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Invertible matrices are closed under multiplication, contain $I$, have associative multiplication, and have inverses by the adjugate formula over the field ${\mathbb{F}}_p$. The first column can be any nonzero vector and the second any vector outside its span, so

For $p=2$, the matrix $\left(\begin{array}{cc}0& 1\\ 1& 1\end{array}\right)$ has order three. There is no element of order six: $G{L}_2({\mathbb{F}}_2)\cong S_3$, whose element orders are $1,2,3$.

For $p>2$, $S{L}_2({\mathbb{F}}_p)=\ker \det$ is a proper normal subgroup. For $p=2$, the order-three subgroup in the copy of $S_3$ is proper and normal.

In $G{L}_2({\mathbb{F}}_{11})$, takewhere $A\le {\mathbb{F}}_{11}^{\times }$ is the subgroup of order five. This semidirect product has order $55$ and is nonabelian because a nontrivial diagonal element does not commute with translations.

For $f(z)=(az+b)/(cz+d)$, finite fixed points obeywith the point at infinity included in the usual way when appropriate. The fundamental theorem of algebra on the Riemann sphere gives at least one fixed point. Unless $f$ is the identity, the equation is nonzero of degree at most two, so there are one or two distinct fixed points.

Let $\zeta$ be a primitive $m$th root of unity. For every $u\in \mathbb{C}$,has order $m$, and these transformations are distinct as $u$ varies.

Projectivising matrices is a homomorphism. Thus $B=CA{C}^{-1}$ in $S{L}_2(\mathbb{C})$ implies $g=[C]f[C{]}^{-1}$ in the Möbius group.

The converse as stated is false because matrix representatives may be rescaled: $A=I$ and $B=2I$ define the same Möbius transformation, while neither $B$ nor $-B$ is conjugate to $A$ (their traces differ).

Green's theorem states, for a positively oriented simple boundary $C=\partial R$,The defining polynomial becomesso the region is the cardioid

The parametrisation $r(x,y)=(x,y,F(x,y))$ has area factorso $\mathrm{area}(S)={\iint }_D\sqrt{1+F_x^2+F_y^2}dxdy$.

For the unbounded saddle $F=(x^2-y^2)/2$, this factor is $\sqrt{1+x^2+y^2}$. Thus the requested integral iswhich converges because its radial tail is $O(r^{-2})$.