On 3-designs from $PGL(2,q)$

$G_{B}$ must be a subgroup from the list of Theorem 2. Moreover, $x\mapsto 1/x$ and $x\mapsto\theta^{r}x$ stabilize $B$ so $\langle x\mapsto 1/x,x\mapsto\theta^{r}x\rangle=D_{2k}\leq G_{B}$. Then $G_{B}$ can not be any of $A_{4}$, $C_{d}$, $Z_{p}^{m}$ or $Z_{p}^{m}\rtimes C_{d}$ because they do not have a subgroup $D_{2k}$ where $k\geq 4$.

Consider the possibility $G_{B}\simeq S_{4}$. The largest dihedral subgroup of $S_{4}$ is $D_{8}$, so $D_{2k}=D_{8}$ and $k=|B|=4$. Since $B$ is a union of some orbits of its stabilizer, from 5 we know that $q$ must be a power of 3. Since $A_{4}\simeq PGL(2,3)$, this situation corresponds to $G_{B}\simeq PGL(2,p^{m})$ that we will consider later.

Next we consider the possibility $G_{B}\simeq A_{5}$. The largest dihedral subgroup of $A_{5}$ is $D_{10}$, so since $D_{2k}\leq G_{B}$, we have $|B|=k\leq 5$. From 6 the orbits of $G_{B}$ have length at least 10, so $|B|\geq 10$, which gives a contradiction.

Now suppose that $G_{B}\simeq PSL(2,p^{m})$ where $m|n$. Since $D_{2k}\leq G_{B}$, using 3 we have $k|\frac{p^{m}\pm 1}{2}$, so $k\leq\frac{p^{m}+1}{2}$. Then using the divisibility condition of 11 we have $\frac{1}{2}p^{m}(p^{2m}-1)\leq k(k-1)(k-2)$. Combining the two inequalities:

which contradicts the assumption that $k\geq 4$. Thus $G_{B}$ can not be $PSL(2,p^{m})$.

Suppose $G_{B}\simeq D_{2d}$ where $d|q\pm 1$. Since $D_{2k}\leq G_{B}$, we have $k|d$ so $k\leq d$. Since $k$ must be the sum of some length of orbits of $G_{B}$, using 8 we can see that the only possibility is $k=d$, so $G_{B}\simeq D_{2k}$.

Finally, suppose $G_{B}\simeq PGL(2,p^{m})$ where $m|n$. Similarly, using Theorem 2 and 11 we get $k\leq p^{m}+1$ and $p^{m}(p^{2m}-1)\leq k(k-1)(k-2)$. The latter inequality implies $p^{m}+1\leq k$, so this time the only possibility is $k=p^{m}+1$.

Therefore when $k-1$ does not divide $q$, we have $G_{B}\simeq D_{2k}$. Now we consider the case where $k-1$ does divide $q$, so $k=p^{m}+1$ for some $m\leq n$. In this case $G_{B}\simeq D_{2k}$ or $PGL(2,p^{m})$.

First we need to show that $m|n$ so that $PGL(2,p^{m})$ can be considered. The Euclidean division of $n$ by $m$ can be written $n=xm+y$ (where $0\leq y\leq m-1$). Then $r(p^{m}+1)+1=p^{n}=p^{xm+y}$, so

Thus either $(-1)^{x}p^{y}=1$ or $p^{m}+1\leq|(-1)^{x}p^{y}-1|\leq p^{y}+1$ which is impossible since $0\leq y\leq m-1$. We deduce that $y=0$ and $x$ is even, thus $2m|n$.

We will show that the orbit $G(B)$ is the same as the orbit of $B^{\prime}:=GF(p^{m})\cup\{\infty\}$. Denote $\beta:=\theta^{r}$, and consider $f:x\mapsto\frac{x+\beta}{\beta x+1}\in G$. Then $B=\{1,\beta,\dots,\beta^{k-1}\}$, $\beta^{k}=1$ and $\beta^{\frac{k}{2}}=-1$. Thus

Furthermore, we can show that the remaining points of $B$ are mapped to $GF(p^{m})=\{x\in GF(p^{n})\mid x^{p^{m}}=x\}$. Note that $GF(p^{m})$ has characteristic $p$, so $(x+y)^{p^{m}}=x^{p^{m}}+y^{p^{m}}$ for any $x,y\in GF(p^{m})$. Then, for $\alpha\in\{0,\dots,k-1\}\setminus\{\frac{k}{2}-1\}$,

This shows that $f(B)=B^{\prime}$, and therefore $G(B)=G(B^{\prime})$. Using the orbit-stabilizer theorem, this implies that $|G_{B}|=|G_{B^{\prime}}|$.

Since $GF(p^{m})$ is a subfield of $GF(p^{n})$, we can extend the action of $PGL(2,p^{m})$ to $X=GF(p^{n})\cup\{\infty\}$. Then $PGL(2,p^{m})\leq PGL(2,p^{n})=G$ and $PGL(2,p^{m})$ stabilize $B^{\prime}$ so $PGL(2,p^{m})\leq G_{B^{\prime}}$. Thus looking at the list of Theorem 2 and 4, $G_{B^{\prime}}$ must be $PGL(2,p^{m^{\prime}})$ with $m|m^{\prime}|n$ or $PSL(2,p^{m^{\prime}})$ with $2m|m^{\prime}|n$ ($p>2$). Then since $B^{\prime}$ is a union of some orbits of its stabilizer, 7 insure that we have in fact $PGL(2,p^{m})=G_{B^{\prime}}$. Finally, since $|G_{B}|=|G_{B^{\prime}}|$ and the possibilities for $G_{B}$ were reduced to $D_{2k}$ or $PGL(2,p^{m})$, we can conclude that $G_{B}\simeq PGL(2,p^{m})$. ∎

The orbit of $B$ is a $3$-$(q+1,k,\lambda)$ design where $\lambda=k(k-1)(k-2)|G_{B}|^{-1}$, and the order of $G_{B}$ is given by 12. ∎

We proceed again by elimination from the list of Theorem 2. $x\mapsto\theta^{r}x$ stabilizes $B$ so $\langle x\mapsto\theta^{r}x\rangle=C_{k}\leq G_{B}$. Then $G_{B}$ can not be $Z_{p}^{m}$ because it does not have a subgroup $C_{k}$.

Suppose $G_{B}\simeq S_{4}$. Then $C_{k}$ must be $C_{3}$ or $C_{4}$, so $k=3$ or $4$ and $|B|=k+1=4$ or $5$. Since $B$ is a union of some orbits of $G_{B}$, 5 gives a contradiction (Note that $k=3$ implies that $q$ is not a power of $3$).

Suppose $G_{B}\simeq A_{5}$. Then $C_{k}$ must be $C_{3}$ or $C_{5}$, so $k=3$ or $5$ and $|B|=4$ or $6$. This time 6 gives a contradiction.

Suppose $G_{B}\simeq D_{2d}$. Then $C_{k}\leq D_{2d}$ so $k$ must divide $d$, and 8 gives a contradiction.

Suppose $G_{B}\simeq PSL(2,p^{m})$ for some $p^{m}|q$. Then from 11 (with this time $|B|=k+1$), $|G_{B}|$ divides $(k+1)k(k-1)$, so $\frac{1}{2}p^{m}(p^{2m}-1)\leq k(k^{2}-1)$. On the other hand, since $C_{k}\leq G_{B}$, from 3 we have $k|\frac{p^{m}\pm 1}{2}$ so $k\leq\frac{p^{m}+1}{2}$. Combining the two inequalities:

which contradicts the assumption that $k\geq 3$.

Suppose $G_{B}\simeq PGL(2,p^{m})$ for some $p^{m}|q$. Similarly, using Theorem 2 and 11 we get $k|p^{m}\pm 1$ and $p^{m}(p^{2m}-1)\leq k(k^{2}-1)$ so $p^{m}\leq k$. This time the only possibility is $k=p^{m}+1$, so $|B|=p^{m}+2$. Then 7 gives a contradiction.

Suppose $G_{B}\simeq A_{4}$. Then $C_{k}$ must be $C_{3}$, so $k=3$.

Suppose $G_{B}\simeq C_{d}$. Since $C_{k}\leq C_{d}$, we have $k|d$. Then looking at the orbit lengths given in 9, the only possibility is $k=d$ so $G_{B}\simeq C_{k}$.

Suppose $G_{B}\simeq Z_{p}^{m}\rtimes C_{d}$ where $m\leq n$, $d|p^{n}-1$ and $d|p^{m}-1$. Since $C_{k}\leq Z_{p}^{m}\rtimes C_{d}$, we have $k|dp^{m}$. As $k$ does not divide $q$ and thus not $p^{m}$, we have $k|d|p^{m}-1$, so $k\leq p^{m}-1$. Then since $B$ is a union of orbits of $G_{B}$ and $|B|=k+1$, from 10 the only possibility is $k=d=p^{m}-1$. Thus $G_{B}\simeq Z_{p}^{m}\rtimes C_{p^{m}-1}$, $k\neq 3$ and $k+1$ must divide $q$.

Therefore in the case (2) where $k\geq 4$ and $k+1$ does not divide $q$, we have $G_{B}\simeq C_{k}$. Now we consider the case (1) where $k=3$. In this case, $G_{B}\simeq C_{3}$ or $A_{4}$, where $C_{3}=\langle x\mapsto\theta^{r}x\rangle$. Thus in order to show that $G_{B}$ is $A_{4}$, it is enough to find an element of $G_{B}\setminus C_{3}$. Consider $f:x\mapsto\frac{x-1}{(\theta^{2r}+\theta^{r}-1)x-1}$. Then $f$ restricted to $B$ is the permutation $(0,1)(\theta^{r},\theta^{2r})$, so $f\in G_{B}\setminus C_{3}$.

The last case to consider is (3) where $k\geq 4$ and $k+1$ does divide $q$, so $k+1=p^{m}$ for some $m\leq n$. In this case, $G_{B}\simeq C_{k}$ or $Z_{p}^{m}\rtimes C_{p^{m}-1}$.

First we show that $m|n$. The Euclidean division of $n$ by $m$ can be written $n=xm+y$ (where $0\leq y\leq m-1$). Then $r(p^{m}-1)+1=p^{n}=p^{xm+y}$, so

Thus either $p^{y}=1$ or $p^{m}-1\leq p^{y}-1$ which is impossible since $0\leq y\leq m-1$. We deduce that $y=0$, so $m|n$.

Then, $B$ is the subfield $GF(p^{m})$ of $GF(q)$. Thus the stabilizer of $B$ in $G^{\prime}:=PGL(2,p^{m})$ is the same as the stabilizer of infinity,

Since $G^{\prime}_{B}\leq G_{B}$ and $G_{B}$ was reduced to the two possibilities $C_{p^{m}-1}$ or $Z_{p}^{m}\rtimes C_{p^{m}-1}$, we can conclude that $G_{B}\simeq Z_{p}^{m}\rtimes C_{p^{m}-1}$.

∎

The orbit of $B$ is a $3$-$(q+1,k+1,\lambda)$ design where $\lambda=(k+1)k(k-1)|G_{B}|^{-1}$, and the order of $G_{B}$ is given by 14. ∎