6
$\begingroup$

Let $G$ be a minimal non-abelian group with cyclic Sylow $p$-subgroup $P$ and normal Sylow $q$-subgroup $Q$ , see [ Huppert, Endlich Gruppen I, Aufgaben III, 5.14].

My quesion is, if there is another minimal non-abelian group $G_1$ which has the same order with $G$, then can we get $G\cong G_1$ ?

Thank you.

$\endgroup$
6
$\begingroup$

The answer is no, but there are at most two isomorphism types of minimal nonabelian groups of a given order that is divisible by $2$ distinct primes. (There are other examples that are $p$-groups such as $D_8$ and $Q_8$, but we are not talking about those here.)

Suppose $G$ is minimal nonabelian with $|G|=p^mq^n$, where $G$ has a cyclic Sylow $p$-subgroup $P$ and a normal Sylow $q$-subgroup $Q$. Then from the result that you quote, $Q$ is elementary abelian, and $P = \langle g \rangle$ is cyclic, with $g^p$ centralizing $Q$. Also $g$ acts irreducibly on $Q$, and so $p|(q^n-1)$ but $p$ does not divide $q^k-1$ for any $k<n$. So the Sylow $p$-subgroups of ${\rm GL}(n,q)$ are cyclic, and any two subgroups of ${\rm GL}(n,q)$ of order $p$ are conjugate. This implies that the isomorphism type of the semidirect product $Q \rtimes P$ is uniquely determined.

However, for some such orders we could reverse the roles of $p$ and $q$, giving two minimal nonabelian groups of that order. This happens for example with $|G| = 2^23$ or $|G|=3^45^2$.

$\endgroup$

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.