Given $M= left[ {begin{array}{cc} a & b \ c & d \ end{array} } right]$ where $a,b,c,dinmathbb{R}$ and...












0












$begingroup$


For $M= left[ {begin{array}{cc} a & b \ c & d \ end{array} } right]$ where $a,b,c,dinmathbb{R}$ and $f_M(z)=frac{az+b}{cz+d}$. Let $text{SL}(2,mathbb{R})={M:det M=1}$ and $I_{2}$ be the $2times2$ identity matrix.Prove that if $M,M'$ are two $2times2$ matrices with real entries then $f_{MM'}(z)=f_{M}(f_{M'}(z))$.Also, prove that the map $Gamma(M)=f_{M}$ is a group homomorphism from $text{SL}(2,mathbb{R})$ to $text{Aut}(mathbb{H})$ and find its kernel. Use this to prove that $text{Aut}(mathbb{H})$ as a group is isomorphic to $text{SL}(2,mathbb{R})/{pm I_{2}}$ (where the latter is a quotient group.) You may assume that as a set $text{Aut}(mathbb{H})={f_{M}:Mintext{SL}(2,mathbb{R})}$



I tried to find many different sources to help me solve this but I'm honestly clueless. Any help would be great.






abstract-algebra group-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you get for $f_{MM'}(z)$ and $f_{M}(f_{M'}(z))$
    $endgroup$
    – reuns
    Dec 12 '18 at 17:41
















0












$begingroup$


For $M= left[ {begin{array}{cc} a & b \ c & d \ end{array} } right]$ where $a,b,c,dinmathbb{R}$ and $f_M(z)=frac{az+b}{cz+d}$. Let $text{SL}(2,mathbb{R})={M:det M=1}$ and $I_{2}$ be the $2times2$ identity matrix.Prove that if $M,M'$ are two $2times2$ matrices with real entries then $f_{MM'}(z)=f_{M}(f_{M'}(z))$.Also, prove that the map $Gamma(M)=f_{M}$ is a group homomorphism from $text{SL}(2,mathbb{R})$ to $text{Aut}(mathbb{H})$ and find its kernel. Use this to prove that $text{Aut}(mathbb{H})$ as a group is isomorphic to $text{SL}(2,mathbb{R})/{pm I_{2}}$ (where the latter is a quotient group.) You may assume that as a set $text{Aut}(mathbb{H})={f_{M}:Mintext{SL}(2,mathbb{R})}$



I tried to find many different sources to help me solve this but I'm honestly clueless. Any help would be great.






abstract-algebra group-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    What do you get for $f_{MM'}(z)$ and $f_{M}(f_{M'}(z))$
    $endgroup$
    – reuns
    Dec 12 '18 at 17:41














0












0








0





$begingroup$


For $M= left[ {begin{array}{cc} a & b \ c & d \ end{array} } right]$ where $a,b,c,dinmathbb{R}$ and $f_M(z)=frac{az+b}{cz+d}$. Let $text{SL}(2,mathbb{R})={M:det M=1}$ and $I_{2}$ be the $2times2$ identity matrix.Prove that if $M,M'$ are two $2times2$ matrices with real entries then $f_{MM'}(z)=f_{M}(f_{M'}(z))$.Also, prove that the map $Gamma(M)=f_{M}$ is a group homomorphism from $text{SL}(2,mathbb{R})$ to $text{Aut}(mathbb{H})$ and find its kernel. Use this to prove that $text{Aut}(mathbb{H})$ as a group is isomorphic to $text{SL}(2,mathbb{R})/{pm I_{2}}$ (where the latter is a quotient group.) You may assume that as a set $text{Aut}(mathbb{H})={f_{M}:Mintext{SL}(2,mathbb{R})}$



I tried to find many different sources to help me solve this but I'm honestly clueless. Any help would be great.






abstract-algebra group-theory







share|cite|improve this question









$endgroup$




For $M= left[ {begin{array}{cc} a & b \ c & d \ end{array} } right]$ where $a,b,c,dinmathbb{R}$ and $f_M(z)=frac{az+b}{cz+d}$. Let $text{SL}(2,mathbb{R})={M:det M=1}$ and $I_{2}$ be the $2times2$ identity matrix.Prove that if $M,M'$ are two $2times2$ matrices with real entries then $f_{MM'}(z)=f_{M}(f_{M'}(z))$.Also, prove that the map $Gamma(M)=f_{M}$ is a group homomorphism from $text{SL}(2,mathbb{R})$ to $text{Aut}(mathbb{H})$ and find its kernel. Use this to prove that $text{Aut}(mathbb{H})$ as a group is isomorphic to $text{SL}(2,mathbb{R})/{pm I_{2}}$ (where the latter is a quotient group.) You may assume that as a set $text{Aut}(mathbb{H})={f_{M}:Mintext{SL}(2,mathbb{R})}$



I tried to find many different sources to help me solve this but I'm honestly clueless. Any help would be great.






abstract-algebra group-theory




abstract-algebra group-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 12 '18 at 17:30









Ya GYa G

469211




469211












  • $begingroup$
    What do you get for $f_{MM'}(z)$ and $f_{M}(f_{M'}(z))$
    $endgroup$
    – reuns
    Dec 12 '18 at 17:41


















  • $begingroup$
    What do you get for $f_{MM'}(z)$ and $f_{M}(f_{M'}(z))$
    $endgroup$
    – reuns
    Dec 12 '18 at 17:41
















$begingroup$
What do you get for $f_{MM'}(z)$ and $f_{M}(f_{M'}(z))$
$endgroup$
– reuns
Dec 12 '18 at 17:41




$begingroup$
What do you get for $f_{MM'}(z)$ and $f_{M}(f_{M'}(z))$
$endgroup$
– reuns
Dec 12 '18 at 17:41










1 Answer
1






active

oldest

votes


















2












$begingroup$

Well, it is simple to prove that
$$f_{MM'}=f_{M}circ f_{M'}$$
just by tedious algebra and simplifications, which I will leave to you. To prove that $Gamma: SL(2,mathbb{R})to text{Aut}(mathbb{H}): Mmapsto f_M$ is a homeomorphism we need to show it sends the identity to the identity and that it respects group operations. From the above we see that
$$Gamma(MM')=f_{MM'}=f_Mcirc f_{M'}=Gamma(M)circ Gamma(M').$$
To show it maps the identity to the identity we simply have
$$Gammaleft(begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)=frac{1cdot z+0}{0cdot z+1}=z.$$
To check the kernel we look for which $M$ we have that $Gamma(M)=z$. Since
$$Gammaleft(begin{bmatrix} a & b \ c & dend{bmatrix}right)=frac{a z+b}{c z+d}$$
we are looking for when
$$frac{a z+b}{c z+d}=z.$$
This is the same as $az+b=cz^2+dz$. For these polynomials to be the same, their coefficients must match, so $c=0$, $d=a$ and $b=0$. Thus $Gamma(M)=z$ iff $M=aI$. Since $det(aI)=a^2$, $aIin SL(2,mathbb{R})$ iff $a=pm 1$. Thus $ker(Gamma)=pm I_2$. The fact that
$$text{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(Gamma)=SL(2,mathbb{R})/{pm I_2}$$
then follows.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    With what theorem do we know $$hbox{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(/Gamma)$$?
    $endgroup$
    – Ya G
    Dec 13 '18 at 19:31










  • $begingroup$
    It’s the first isomorphism theorem using the assumption we’re given that $text{Im}(Gamma)=text{Aut}(mathbb{H})$.
    $endgroup$
    – Will Fisher
    Dec 13 '18 at 21:30











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









2












$begingroup$

Well, it is simple to prove that
$$f_{MM'}=f_{M}circ f_{M'}$$
just by tedious algebra and simplifications, which I will leave to you. To prove that $Gamma: SL(2,mathbb{R})to text{Aut}(mathbb{H}): Mmapsto f_M$ is a homeomorphism we need to show it sends the identity to the identity and that it respects group operations. From the above we see that
$$Gamma(MM')=f_{MM'}=f_Mcirc f_{M'}=Gamma(M)circ Gamma(M').$$
To show it maps the identity to the identity we simply have
$$Gammaleft(begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)=frac{1cdot z+0}{0cdot z+1}=z.$$
To check the kernel we look for which $M$ we have that $Gamma(M)=z$. Since
$$Gammaleft(begin{bmatrix} a & b \ c & dend{bmatrix}right)=frac{a z+b}{c z+d}$$
we are looking for when
$$frac{a z+b}{c z+d}=z.$$
This is the same as $az+b=cz^2+dz$. For these polynomials to be the same, their coefficients must match, so $c=0$, $d=a$ and $b=0$. Thus $Gamma(M)=z$ iff $M=aI$. Since $det(aI)=a^2$, $aIin SL(2,mathbb{R})$ iff $a=pm 1$. Thus $ker(Gamma)=pm I_2$. The fact that
$$text{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(Gamma)=SL(2,mathbb{R})/{pm I_2}$$
then follows.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    With what theorem do we know $$hbox{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(/Gamma)$$?
    $endgroup$
    – Ya G
    Dec 13 '18 at 19:31










  • $begingroup$
    It’s the first isomorphism theorem using the assumption we’re given that $text{Im}(Gamma)=text{Aut}(mathbb{H})$.
    $endgroup$
    – Will Fisher
    Dec 13 '18 at 21:30
















2












$begingroup$

Well, it is simple to prove that
$$f_{MM'}=f_{M}circ f_{M'}$$
just by tedious algebra and simplifications, which I will leave to you. To prove that $Gamma: SL(2,mathbb{R})to text{Aut}(mathbb{H}): Mmapsto f_M$ is a homeomorphism we need to show it sends the identity to the identity and that it respects group operations. From the above we see that
$$Gamma(MM')=f_{MM'}=f_Mcirc f_{M'}=Gamma(M)circ Gamma(M').$$
To show it maps the identity to the identity we simply have
$$Gammaleft(begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)=frac{1cdot z+0}{0cdot z+1}=z.$$
To check the kernel we look for which $M$ we have that $Gamma(M)=z$. Since
$$Gammaleft(begin{bmatrix} a & b \ c & dend{bmatrix}right)=frac{a z+b}{c z+d}$$
we are looking for when
$$frac{a z+b}{c z+d}=z.$$
This is the same as $az+b=cz^2+dz$. For these polynomials to be the same, their coefficients must match, so $c=0$, $d=a$ and $b=0$. Thus $Gamma(M)=z$ iff $M=aI$. Since $det(aI)=a^2$, $aIin SL(2,mathbb{R})$ iff $a=pm 1$. Thus $ker(Gamma)=pm I_2$. The fact that
$$text{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(Gamma)=SL(2,mathbb{R})/{pm I_2}$$
then follows.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    With what theorem do we know $$hbox{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(/Gamma)$$?
    $endgroup$
    – Ya G
    Dec 13 '18 at 19:31










  • $begingroup$
    It’s the first isomorphism theorem using the assumption we’re given that $text{Im}(Gamma)=text{Aut}(mathbb{H})$.
    $endgroup$
    – Will Fisher
    Dec 13 '18 at 21:30














2












2








2





$begingroup$

Well, it is simple to prove that
$$f_{MM'}=f_{M}circ f_{M'}$$
just by tedious algebra and simplifications, which I will leave to you. To prove that $Gamma: SL(2,mathbb{R})to text{Aut}(mathbb{H}): Mmapsto f_M$ is a homeomorphism we need to show it sends the identity to the identity and that it respects group operations. From the above we see that
$$Gamma(MM')=f_{MM'}=f_Mcirc f_{M'}=Gamma(M)circ Gamma(M').$$
To show it maps the identity to the identity we simply have
$$Gammaleft(begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)=frac{1cdot z+0}{0cdot z+1}=z.$$
To check the kernel we look for which $M$ we have that $Gamma(M)=z$. Since
$$Gammaleft(begin{bmatrix} a & b \ c & dend{bmatrix}right)=frac{a z+b}{c z+d}$$
we are looking for when
$$frac{a z+b}{c z+d}=z.$$
This is the same as $az+b=cz^2+dz$. For these polynomials to be the same, their coefficients must match, so $c=0$, $d=a$ and $b=0$. Thus $Gamma(M)=z$ iff $M=aI$. Since $det(aI)=a^2$, $aIin SL(2,mathbb{R})$ iff $a=pm 1$. Thus $ker(Gamma)=pm I_2$. The fact that
$$text{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(Gamma)=SL(2,mathbb{R})/{pm I_2}$$
then follows.






share|cite|improve this answer









$endgroup$



Well, it is simple to prove that
$$f_{MM'}=f_{M}circ f_{M'}$$
just by tedious algebra and simplifications, which I will leave to you. To prove that $Gamma: SL(2,mathbb{R})to text{Aut}(mathbb{H}): Mmapsto f_M$ is a homeomorphism we need to show it sends the identity to the identity and that it respects group operations. From the above we see that
$$Gamma(MM')=f_{MM'}=f_Mcirc f_{M'}=Gamma(M)circ Gamma(M').$$
To show it maps the identity to the identity we simply have
$$Gammaleft(begin{bmatrix} 1 & 0 \ 0 & 1end{bmatrix}right)=frac{1cdot z+0}{0cdot z+1}=z.$$
To check the kernel we look for which $M$ we have that $Gamma(M)=z$. Since
$$Gammaleft(begin{bmatrix} a & b \ c & dend{bmatrix}right)=frac{a z+b}{c z+d}$$
we are looking for when
$$frac{a z+b}{c z+d}=z.$$
This is the same as $az+b=cz^2+dz$. For these polynomials to be the same, their coefficients must match, so $c=0$, $d=a$ and $b=0$. Thus $Gamma(M)=z$ iff $M=aI$. Since $det(aI)=a^2$, $aIin SL(2,mathbb{R})$ iff $a=pm 1$. Thus $ker(Gamma)=pm I_2$. The fact that
$$text{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(Gamma)=SL(2,mathbb{R})/{pm I_2}$$
then follows.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 17:44









Will FisherWill Fisher

4,0381032




4,0381032












  • $begingroup$
    With what theorem do we know $$hbox{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(/Gamma)$$?
    $endgroup$
    – Ya G
    Dec 13 '18 at 19:31










  • $begingroup$
    It’s the first isomorphism theorem using the assumption we’re given that $text{Im}(Gamma)=text{Aut}(mathbb{H})$.
    $endgroup$
    – Will Fisher
    Dec 13 '18 at 21:30


















  • $begingroup$
    With what theorem do we know $$hbox{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(/Gamma)$$?
    $endgroup$
    – Ya G
    Dec 13 '18 at 19:31










  • $begingroup$
    It’s the first isomorphism theorem using the assumption we’re given that $text{Im}(Gamma)=text{Aut}(mathbb{H})$.
    $endgroup$
    – Will Fisher
    Dec 13 '18 at 21:30
















$begingroup$
With what theorem do we know $$hbox{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(/Gamma)$$?
$endgroup$
– Ya G
Dec 13 '18 at 19:31




$begingroup$
With what theorem do we know $$hbox{Aut}(mathbb{H})cong SL(2,mathbb{R})/ker(/Gamma)$$?
$endgroup$
– Ya G
Dec 13 '18 at 19:31












$begingroup$
It’s the first isomorphism theorem using the assumption we’re given that $text{Im}(Gamma)=text{Aut}(mathbb{H})$.
$endgroup$
– Will Fisher
Dec 13 '18 at 21:30




$begingroup$
It’s the first isomorphism theorem using the assumption we’re given that $text{Im}(Gamma)=text{Aut}(mathbb{H})$.
$endgroup$
– Will Fisher
Dec 13 '18 at 21:30


















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