# Odd-Order Groups with Cyclic p-Sylow Subgroups (for smallest p | G)

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.

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.

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)

• Isn't that isomorphic to the example the OP gave? – Myself Dec 20 '14 at 13:24
• @Myself I can't say, yet I don't see prime but $\;q\;$ in the OP's example – Timbuc Dec 20 '14 at 13:49
• @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. – Ofir Dec 20 '14 at 13:50
• 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.... – Timbuc Dec 20 '14 at 13:55
• @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 $\;\Bbb F_q\;$ ? – Timbuc Dec 20 '14 at 13:59