# Solvable group of certain order

‎We ‎know ‎that ‎in a non-abelian ‎group ‎of ‎order ‎‎$p^2q‎$ ($‎‎p$ ‎and ‎$q‎‎$‎ ‎are distinct primes)‎‎, ‎ if ‎$p>q‎‎$ and it's Sylow $p$-subgroup is elementary abelian $p$-group‎‎, ‎then ‎one ‎of ‎the ‎presentations ‎for ‎‎$G‎‎$‎‎ is ‎$$‎\langle a , b ‎,c \mid a^p=b^p=c^q=1, ab=ba, cac^{-1}=a^{i}b^{j}, cbc^{-1}=a^{k}b^{l}‎‎‎‎\rangle‎‎‎,$$ where ‎‎‎$$\begin{pmatrix}‎ i & j\\‎ k & l \end{pmatrix}$$‎has order $q$ in ‎‎$\operatorname{GL}(2,p)‎‎$‎. ‎ ‎(According to Burnside's classification in his book "Theory of groups of finite order")‎. ‎ Is it true that this group has NO subgroup of order ‎$‎‎pq$‎?‎‎ ‎‎‎‎

Might well have one. Consider the case where $p$ is any odd prime, $q = 2$, and $$\begin{pmatrix} i & j \\‎ k & l \end{pmatrix} = \begin{pmatrix} -1 & 0 \\‎ 0 & -1 \end{pmatrix}.$$

Here any $\langle a^{u} b^{v}, c \rangle$ has order $pq$, for $a^{u} b^{v} \ne 1$.

In general, if $$M = \begin{pmatrix} i & j \\‎ k & l \end{pmatrix}$$ has eigenvalues in $\mathbb{F}_{p}$, there is at least one subgroup of order $p q$, generated by $c$ and an eigenvector. (Actually, there are at least two such subgroups, see below.) Conversely, let $H$ be a subgroup of order $pq$, then clearly $H \cap \langle a, b \rangle$ has order $p$, and it is a normal subgroup of $H$, so if $H \cap \langle a, b \rangle = \langle a^{u} b^{v} \rangle$, then
$$c (a^{u} b^{v}) c^{-1}= (a^{u} b^{v})^{\lambda}$$ for some $\lambda \ne 0$, so the vector $(u, v)^{t}$ is an eigenvector with respect to the eigenvalue $\lambda \in \mathbb{F}_{p}$.

Now for $M$ to have eigenvalues in $\mathbb{F}_{p}$, you need $q$ to divide $p-1$. First note that if $M$ has only the eigenvalue $1$, then $M^{p} = I$, against the assumptions. If $\lambda \in \mathbb{F}_{p}^{\star}$ is an eigenvalue of $M$, it satisfies both $\lambda^{q} = 1$ and $\lambda^{p-1} = 1$. If $\lambda \ne 1$, we have $\gcd(q, p-1) \ne 1$, and thus $q \mid p - 1$.

So if $q \nmid p-1$, there is no subgroup of order $pq$. If $q \mid p-1$, then $M^{p-1} = I$, the identity, so the minimal polynomial of $M$ divides $x^{p-1} - 1$, which has $p-1$ distinct roots in $\mathbb{F}_{p}$, so that $M$ is diagonalisable. It follows that

there is a subgroup of order $pq$ in $G$ iff $q \mid p - 1$.

• Dear professor Andreas Caranti, many thanks for your exact and invaluable answer! – shankfei Oct 1 '13 at 8:00
• You're welcome. It can surely be streamlined a bit, but it should do. – Andreas Caranti Oct 1 '13 at 8:26
• Is it true that there is a correspondence between the subgroups of order $pq$ of the group $G$ and the eigenspaces of the matrix $M$? Thanks in advance. – shankfei Oct 11 '13 at 21:53