An abstract algebra problem

Problem:

Solution:

For part (a), we simply verify all the group axioms:

Closure: For any two matrices $ 2 \times 2 $ with determinant 1, their product must also be a $ 2 \times 2 $ matrix with determinant 1, so closure is satisfied.

Identity: we designate $ \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) $ be the identity of the group, apparently, it is an element of the group and any matrix multiplied with it must stay the same, so identity is satisfied.

Associative: Associativity of this group simply follows from matrix multiplication.

Inverse: $ \left(\begin{array}{cc} a & b \\ c & d \end{array}\right)\left(\begin{array}{cc} d & -b \\ -c & a \end{array}\right) = \left(\begin{array}{cc} ad-bc & -ab+ab \\ cd-cd & -bc+ad \end{array}\right) = \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) $, therefore inverse is satisfied.

We will prove part (b) and part (c) together by considering the mapping

$ \left(\begin{array}{cc} a & b \\ c & d \end{array}\right) \to \left(\begin{array}{cc} a \mod 2 & b \mod 2 \\ c \mod 2 & d \mod 2 \end{array}\right) $.

It is routine to check this is a homomorphism, the kernel for this mapping are matrices with $ a, d $ odd and $ b, c $ even and $ ad - bc = 1 $, therefore it is a normal subgroup (as all kernels are), its image is $ SL(2, \mathbf{F_2}) $, therefore the quoient group is isomorphic to it.

part (d) is an interesting part, I owe this solution to Raymond Chow.

Consider the elements of  $ SL(2, \mathbf{F_2}) $ as invertible linear transformations $ \mathbf{F_2}^2 $, there are four points in  the space, because the mapping is linear, it must map $ (0, 0) $ to $ (0, 0) $. Because the mapping is invertible, it can only permute the points, so we can build an homomorphism from $ SL(2, \mathbf{F_2}) $ to $ S_3 $. Now it is routine to just enumerate the elements and show it is an isomorphism.

It would be really tedious to just write out all of them, for example, here is the computation of one of the matrix

$\left(\begin{array}{cc} 1 & 0 \\ 1 & 1\end{array}\right)\left(\begin{array}{c} 0 \\ 1\end{array}\right) = \left(\begin{array}{c} 0 \\ 1\end{array}\right)$

$\left(\begin{array}{cc} 1 & 0 \\ 1 & 1\end{array}\right)\left(\begin{array}{c} 1 \\ 0\end{array}\right) = \left(\begin{array}{c} 1 \\ 1\end{array}\right)$

$\left(\begin{array}{cc} 1 & 0 \\ 1 & 1\end{array}\right)\left(\begin{array}{c} 1 \\ 1\end{array}\right) = \left(\begin{array}{c} 1 \\ 0\end{array}\right)$

The computation indicate we should map $\left(\begin{array}{cc} 1 & 0 \\ 1 & 1\end{array}\right) $ to $ (1, 3, 2) $.

The final mapping is

$ \left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right) \to (1, 2, 3) $

$ \left(\begin{array}{cc} 1 & 0 \\ 1 & 1\end{array}\right) \to (1, 3, 2) $

$ \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right) \to (2, 1, 3) $

$ \left(\begin{array}{cc} 1 & 1 \\ 1 & 0\end{array}\right) \to (2, 3, 1) $

$ \left(\begin{array}{cc} 0 & 1 \\ 1 & 1\end{array}\right) \to (3, 1, 2) $

$ \left(\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array}\right) \to (3, 2, 1) $