Describe the left cosets of $SL(2, 3)$ in $GL(2, 3)$.
Describe the left cosets of $SL(2, R)$ in $GL(2, R)$.
I think I have an idea of how to do the second part but I don't know how to describe the first.
Let $G = GL(2, R)$ and $S = SL(2, R)$. Let $g \in GL(2, R)$.
The left coset of $SL(2, R)$ in $GL(2, R)$ can be represented $gSL(2, R) = [gs: s \in SL(2, R), \det(s) = 1]$. I know that $\det(g) \neq 0$ because it is invertible.
I don't know how to proceed further for either part.
On
If $G$ is a group, and $H$ a subgroup, a way of understanding this question (the way I suggested in the comments above) is that it is asking you to write $G$ as a disjoint union, i.e., $$G = \bigcup_g\, gH, $$ for a choice of 'nice' $g$, where $g_1H \cap g_2H = \emptyset$ iff $g_1 \ne g_2$.
Another way of understanding the question (@Omnomnomnom's suggestion - sometimes a better way) is to have some criterion that characterizes the cosets $gH$.
But from your conversation with @Omnomnomnom it doesn't feel you are comfortable with all this... Let me first take another warm-up example:
Suppose instead your group $G = \mathbb R^*$, the set of non-zero real numbers. It's a group, viewed multiplicatively.
Let $H = \mathbb R^+$, the strictly positive (non-zero) real numbers - it's a subgroup: the product of two positive numbers is positive, etc.
Now, $xH = yH$ iff $x = yh$, where $h\in H$, i.e., $h$ is a positive number. This happens iff $x$ and $y$ have the same sign. Agree? Arguably 'nice' representative choices to distinguish between the positives and negatives are $1$ and $-1$, and we can write (a coset decomposition) $$ G = 1 \cdot H \ \cup \ (-1)H,$$ where the union is disjoint: if $x$ is positive, then by definition $x\in H$, and $x= 1 \cdot x \in 1H$, and $x$ belongs to the first coset; on the other hand, if $x$ is negative, then $x = -1 \cdot |x| \in -1\cdot H$, and $x$ belongs to the second coset.
So the above (disjoint) union is a 'nice' presentation of $G/H$, i.e of the decomposition of $G$ into cosets. @Omnomnomnom's way is to say that we broke $G$ up into the set of positive and negative numbers.
Back to the real question:
Here $G$ = invertible $2\times 2$ matrices, and $H$ the set of matrices of determinant $1$. As in your conversation with @Omnomnomnom, the point is that $\det g' = \det g$ if and only if there exists $h$, of determinant $1$, such that $g'= g h$ [explicitly solving for $h$: $h = g^{-1}g'$ has determinant $1$ iff $\det g' = \det g$.], i.e. $g'H = g H$. (The role of 'same determinant' here is therefore the same as 'same sign' before - Check that you understand that! )
So to decompose $G$ into $H$ cosets is to break $G$ up into subsets of elements with the same determinant - @Omnomnomnom's suggestion - which is absolutely fine here - because all such subsets are of the form $gH$, for ('nice' I suggested, but it doesn't really matter) representative choices of $g$. I was suggesting, if your arbitrary $g'$ has determinant $d$, you find a 'nice' diagonal matrix $g$ with determinant $d$. If so $g'\in gH$, and one has a 'nice' description of the cosets...
So - given $d\ne 0$, can you find a diagonal matrix with determinant $d$?
I hope this helps!
My guess is that they're expecting an answer along the lines of $$ g\,S = \{h \in G: \det(h) = \det(g)\} $$ I think it is worth trying to convince yourself that this is the case. It turns out that the same is true for matrices over $\Bbb F_3$. Note that this means that $SL(2,3)$ has exactly two cosets.