclic.

As so many famous non-abelian groups are of even order, I find it hard to find such groups. I would be glad to see some more examples.

EDIT: I've just realized that since every group of prime order is cyclic, the 2nd property is satisfied once $p$ divides $|G|$ exactly once.

$\endgroup$
1
$\begingroup$

One thing that you may have overlooked is that for any prime $q$ there is an enormous amount of non-abelian $q$-groups, so take one, call it $Q$, and all direct products $\mathbf C_{p^i} \times Q$ will qualify provides $p<q$.

In fact this is just one instance of the following generalization of your construction.

Let's say a group satisfies $\star_p$ if it satisfies your conditions with the exception of being non-abelian.

  • Take any group $G$ such that the least prime dividing its order is $q>p$.
  • Let $A$ be any group satisfying $(\star_p)$.
  • Take any morphism $\theta: A\to\mathrm{aut}(G)$.

Then $G\rtimes_{\theta} A$ satisfies $(\star_p)$. Moreover, if at least one of these conditions is fulfilled then the group will be non abelian and satisfy your conditions.

  • $G$ is non-abelian
  • $A$ is non-abelian
  • $\theta$ is non-trivial.

So this provides an inductive way of creating more and more of this type of groups.

For instance, it's clear that this is $G_{q,c}$ arises by starting with $G$ and $A$ both cyclic. (In fact every $\star_p$-group arises trivally by setting $G=1$ although this is not very interesting.)

Other example: by choosing $\theta$ trivial you find the example that I started this answer with; in particular if $A=G_{q,c}$ you will for instance find examples of the type $G_{p,c} \times G$ where $G$ is not divisible by primes larger than $p$.

More examples can be found by starting with $A=\mathbf C_p^i$ and $G$ any $q$-group for a prime $p>q$, for instance any abelian group having an automorphism of order a power of $p$, or a Heisenberg group (where $\theta$) can be trivial, etc, etc.

$\endgroup$
1
$\begingroup$

Take any two different odd primes $\;p\,,\,q\;$ . s.t. $\;p>q\;,\;\;q\mid(p-1)\;$ . Take two cyclic groups $\;C_p=\langle\,y\,\rangle\,,\,\,C_q=\langle\,x\,\rangle\;$ of order $\;p\,,\,\,q\;$ , resp.

Then you can build a(n exterior) semidirect product $\;C_q\rtimes C_p\;$ by means of the homomorphism

$$\;f: C_q\to \text{Aut}\,(C_p)\;,\;\;f(x):=\phi_q$$

where $\;\phi_q\;$ is the automorphism of order $\;q\;$ of $\;C_p\;$

You get a non-abelian group of odd order and such that all its Sylow subgroups are cyclic (and even of prime order)

$\endgroup$
  • 1
    $\begingroup$ Isn't that isomorphic to the example the OP gave? $\endgroup$ – Myself Dec 20 '14 at 13:24
  • $\begingroup$ @Myself I can't say, yet I don't see prime but $\;q\;$ in the OP's example $\endgroup$ – Timbuc Dec 20 '14 at 13:49
  • $\begingroup$ @Myself Yes, those examples are the same. This is $G_{p,\frac{p-1}{q}}$. Can be seen by writing the group operation explicitly. In particular, this means we can't get more examples by taking non-abelian groups whose order is a product of 2 odd primes. $\endgroup$ – Ofir Dec 20 '14 at 13:50
  • $\begingroup$ Shift then to $\;C_{p^2}\;$ or to $\;C_p\times C_p\;$ in the above , and now only the Sylow subgroups of the smallest prime,. $\;q\;$ , are cyclic for sure.... $\endgroup$ – Timbuc Dec 20 '14 at 13:55
  • $\begingroup$ @Ofir In your example, now that I look at it more slowly, I don't understand what do you mean by $\;\left(\Bbb F_q^*\right)^c\; $ ? Is $\;a\;$ then a $\;c$-dimensional vector over $\;\Bbb F_q\;$ ? Then how $\;f\;$ is a function with domain and image in $\;\Bbb F_q\;$ ? $\endgroup$ – Timbuc Dec 20 '14 at 13:59

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.