Every element in the group has eigenvalue 1 Let $V$ be a two dimensional vector space over a field $k$ and $H$ a subgroup of $\text{Aut}_k(V)$ such that $\det(h - I) = 0$ for all $h\in H$. In other words, every element of $H$ leaves some vector fixed.
Show that $H$ is contained (up to conjugation) in the subgroup $\begin{bmatrix}
    1      & 0  \\
    *       & * 
\end{bmatrix}$ or $\begin{bmatrix}
    1      & *  \\
    0       & * 
\end{bmatrix}$. 
The second case is clearly when every element of $H$ leaves the same vector fixed. I am having trouble proving the other possibility. I tried to show that the dual representation leaves some liner form fixed but to no avail...
 A: Here is an outline of one way to show this. There might be easier ways.
First note that if the matrix $a \in {\rm GL}_2(k)$ has an eigenvalue 1, then ${\rm tr}(A) = 1 + \det(A)$.
Suppose first that $H \le {\rm SL}_2(k)$. Then all elements of $H$ have eigenvalue $1$ with multiplicity $2$. If the conclusion does not hold for $H$ then, with respect to a suitable basis, we can find elements $\left( \begin{array}{cc}1&0\\\lambda & 1 \end{array}\right)$ and 
$\left( \begin{array}{cc}1&\mu\\0 & 1 \end{array}\right)$ in $H$ with $\lambda \ne 0 \ne \mu$, but then the above condition for their product to have an eigenvalue $1$ gives $\lambda\mu=0$, contradiction.
So in general, the desired conclusion holds for $H \cap {\rm SL}_2(k)$.
There are now two cases. If $H \cap {\rm SL}_2(k) \ne 1$ then by changing basis we may suppose that all matrices in $H$ are upper triangular, and the normalizer of $H$ in ${\rm GL}_2(k)$, which contains $H$, is also contained in the group of upper triangular matrices. It then follows easily that $H$ has one of the two specified forms.
On the other hand, if $H \cap {\rm SL}_2(k)=1$, then $H$ is abelian, and again it is not hard to complete the proof.
