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## Homework Statement

Prove that [itex]SL_{2}(ℝ)[/itex] is generated by the set:

[1 a], [1 0]

[0 1], [b 1], [itex]a,b \in ℝ[/itex]

## Homework Equations

- GCD (Greatest common divisor)
- The property of special linear group
- Some basic linear algebra, like determinant

## The Attempt at a Solution

[itex]SL_{2}(ℝ)[/itex] is the group consisting of invertible matrices with the determinant 1. Let [itex]A[/itex] be this matrix:

[a b]

[c d]

Then, [itex]det(A) = ad - bc = 1[/itex]

In order for the determinant to be 1, [itex]gcd(a,b) = gcd(a,c) = gcd(d,b) = gcd(d,c) = 1[/itex].

I'm not sure if my reasoning is right. I treat ad - bc = 1 as the linear combination. If 1 divides a and 1 divides b, then 1 divides (ad - bc). But -1 also divides a, b and (ad - bc). Thus a = ±1.

If [itex]a = 1[/itex], then [itex]det(A) = d - bc = 1[/itex]. Here, we need to split into some cases:

If [itex]c = 0[/itex], then [itex]d = 1[/itex], so we are done, having this matrix:

[1 b]

[0 1]

Similar argument shows that if [itex]a = 1[/itex] and [itex]b = 0[/itex], then d = 1, so we are done, leaving off this matrix:

[1 0]

[c 1]

If [itex]a = -1[/itex], then [itex]d = -1[/itex]. Similar arguments show that if either [itex]b = 0[/itex] or [itex]c = 0[/itex], then we obtain these similar matrices that belong to the special linear group:

[-1 0]

[c -1]

[-1 b]

[0 -1]

So these matrices form the set as given.

What if [itex]c ≠ 0[/itex]? Would I need to use Euclidean Algorithm to work out this problem?