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Let $F$ be a field and $G = SL_2(F)$. Let us define the following matrices $$X(b) = \begin{pmatrix} 1 & b\\ 0 & 1 \end{pmatrix}, \quad Y(c) = \begin{pmatrix} 1 & 0\\ c & 1 \end{pmatrix},$$ where $b,c\in F$. It a well-known result that these matrices generate $G$ (Lemma 8.1, Ch. XIII in 3rd edition of Lang's Algebra).

After doing some short calculations, I noticed that the set consisting of all matrices $X(b)$ and the matrix $Y(c_0)$ (for some $c_0\in F^\times$) is already a generating set.

Is there some other way to obtain this result?

AYK

PS. These are the calculations that I mentioned above: $$X(-1/x)Y(x)\cdot X(y)\cdot Y(-x)X(1/x) = Y(-x^2y).$$

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    Can we make any assumptions about $F$? Can we say it has characteristic $0$? – 2017-02-14
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    The statement holds in every characteristic. But all arguments are welcome of course.. – 2017-02-14
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    wouldn't have expected that. Good to know. – 2017-02-14
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    By the way: I've added some more generic tags with the hopes of increasing your question's visibility – 2017-02-14
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    @AYK: Are you sure your result is correct. Can you outline the proof? – 2017-02-14
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    I am sorry, I forgot to include the inverse of $Y(c_0)$ which I've also used in the calculations. – 2017-02-14
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    @AYK: Suppose $F=\mathbb{Q}$. Using, as generators, the set of matrices of the form $X(b)$, together with $Y(2)$, how would you generate $Y(1)$? – 2017-02-14
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    @quasi I've edited the question. – 2017-02-14
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    But $Y(-c_0)$ is just the inverse of $Y(c_0)$, hence if you have $Y(c_0)$, you get $Y(-c_0)$ for free. So my question remains: How can you generate $Y(1)$ from $Y(2)$ (and the matrices $X(b)$)? – 2017-02-14
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    Ok, I see the edited post -- no need to answer my question. – 2017-02-14
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    @quasi You are right on the 'inverse comment', my bad – 2017-02-14
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    @AYK: Your construction is very nice, and fully answers my question. – 2017-02-14

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I think, doing "these short calculations" is the best method to obtain this result. We may even use only two unipotent generators for finite fields, with a few exceptions. A precise statement is given in Dickson's Theorem. For a finite field $F$, and a generator $\lambda$ for $F$ over $\mathbb{F}_p$, $$ L:=\langle \left( \begin{array}{clcr} 1 & 0\\1 &1 \end{array} \right),\left( \begin{array}{clcr} 1 & \lambda\\0 &1 \end{array} \right) \rangle $$ is indeed $SL_2(F)$, except when $|F| =9$, in which case $L\cong SL_2(5)$. For a reference see here.

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    I did not assume the field $F$ is finite... – 2017-02-14
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    Yes, but this is the only interesting case left, I think. – 2017-02-14