# Does $GL(n,K)$ act transitively on $1$-dim subspaces of $K$

If we let $K$ be a field and $GL(n,K)$ act by right multiplication on the $1$-dim subspaces of $K^n$.

Then if we take $\langle v_1 \rangle, \ldots \langle v_n \rangle \in K^n$ distinct and $\langle w_1 \rangle, \ldots \langle w_n \rangle \in K^n$ again distinct.

Then $v_1, \ldots , v_n$ forms a basis of $K^n$ as does $w_1, \ldots , w_n$.

So we know that there is an invertible map $f:K^n \rightarrow K^n$ for which $v_i^m = w_i$ for $1 \leq i \leq n$ with $m \in GL(n,K)$ and moreover $\langle v_n \rangle ^m = \langle w_i \rangle$ hence $GL(n,K)$ acts n-transitively.

Is the above correct?

Thanks for any help.

• Distinct $1$-dimensional subspaces are not necessarily linearly independent. E.g. consider the lines $(1,0)$, $(0,1)$, and $(1,1)$ in $\mathbb R^2$. – Dustan Levenstein Mar 4 '14 at 10:44
• ^ Or $(1,0,0)$, $(0,1,0)$ and $(1,1,0)$ in $\mathbb{R}^3$. – Casteels Mar 4 '14 at 10:48
• That's kind of the right idea, but there's a small problem. Just because the $1$-dimensional subspaces are distinct doesn't necessarily imply that they are linearly independent: e.g. take $\langle (1, 0, 0) \rangle, \langle (0, 1, 0) \rangle, \langle (1, 1, 0) \rangle < \Bbb{R}^3$. – Sammy Black Mar 4 '14 at 10:49
• @DustanLevenstein of course. So I could prove that it is $2$-transitive this way. Can I modify this to show that it acts n-transitively or is this just false? – hmmmm Mar 4 '14 at 10:53
• It's false because a basis in $n$-dimensional space cannot be sent to a linearly dependent distinct set of $n$ lines, which exists for $n>2$. – Dustan Levenstein Mar 4 '14 at 11:58

From the comments above the proof is wrong because if $\langle v_1 \rangle \ldots \langle v_n\rangle$ are distinct it does not follow that all of these spaces are linearly independent.

Moreover the claim is false. We have that $GL(n,K)$ acts 3-transitively iff $n=2$ and hence $GL(3,K)$ does not act $3$ transitively.

To show that $GL(n,K)$ does not act $3$-transitively for $n\geq 3$ suppose that it was.

Then we can take $\langle v_1\rangle, \langle v_2\rangle , \langle v_3 \rangle \in K^n$ all distinct and linearly independent (as $n\geq 3$)

We can then also choose take $\langle v_1\rangle, \langle v_2\rangle, \langle v_1+v_2\rangle$ which are all distinct but no longer independent.

Then we have to have $g\in GL(n,K)$ with $v_1^g=\lambda_1 v_1$, $v_2^g=\lambda_2$ and $v_3^g=\lambda_3(v_1+v_2)$ but as $g$ is invertible this will give a contradiction

• @DerekHolt I'm sorry what can? – hmmmm Mar 4 '14 at 13:27
• That was a comment on a (now deleted) erroneous comment by me. – Tobias Kildetoft Mar 4 '14 at 13:29
• @TobiasKildetoft ah ok cool thanks – hmmmm Mar 4 '14 at 13:29