3
$\begingroup$

Let $p<q$ be prime numbers and let $G$ be a group of order $p^2q^2$. I wish to determine up to isomorphism how many groups $G$ are there.

What I know:

The abelian case is very clear.

Moreover, let assume that $pq\neq 6$ then it can be shown that $$G=Q\rtimes P,$$ where $P,Q$ are the corresponding Sylow subgroups. For some $p,q$ the only groups $G$ are abelian, but lets focus on these $p,q$ such that $G$ is non-abelian.

I believe that if $Q$ is cyclic, then for any $P$ (cyclic or of rank $2$) there exist exactly one isomorphism class.

However, in the case where $Q=C_q\times C_q$ I am not sure about the number of isomorphism classes.

Any help will be appreciated.

$\endgroup$
  • $\begingroup$ You are looking for subgroups of ${\rm GL}(2,q)$ isomorphic to $C_p$, $C_{p^2}$ or $C_p \times C_p$. The cases to consider are $p|q-1$, $p^2|q-1$, $p|q+1$ and $p^2|q+1$. Note that $C_p \times C_p \le {\rm GL}(2,q) \Leftrightarrow p|q-1$. $\endgroup$ – Derek Holt Oct 21 '15 at 9:56
  • $\begingroup$ @DerekHolt Thank you, but I wanted to know is for two such isomorphic subgroups of GL$(2,q)$ (lets say $C_p$) whether the action they induce on $Q$ induce isomorphic groups $G$ or not? $\endgroup$ – Ofir Schnabel Oct 21 '15 at 10:01
  • $\begingroup$ $p=2$ may be a bit different, so you need to do that separately. When $p$ is odd and $p|q-1$, then ${\rm GL}(2,q)$ has $2 + (p-1)/2$ conjugacy classes of subgroups of order $p$, and they all give rise to separate nonabelian groups of order $pq^2$. I think in all cases there is a unique conjugacy class of subgroups of ${\rm GL}(2,q)$ of the order concerned. $\endgroup$ – Derek Holt Oct 21 '15 at 15:31
  • $\begingroup$ Sorry, in the last comment, I meant in all other cases there is a unique conjugacy class of subgroups i.e. subgroups $C_{p^2}$ when $p^2|q-1$ or $p^2|q+1$, subgroups of order $p$ when $p|q+1$, and subgroups isomorphic to $C_p^2$ when $p|q-1$. $\endgroup$ – Derek Holt Oct 21 '15 at 16:03
  • $\begingroup$ @DerekHolt thanks a lot, this is what iv'e been looking for, lets see if I got it; Lets take $p=3$ and $q=19$. Then $p^2|q-1$. So up to an isomorphism the non-abelian groups of order $p^2q^2$ are as follows. When $Q$ is cyclic then there are two such groups, one for $P\cong C_p^2$ and one for $P\cong C_{p^2}$. When $Q$ is of rank $2$ we got one group corresponding to $P$ being cyclic and acting without a kernel and $3$ groups which correspond to a $C_p$ action. Similarly for $P$ being of rank $2$. $\endgroup$ – Ofir Schnabel Oct 22 '15 at 8:21
3
$\begingroup$

Let's study the case $p=3$, $q=19$. Let $P \in {\rm Syl}_{19}(Q)$, $Q \in {\rm Syl}_3(G)$. (Sorry, I have managed to swap $P$ and $Q$!)

Case 1. $P,Q$ cyclic. $Q$ can induce an automorphism of order $3$ or $9$ on $Q$, giving $2$ groups.

Case 2. $P$ cyclic, $Q$ non-cyclic. $Q$ must induce automorphism of order $3$ of $P$, giving $1$ group.

Case 3. $P$ non-cyclic, $Q$ cyclic. Let $\omega$ be an element of order $9$ in ${\mathbb F}_{19}^*$; for example $\lambda=4$.

a) If $Q$ induces automorphism of order $3$ of $P$, then there are $3$ groups, in which the eigenvalues of the action of $P$ on $Q$ are respectively $(1, \omega^3)$, $(\omega^3,\omega^3)$, and $(\omega^3,\omega^6)$.

b) If $Q$ induces automorphism of order $9$ of $P$, then there are $7$ groups, in which the eigenvalues of the action of $P$ on $Q$ are respectively $(1, \omega)$, $(\omega,\omega)$, $(\omega,\omega^2)$, $(\omega,\omega^3)$. $(\omega,\omega^4)$, $(\omega,\omega^6)$, $(\omega,\omega^8)$. (Note that $(\omega,\omega^5)$ would give a group isomorphic to $(\omega,\omega^2)$ and $(\omega,\omega^7)$ isomorphic to $(\omega,\omega^4)$.)

Case 4. $Q$ and $P$ both non-cyclic.

a) If $Q$ induces automorphism of order $3$ of $P$, then there are $3$ groups, just as in Casse 3 a).

b) If $Q$ acts faithfully on $P$, then there is a unique group.

So we get $17$ nonabelian groups altogether which, together with the $4$ abeliabn groups, makes $21$ groups of this order. This agrees with the number given by GAP.

$\endgroup$
  • $\begingroup$ Thanks a lot, just a comment and a question. When you write "in which the eigenvalues of the action of $P$ on $Q$ are respectively", you mean the action of $Q$ on $P$ I believe. And it seems that you assume that the $Q$ action (or the matrix correspond to in Aut$(P)$) is always diagonalizable, in other words the $Q$ action acts on the different copies of $C_p\times C_p$ without mixing them. why we can assume that? $\endgroup$ – Ofir Schnabel Oct 23 '15 at 8:18
  • $\begingroup$ Yes, sorry I kept $P$ and $Q$ confused! The diagonalizability of the action of the $3$-group on the $19$-group follows from Maschke's Theorem in Group Representation Theory. Think of it as a $2$-dimensional representation of $Q$ on the vector space of dimension $2$ over ${\mathbb F}_{19}$. $\endgroup$ – Derek Holt Oct 23 '15 at 14:29
  • $\begingroup$ Thanks again, so what is the condition on $p,q$ in order that any action is diagonalizable? Clearly this is not for any $p,q$, for example the action of $C_3\times C_3$ (with kernel isomorphic to $C_3$) on $C_2\times C_2$ by permutation of order $3$ of the elements of order $2$, or the action of $C_7\times C_7$ on $C_{13}\times C_{13}$ are not diagnosable (I think). So I believe the condition is that $q$ is a divisor of $p-1$, is that right? $\endgroup$ – Ofir Schnabel Oct 26 '15 at 8:36

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.