Determining whether these two groups are isomorphic Consider the following group of matrices with multiplication:
$$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\ A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \ B = \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}, \\ C = \begin{pmatrix} -1 & -1 \\ 0 & 1 \end{pmatrix}, \ D = \begin{pmatrix} -1 & -1 \\ 1 & 0 \end{pmatrix}, \ K = \begin{pmatrix} 1 & 0 \\ -1 & -1 \end{pmatrix}$$
I have to determine if this group is isomorphic to $S_3$. The two groups look pretty similar: they both divide into two halves: one is cyclic, and the other consists of elements that are their own inverse. I have been trying for a while to match up their group tables, but there's always something that doesn't fit.
Is there an isomorphism between these groups, and if there isn't, how can I prove it?
 A: @DanielFischer has given a very clear and satisfactory answer in his comment.
For an explicit one, consider that $A^2 = I$, $B^3 = I$, $A B A B = I$, so $A, B$ satisfy the standard relations that define $S_{3}$. 
Note also
$$
B^2 = D, A B = C, A B^2 = K,
$$
so $\langle A, B \rangle = \{I, A, B, C, D, K \}$.
This also allows you to write down explicitly an isomorphism as $A \mapsto (12), B \mapsto (123)$. 
A: If you are familiar to $S_3$'s presentation:
 $$S_3=\langle a,b\mid a^2=b^3=(ab)^2=1\rangle=\{e,a, b, ab, b^2, ba \}$$ then you see that $$A^2=I,~~B^3=I, B^2=D,...$$
A: A good way to learn things about a given group is to let it act. Let this group $G$ act on the complex projective line (any projective line over a field would work)
$$
\mathbb{P}^1(\mathbb{C})=\{[x:y]\,;\,(x,y)\in\mathbb{C}^2\setminus\{(0,0)\}\}
$$ 
by $g\cdot[x:y]:=[g(x,y)]$, the natural action of $GL(2,\mathbb{C})$ on $\mathbb{P}^1(\mathbb{C})$.
With the three particular points
$$
1=[1:0]\quad 2=[0:1]\quad 3=[1:-1]
$$
you can easily check that $G$ acts faithfully on $\{1,2,3\}$. With this identification:
$$
I=Id\;\; A=(1,2)\; B=(1,2,3)\; C=(2,3)\; D=(1,3,2)\; K=(1,3)
$$
whence $G\simeq S_3$.
