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It seems that there is a action by which $SL_2(\mathbb{F}_p)$ and $GL_2(\mathbb{F}_p)$ permute the $p^2$ ordered tuples in $\mathbb{F}_p^2$. What is the map from the $2 \times 2$ matrices over $\mathbb{F}_p$ versions of the elements of these groups to the elements of $S_{p^2}$ that this action picks out?

Similarly there seems to be an injective homomorphism map from $PSL_2(\mathbb{F}_p)$ to $S_{p+1}$ by virtue of the natural action on $\mathbb{F}_pP^1$ (the projective space obtained from $\mathbb{F}_p^2$) What is this mapping?

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    $\begingroup$ I don't understand the question. You've already described these maps. $\endgroup$ – Qiaochu Yuan Apr 30 '15 at 16:52
  • $\begingroup$ @QiaochuYuan See my comments to Herbert's answer. $\endgroup$ – user6818 Apr 30 '15 at 17:10
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A matrix $A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \in \text{GL}_2(\mathbb{F}_p)$ acts on a vector $v = \left[ \begin{array}{cc} v_0 \\ v_1 \end{array} \right] \in \mathbb{F}_p^2$ by matrix multiplication:

$$Av = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left[ \begin{array}{cc} v_0 \\ v_1 \end{array} \right] = \left[ \begin{array}{cc} a v_0 + b v_1 \\ c v_0 + d v_1 \end{array} \right].$$

This is far and away the best description of this action I can give. You can figure out everything you could possibly want to know about this action from linear algebra. It is even better than knowing a permutation representation because linear algebra is easier than set theory, even finite set theory.

For example, here is how you figure out the cycle decomposition of the permutation corresponding to $A$. It splits into cases depending on the conjugacy class of $A$ as follows.

Case: $A$ is diagonalizable over $\mathbb{F}_p$. Then up to conjugacy it is a diagonal matrix with diagonal entries $\lambda_0, \lambda_1 \in \mathbb{F}_p^{\times}$. Its order is the lcm of the orders $d_i$ of $\lambda_i \in \mathbb{F}_p^{\times}$, which can be any divisor of $p - 1$. The corresponding permutation of $\mathbb{F}_p^2$ fixes zero and splits the nonzero vectors into cycles as follows:

  • The nonzero vectors of the form $(v_0, 0)$ form cycles of length $d_0$.
  • The nonzero vectors of the form $(0, v_1)$ form cycles of length $d_1$.
  • The remaining nonzero vectors form cycles of length $\text{lcm}(d_0, d_1)$.

Case: $A$ is diagonalizable over $\mathbb{F}_{p^2}$ but not over $\mathbb{F}_p$. Then up to conjugacy its action on $\mathbb{F}_p^2$ is the action of multiplication by an element $\lambda \in \mathbb{F}_{p^2}^{\times}$ (one of the eigenvalues of $A$) on $\mathbb{F}_{p^2}$. Its order is the order $d$ of $\lambda$ in $\mathbb{F}_{p^2}^{\times}$, which can be any divisor of $p^2 - 1$. The corresponding permutation of $\mathbb{F}_{p^2}$ fixes zero, and the nonzero vectors form cycles of length $d$.

Case: $A$ is not diagonalizable even over $\mathbb{F}_{p^2}$. From the theory of rational canonical form we deduce that up to conjugacy $A$ is a Jordan block

$$A = \left[ \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array} \right]$$

where $\lambda \in \mathbb{F}_p^{\times}$. Its order is $p$ times the order $d$ of $\lambda$, which can be any divisor of $p - 1$. The corresponding permutation of $\mathbb{F}_p^2$ fixes zero and splits the nonzero vectors into cycles as follows:

  • The nonzero vectors of the form $(v_0, 0)$ form cycles of length $d$.
  • The remaining nonzero vectors form cycles of length $pd$.

The analysis for the action on the projective line is similar. Writing a point on the projective line in projective coordinates $(v_0 : v_1)$, the action takes the form

$$A (v_0 : v_1) = (av_0 + bv_1 : cv_0 + dv_1)$$

or, thinking of an element of the projective line as a fraction $z = \frac{v_0}{v_1}$ (which may take the value $\infty$),

$$A z = \frac{az + b}{cz + d}.$$

Again you can figure out everything you could possibly want to know about this action from linear algebra as above. In the diagonalizable case, for example, projectively the action of a diagonal matrix with entries $d_0, d_1$ is multiplication by $\frac{d_0}{d_1}$, and hence its cycle decomposition is determined by the order of $\frac{d_0}{d_1}$. So things are even simpler.

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  • $\begingroup$ @Qiachu Yuan Thanks for the help! Do you know what the irrep decomposition is of these permutation representations? $\endgroup$ – user6818 Apr 30 '15 at 19:34
  • $\begingroup$ @Qiachu Yuan What I need is an enumeration of all the group elements of these groups, SL_2(F_q) or PSL_2(F_q) or GL_2(F_q) or PGL_2(F_q), so that I can do these calculations by hand for specific cases. Is this somehow available? $\endgroup$ – user6818 Apr 30 '15 at 20:05
  • $\begingroup$ @user6818: I don't know what you mean by the irrep decomposition of a permutation representation. The decomposition of $\mathbb{F}_p^2$ into orbits is that $0$ is an orbit and the nonzero vectors are the other orbit. The decomposition of $\mathbb{F}_p^2$ as a linear representation is that it's irreducible. I also don't know what you mean by an enumeration of all the group elements. You can just write down matrices and chuck out the ones that aren't invertible. $\endgroup$ – Qiaochu Yuan Apr 30 '15 at 20:11
  • $\begingroup$ I am not understanding what you are not understanding :D Here is an action of these groups on $\mathbb{F}_p^2$ (or on $\mathbb{P}(\mathbb{F}_p^2 )$) so we have a representation of these groups on these vector spaces. (so we have here basically listed out 6 representations) Are any of these irreducible? If not then what is the irrep decomposition of these representations? $\endgroup$ – user6818 Apr 30 '15 at 20:23
  • $\begingroup$ @user6818: the grammar was throwing me off. "Irrep decomposition" is a term people apply to linear representations, not permutation representations (which I took to mean actions on sets). One thing you might have meant by "permutation representation" is the linear representation induced by an action on a set, which is different from the linear representation on $\mathbb{F}_p^2$ that we're talking about, so I found that ambiguous. Anyway, the projective line is not a vector space, but the action of $\text{GL}_n$ on an $n$-dimensional vector space is always irreducible. $\endgroup$ – Qiaochu Yuan Apr 30 '15 at 20:26
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$\mathrm{GL}_2 (\mathbb{F}_p)$ have a natural action on $\mathbb{P}_1(\mathbb{F}_p)$ : the action of $g$ on a vector line $d$ is just the vector line $g\cdot d$. Its kernel is the set of elements in $\mathrm{GL}_2 (\mathbb{F}_p)$ which stabilize every one-dimensional vector space and it is easy to show that this is the center of $\mathrm{GL}_2 (\mathbb{F}_p)$ (homotheties). Therefore, the quotient of $\mathrm{GL}_2 (\mathbb{F}_p)$ (which is $\mathrm{PGL}_2 (\mathbb{F}_p)$ ) is mapped to the permutation group of $\mathbb{P}_1(\mathbb{F}_p)$ and this is an injection.

As $\mathbb{P}_1(\mathbb{F}_p)$ has $1+q$ elements, this provides an injection $$\mathrm{PGL}_2 (\mathbb{F}_p) \hookrightarrow \mathfrak{S}_{p+1}$$

For low $p$, this leads to so-called "exceptional isomorphisms", such as $$ \mathrm{PGL}_2(\mathbb{F}_5) \simeq \mathfrak{S}_5 $$

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  • $\begingroup$ I guess my question isn't clear! I want to know what these maps are! Like given a $2\times 2$ matrix with entries in $\mathbb{F}_p$ for any of these three groups, I want to know what is the permutation matrix that corresponds to it in this action. $\endgroup$ – user6818 Apr 30 '15 at 17:00
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    $\begingroup$ It will depend on the basis of the underlying vector space. $\endgroup$ – Tlön Uqbar Orbis Tertius Apr 30 '15 at 17:04
  • $\begingroup$ Yes, obviously. But can you give me the action? As in given a $2 \times 2$ matrix over $\mathbb{F}_p$ an element of say $SL_2(F_p)$ what is its action on some element say $(a,b) \in \mathbb{F}_p^2$ ? $\endgroup$ – user6818 Apr 30 '15 at 17:06
  • $\begingroup$ And equivalently is there a way to explicitly enumerate all the elements of any of these groups? $\endgroup$ – user6818 Apr 30 '15 at 17:10
  • $\begingroup$ @user6818: the way you phrase this question makes it unclear to me whether you know how a matrix acts on a vector. Is that all you're asking about? $\endgroup$ – Qiaochu Yuan Apr 30 '15 at 17:33

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