Normal subgroup of $\operatorname{SL}_2(K)$ The setup:

Let $K$ be a field which is not of characteristic 2 and which contains at least seven elements. Let $N$ be a normal subgroup of $\operatorname{SL}_2(K)$ which contains a matrix $A \neq \pm I$.

The problem:

(a) Show that $N$ contains an upper triangular matrix other than $\pm I$.
(b) Show that $N$ contains a unit upper triangular matrix other than $\pm I$.

Now the problem goes on, and the end goal is to show that there are infinitely many nonabelian simple groups. I've got part (c) through (f), but I simply can't get anywhere with these two.
There is a hint attached to part (a), suggesting we conjugate to get a 0, then compute a commutator with a diagonal matrix to place the 0.
Conjugation isn't exactly the most elegant operation here. For any $\begin{pmatrix} a&b\\c&d \end{pmatrix}$ in $N$, and any $\begin{pmatrix} r&s\\t&u\end{pmatrix}$ in $\operatorname{SL}_2(K)$, we have
$$ \begin{pmatrix} u&-s\\-t&r\end{pmatrix}\begin{pmatrix} a&b\\c&d\end{pmatrix}\begin{pmatrix} r&s\\t&u\end{pmatrix} = \begin{pmatrix} u(ar+bt)-s(cr+dt)&u(as+bu)-s(cs+du)\\r(cr+dt)-t(ar+bt)&r(cs+du)-t(as+bu)\end{pmatrix},$$
and none of those look particularly easy to force to be 0. (I can get the bottom left to $-b$ or $-c$, but those haven't been helpful on their own.)
I don't know what to do for the commutator yet, but to be fair, I don't know where my 0 is supposed to be, so that will hopefully resolve with the issue above.
I'm hoping my problems with part (b) also stem from my lack of progress with part (a), so any hints would be greatly appreciated. Thanks!
 A: Assuming your computations are correct, here's an easy way to get a zero:
If $c=0$, you are done. If $c\neq 0$, then consider the matrix
$$\left(\begin{array}{cc}1 & s \\ 0 & 1
\end{array}\right).$$
(That is, $r=u=1$, $t=0$). Then according to your computations, the conjugate would be:
$$\begin{pmatrix} a-sc&as+b-cs^2+d\\c&cs+d\end{pmatrix},$$
so selecting $s=\frac{a}{c}$ guarantees a $0$.
It should be fairly clear that this works, even if your computations were incorrect: think of the matrix
$$\left(\begin{array}{cc}
1 & \frac{a}{c} \\
0 & 1
\end{array}\right)$$
as a change-of-basis matrix. You are replacing the standard basis with $[(1,0), (\frac{a}{c},1)]$; now look at how the image of $(1,0)$ is expressed in terms of this new basis:
$$\left(\begin{array}{cc}
a&b\\
c&d
\end{array}\right)\left(\begin{array}{c}1\\0\end{array}\right) = \left(\begin{array}{c}a\\c\end{array}\right) = 0\left(\begin{array}{c}1\\0\end{array}\right) + c\left(\begin{array}{c}a/c\\1\end{array}\right),$$
so under the new basis, the matrix of the linear transformation will have first column $(0,c)^T$. This is the result of your conjugation.
A: Here is a slightly different way to think about this, that avoids computation or explicit discussion of normal forms:
Let $K$ be a field, and let $\phi:K^2 \to K^2$ be a linear operator.
Let $v$ be any non-zero vector in $K^2$.  If $\phi(v)$ is a multiple of $v$, then choose any $w$ linearly independent from $v$, and with respect to the basis $v, w$, the linear operator $\phi$ will have upper triangular basis.  On the other hand, if $\phi(v)$ is not a scalar multiple of $v$, then one can take $v$ and $\phi(v)$ to be a basis, and then the matrix of $\phi$ will have a zero in the upper left corner.
Changing basis corresponds to conjugation by an element in $GL_2(K)$, but as Ted notes, one can always replace such a conjugation by conjugation by an element of $SL_2(K)$ by conjugating by an appropriate diagonal matrix, which won't affect the $0$.
A: The way to think of conjugation in $SL_2(K)$ is as a change of basis for a linear transformation on a $K$-vector space of dimension 2.  According to a result of linear algebra, every matrix in $GL_2(K)$ is conjugate over $GL_2(K)$ to a matrix in rational canonical form, which (for 2 by 2 matrices) is either a diagonal matrix, or a matrix with a 0 in the upper left corner.  There is a minor complication, in that your question is about $SL_2(K)$ rather than $GL_2(K)$, but this turns out not to matter too much: once you have a 0, you can conjugate by a diagonal matrix of arbitrary determinant without affecting the 0.
