Let $G=GL(2, \mathbb{R})$, and $H=\{A\in G\mid \det A=2^{d}, d\in \mathbb{Z}\}$. Prove that $H$ is a subgroup of $G$. So I know I need to prove first that H is a subset of G which is shown because A is in H and has a nonzero determinant so it must also be in G. Now I need to show that H is closed under matrix multiplication. I was thinking I could say that if $A, B\in H$ then $\det(A)=2^{d}$, $\det(B)=2^{n}$ where $d, n\in \mathbb{Z}$. So, $\det(AB)=\det(A)\det(B)=2^{d}2^{n}=2^{d+n}$. Then $d+n$ must be some integer and so it holds that the determinant is $2^{d}$. Is this a just argument?
After this I will be able to say H is associative because G is a group. The identity is in H because the determinant is $1=2^{0}$. 
But then I struggle again with the inverse property. Is it enough to say that $\det(A^{-1})=\frac{1}{\det(A)}=\frac{1}{2^{d}}$ so the inverse must be contained S?
Any help on these two questions is appreciated! 
 A: Your argument is fine and correct. It can be made more concise with the "one-step subgroup test". That is, let $G$ be a group $G$ and $H$ a non-empty subset of $G$. If for all $a,b \in H$ we have $ab^{-1} \in H$, then $H$ is a subgroup of $G$.

Let $A,B \in H$. Then there are integers $d,n$ so that $\det(A)=2^d$ and $\det(B)=2^n$. Now, $\det(AB^{-1})=\det(A)\det(B^{-1})=2^d \times \frac{1}{2^n}=2^{d-n}.$ Since $d-n \in \mathbb{Z}$, we are done. 
A: You can also use arrows (morphisms) : the mapping $\phi : G\mapsto \det(G)$,  $GL(2,\mathbb{R})\rightarrow\mathbb{R}^*$ is a morphism of groups. Then $H=\phi^{-1}(\{2^k\}_{k\in \mathbb{Z}})$ and as $U=\{2^k\}_{k\in \mathbb{Z}}$ 
is a subgroup of $\mathbb{R}^*$, one gets the property. 
A: For the sake of completion I will show that H meets all four group axioms. First of all, associativity is shown, since matrix multiplication is known to be associative. Choosing the $e= \begin{bmatrix} 
1&0 \\
0&1\end{bmatrix}$. It suffices to show that the $\det(e)=2^k $ for some $k\in \mathbb{Z}$. Notice $\det(e)=1=2^0$. Next I will show that any inverse, $A^{-1}$ is also in $H$. By properties of the determinant, $\det(A^{-1})=\frac{1}{\det(A)}=\frac{1}{2^t}=2^{-t}$. Which is of course an integer power of 2. Lastly, I will show closure. Since products of 2 by 2 matrices, are also 2 by 2 matrices, I will show the product $AB$, for both$A,B \in H$ also has a determinant that is an integer power of two. Applying some more properties of the determinant shows $\det(AB)=\det(A)\det(B)=2^l2^p=2^{l+p}$. Since sums of integers, are integers I will rest the proof. $\square$ 
