Does permutation representation of PSL always contain a transposition I am trying to understand how to view $\mathrm{PSL}$ as a subgroup of $\mathrm{S}_n$.
From what I understand till now a way to look at $\mathrm{PSL}(d,q)$ in $\mathrm{S}_n$ is following:
look at the projective space of dimension $d-1$ over $\mathbb{F}_q$. This has $\dfrac{q^d-1}{q-1}$ elements. Now take $n= \dfrac{q^d-1}{q-1}$. Every linear operator will have action on this projective space and as they are invertible it will be injective.
But this I guess in most cases (if not all) will be a strict subgroup of $\mathrm{S}_n$ because it has to be linear (and moreover determinant 1).
Now I want to know how can I see if particular elements of $\mathrm{S_n}$ are in it? For eg. does it contain a transposition when viewed as  a subgroup of $\mathrm{S}_n$. It definitely contains an element of order 2 (as order is even). Is there a way to cook up a transposition that is in in the group?
 A: $\mathrm{PSL}_d(q)$, for $d>2$ or $q>3$, is a simple group. Thus if it is a subgroup of $S_n$ for some $n$ then it is a subgroup of $A_n$, as it has no subgroup of index $2$. In particular, this means that there can be no odd permutations in the image of a permutation representation of $\mathrm{PSL}_d(q)$, such as a transposition.
A: It is impossible for the natural action of $PSL(d,q)$ to contain a 2-cycle when $q>3$.

Consider the following. Let $g\in PSL(d,q)$ be a matrix yielding a 2-cycle. Let
$L_0=\langle v_0\rangle$ and $L_\infty=\langle v_\infty\rangle$ be the two lines through the origin interchanged by $g$. Their generators $v_0$ and $v_\infty$ are obviously linearly independent. So
$$
g\cdot v_0=\mu_\infty v_\infty\qquad\text{and}\qquad g\cdot v_\infty=\mu_0v_0
$$
for some constants $\mu_0,\mu_\infty\in\Bbb{F}_q^*$.
For every $\alpha\in\Bbb{F}_q^*$ the line $L_\alpha:=\langle v_0+\alpha v_\infty\rangle$ is distinct from both $L_0$ and $L_\infty$, so $L_\alpha$ must be a fixed point of $g$.
In other words the vectors
$$
v_0+\alpha v_\infty\qquad\text{and}\qquad g\cdot(v_0+\alpha v_\infty)=\alpha \mu_0v_0+\mu_\infty v_\infty
$$
are linearly dependent. This implies that
$$
\alpha^2\mu_0=\mu_\infty\qquad(*)
$$
for all $\alpha\neq0$. But $(*)$ specified $\alpha^2$ uniquely, so there is a single non-zero square in the field $\Bbb{F}_q$. This happens only when $q=2$ or $q=3$.

The natural actions of both $PSL(2,2)\simeq S_3$ and $PSL(2,3)\simeq S_4$ contain 2-cycles, so when $q=2$ or $q=3$ this is possible. The condition $d=2$ is also forced upon us, for otherwise there will be lines involving one but not both of $v_0$ or $v_\infty$, and such cannot be eigenvectors of $g$. Meaning that $g$ moves more than two lines.
