Generating a particular element of $\mathrm{SL}_2(\Bbb{Z})$ from generators

Question

This should be easy, I hope.

$\mathrm{SL}_2(\mathbb Z) = \{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a,b,c,d \in \mathbb Z, ad-bc=1\}$

A book I was reading said that the elements of $\mathrm{SL}_2(\mathbb Z)$ could be generated by:

$ \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $ and $ \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} $.

I have one element that I think should be in $\mathrm{SL}_2(\mathbb Z)$ that I can't quite figure out how to generate from these. That's this one:

$ \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} $

How do I form this matrix from the others?

Thanks.

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Edit: In Diamond and Shurman's A First Course in Modular Forms $\mathrm{SL}_2(\mathbb Z)$ is called the modular group (on page 1). I am supposing that this is a group under matrix multiplication, since the additive matrix identity has determinant 0 and thus wouldn't be in the group.

Confirming with a second source that the matrices I've listed are generators for $\mathrm{SL}_2(\mathbb Z)$, here's some notes by Conrad that mention this fact immediately.

Answer 1

Well, let's look at what multiplying by each generator does.

We have that: $$Ae_1 = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix} = \begin{pmatrix}a & a+b \\ c & c+d\end{pmatrix}$$ We have that: $$Ae_2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix}b & -a \\ d & -c\end{pmatrix}$$ These are the "moves" we have to work with. Now, lets try to build your matrix.

It looks like we'll want to start with the first generator. Just for fun, lets try applying $e_2$ first: $$e_1e_2 = \begin{pmatrix}1 & -1 \\ 1 & 0\end{pmatrix}$$ Now, it's worth mentioning that $e_2^2 = -I$, where $I$ is the identity matrix. As we weren't just trying to multiply by $-1$, it makes sense to try applying $e_1$ now: $$e_1e_2e_1 = \begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}$$ This is exactly what we wanted, so we're done.

Another way we could have worked with this is by computing $e_1^{-1}$ and $e_2^{-1}$, and tried "working backwards" from the matrix we want.