Show that $G = SL_3(F_p)$ acts transitively on $H = F_p^3\setminus \lbrace0\rbrace$

Question

I need to show that $G = SL_3(F_p)$ acts transitively on $H = F_p^3\setminus \lbrace0\rbrace$. That means to show that, for all $s,t \in H$, there is $g \in G$ such that $gt = s$. I tried to make three equations from $gt = s$ and then find the constraints for $g$ (considering also that $\det, g = 1$), but it didn't work out.

Any help or clue will be appreciated.

Answer 1

Hint To show that a group $G$ acts transitively on a set $X$, it is enough to pick a convenient element $x_0 \in X$ and show that for any $x \in X$ there is some $g \in G$ such that $g \cdot x_0 = x$: Given any $x, x' \in X$, for any corresponding group elements $g, g' \in G$, we have $$(g'g^{-1}) \cdot x = g' \cdot (g^{-1} \cdot x) = g' \cdot x_0 = x'.$$

(I've written this condition for a left action, but everything applies here mutatis mutandis to right actions.)

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In our case, we have $G = SL_3(\Bbb F_p)$ and $X = \Bbb F_p^3 - \{ {\bf 0} \}$, and a natural choice for $x_0$ is the first element of the canonical basis of $\Bbb F_p^3$, namely, $[1, 0, 0]^T$, which we henceforth use.

Computing in matrix entries gives that the action on $x_0$ is $$A \cdot x_0 = \pmatrix{a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}} \cdot \pmatrix{1 \\ 0 \\ 0} = \pmatrix{a_{11} \\ a_{12} \\ a_{13}},$$ that is, the action of the element $A \in SL_3(\Bbb F_p)$ on the first vector in the standard basis simply picks out the first column of $A$. So, we've produced an equivalent problem that is more concrete:

Show that for any vector $x \in \Bbb F_p^3 - \{ {\bf 0} \}$ there is a matrix $A \in SL_3(\Bbb F_p)$ with first column $x$.

Remark There is nothing special to the case of prime fields or the vector length $3$ here: The claim holds true for the standard action of $SL_n(\Bbb F)$ on $\Bbb F^n - \{{\bf 0}\}$ for any $n > 1$ and $\Bbb F$.

Answer 2

Plan of attack:

1. Prove that there exist bases $\{s,b_2,b_3\}$ and $\{t,c_2,c_3\}$ of $\Bbb{F}_p^3$ as a vector space over $\Bbb{F}_p$.
2. Prove that you can select a constant $r\in\Bbb{F}_p^*$ such that the matrix of the linear transformation $f$ determined by $f(s)=t, f(b_2)=c_2)$, $f(b_3)=rc_3$ has determinant $=1$.

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More details:

1. There are plenty of such bases. Anyone will do! Undoubtedly you have seen a result in linear algebra stating that any linearly independent set of vectors can be extended to a basis. Well, a non-zero vector by itself is linearly independent.
2. Let $f_r$ be the linear transformation above. The transformation $f_1$ is surjective (why?). Let $g_\ell$ be the linear transformation defined by $g_\ell(t)=t$, $g_\ell(c_2)=c_2$, $g_\ell{c_3}=\ell c_3$, $\ell\in\Bbb{F}_p^*.$ Using the basis $\{t,c_2,c_3\}$ prove that $\det(g\ell)=\ell$.
3. Show that for some choice of $\ell\in\Bbb{F}_p^*$ the composition $g\circ f_1=f_\ell$ has determinant $=1$.