non-abelian groups of order $p^2q^2$.

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.

• 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 $C_p \times C_p \le {\rm GL}(2,q) \Leftrightarrow p|q-1$. – Derek Holt Oct 21 '15 at 9:56
• @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? – Ofir Schnabel Oct 21 '15 at 10:01
• $p=2$ 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. – Derek Holt Oct 21 '15 at 15:31
• 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 $p|q-1$. – Derek Holt Oct 21 '15 at 16:03
• @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 $2$. – Ofir Schnabel Oct 22 '15 at 8:21

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.

• 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? – Ofir Schnabel Oct 23 '15 at 8:18
• 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 ${\mathbb F}_{19}$. – Derek Holt Oct 23 '15 at 14:29
• 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? – Ofir Schnabel Oct 26 '15 at 8:36