# The order of the cyclic subgroups of $GL(3,q)$

Derek Holt gave an counter example of my question because i had forgotten to write this fact that $p,q$ are distinct primes. Thus i repair my question.

Let $G=P \ltimes Q$ be a finite group scuch that $P$ is a maximal subgroup of $G$ which $P$ is non-normal cyclic and $P=C_G(P)=N_G(P)$ and $G'=Q \cong C_q \times C_q \times C_q$, where $p,q$ are distinct primes. . Also assume that $G$ has the properties that if $H,K$ be two distinct non-normal non-cyclic subgroup of $G$, then $H,K$ are conjugate in $G$.

We see that there exists a homomorphism $\varphi : P \longrightarrow Aut(Q)\cong GL(3,q)$. also clearly, $Ker(\varphi)=C_P(Q)=Z(G)$. I want to know why $P \cong C_p$, where $p=q^2+q+1$ is a prime.

• What leads you to believe that this statement is true? – Derek Holt Sep 2 '16 at 12:01
• I am still not clear what $p$ is. I understand that you want to assume that $p$ is prime, but what is $p$ exactly? Are you defining $p = q^2+q+1$? – Derek Holt Sep 2 '16 at 17:24

Here is a counterexample with $q=11$. The group ${\rm GL}(3,11)$ has a unique conjugacy class of cyclic subgroups of order $133 = 7 \times 19$.

Let $Q$ be elementary abelian of order $11^3$, let $P \cong C_p$ with $p=133$, and let $G = P \ltimes Q$ with the action corresponding to this cyclic subgroup of ${\rm GL}(3,q)$. The action of $P$ on $Q$ is irreducible, so $P$ is maximal in $G$ and hence $P=C_G(P)=N_G(P)$. Clearly $G'=Q$.

So we have to check the condition that all non-normal non-cyclic subgroups are conjugate in $G$. Let $H$ be a non-cyclic subgroup of $G$. Since $G/Q$ is cyclic, $H \cap Q \ne 1$.

But all nontrivial subgroups of $P$ (i.e. $P$ itself and its subgroups of orders $7$ and $19$) act irreducibly on $Q$, so either $Q \le H$ or $H \le Q$, If $Q \le H$ then $H \unlhd G$, so we can assume that $H < Q$. Then $H$ non-cyclic implies that $|H|=q^2$.

Now $Q$ has exactly $q^2+q+1=133$ subgroups of order $q^2$. Since $P$ and its nontrivial subgroups act irreducibly on $Q$, they cannot normalize $H$, so these $133$ subgroups are all conjugate under the action of $P$.

Note that the fact that $Q$ has $q^2+q+1$ subgroups of order $q^2$, which must all be conjugate under the action of $P$, means that $|P|$ must be divisible by $q^2+q+1$.

If $|P| > q^2+q+1$ then by the Orbit-Stabilizer Theorem, some nontrivial element $g$ of $P$ must stabilize (i.e. normalize) some subgroup $R$ of order $q^2$. Now $\langle g,R \rangle$ is a non-normal subgroup of $G$ of order greater than $q^2$, so it cannot be conjugate to the subgroups of order $q^2$, contradicting the hypotheses. So $|P|=q^2+q+1$.

• The fact that $133$ is not a prime means that it is a counterexample. You asked for help in proving that $P \cong C_p$ where $p=q^2+q+1$ is prime. I have produced an example where it is not prime, so what you are trying to prove is false. You may be angry with me, but in fact I have a lot of time (probably nearly an hour) trying to help you. – Derek Holt Sep 2 '16 at 14:41
• @Rima don't yell. It's an inappropriate behaviour. – Santiago Sep 2 '16 at 14:49
• Now you have edited the question. If you change a question after it has been answered, then of course the answer may no longer be correct. You have not said what $p$ is. Is $|P|=p$? – Derek Holt Sep 2 '16 at 14:50
• I am apologize. you right. I I've edited my question and put in site again. I'm sorry and thanks for response – Rima Sep 2 '16 at 15:32
• Dear Derek Holt, Forgive me for my inappropriate behavior? – Rima Sep 2 '16 at 16:55