Mixing
Möbius transformations form a simple group
$begingroup$
How to show the group $M$ of Möbius transformations is a simple group?
I know: $SL_2(mathbb C)/{+I,-I}cong M$ then if $A lhd M implies phi^{-1}(A) lhd SL_2(mathbb C)/{+I,-I}$.
So if I can show $SL_2(mathbb C)/{+I,-I}$ is simple probably that answers the question.
abstract-algebra complex-analysis
edited May 7 '13 at 11:54
azimut
16.6k1052101
asked Mar 2 '13 at 6:52
jimjim
2,786829
$endgroup$
add a comment |
$begingroup$
How to show the group $M$ of Möbius transformations is a simple group?
I know: $SL_2(mathbb C)/{+I,-I}cong M$ then if $A lhd M implies phi^{-1}(A) lhd SL_2(mathbb C)/{+I,-I}$.
So if I can show $SL_2(mathbb C)/{+I,-I}$ is simple probably that answers the question.
abstract-algebra complex-analysis
edited May 7 '13 at 11:54
azimut
16.6k1052101
asked Mar 2 '13 at 6:52
jimjim
2,786829
$endgroup$
$begingroup$
Actually you want a proof for this: PSL$(2,mathbb C)$ is a simple group. Try google!
$endgroup$
– user26857
Mar 3 '13 at 12:38
$begingroup$
What is $phi$?
$endgroup$
– Julian Kuelshammer
Jun 24 '13 at 19:11
2
$begingroup$
I do not understand the votes to close - is it merely because this is an old, yet unanswered question? Personally, I think it should be left open and that someone should, you know, answer it...
$endgroup$
– user1729
Jun 24 '13 at 19:36
1
$begingroup$
@user1729 I completely agree.
$endgroup$
– Potato
Jun 25 '13 at 18:12
1
$begingroup$
Look at this
$endgroup$
– Bumblebee
Apr 5 '18 at 23:29
add a comment |
$begingroup$
How to show the group $M$ of Möbius transformations is a simple group?
I know: $SL_2(mathbb C)/{+I,-I}cong M$ then if $A lhd M implies phi^{-1}(A) lhd SL_2(mathbb C)/{+I,-I}$.
So if I can show $SL_2(mathbb C)/{+I,-I}$ is simple probably that answers the question.
abstract-algebra complex-analysis
edited May 7 '13 at 11:54
azimut
16.6k1052101
asked Mar 2 '13 at 6:52
jimjim
2,786829
$endgroup$
How to show the group $M$ of Möbius transformations is a simple group?
I know: $SL_2(mathbb C)/{+I,-I}cong M$ then if $A lhd M implies phi^{-1}(A) lhd SL_2(mathbb C)/{+I,-I}$.
So if I can show $SL_2(mathbb C)/{+I,-I}$ is simple probably that answers the question.
abstract-algebra complex-analysis
abstract-algebra complex-analysis
edited May 7 '13 at 11:54
azimut
16.6k1052101
asked Mar 2 '13 at 6:52
jimjim
2,786829
edited May 7 '13 at 11:54
azimut
16.6k1052101
asked Mar 2 '13 at 6:52
jimjim
2,786829
edited May 7 '13 at 11:54
azimut
16.6k1052101
edited May 7 '13 at 11:54
azimut
16.6k1052101
edited May 7 '13 at 11:54
azimut
16.6k1052101
16.6k1052101
asked Mar 2 '13 at 6:52
jimjim
2,786829
asked Mar 2 '13 at 6:52
jimjim
2,786829
asked Mar 2 '13 at 6:52
jimjim
2,786829
2,786829
$begingroup$
Actually you want a proof for this: PSL$(2,mathbb C)$ is a simple group. Try google!
$endgroup$
– user26857
Mar 3 '13 at 12:38
$begingroup$
What is $phi$?
$endgroup$
– Julian Kuelshammer
Jun 24 '13 at 19:11
2
$begingroup$
I do not understand the votes to close - is it merely because this is an old, yet unanswered question? Personally, I think it should be left open and that someone should, you know, answer it...
$endgroup$
– user1729
Jun 24 '13 at 19:36
1
$begingroup$
@user1729 I completely agree.
$endgroup$
– Potato
Jun 25 '13 at 18:12
1
$begingroup$
Look at this
$endgroup$
– Bumblebee
Apr 5 '18 at 23:29
add a comment |
$begingroup$
Actually you want a proof for this: PSL$(2,mathbb C)$ is a simple group. Try google!
$endgroup$
– user26857
Mar 3 '13 at 12:38
$begingroup$
What is $phi$?
$endgroup$
– Julian Kuelshammer
Jun 24 '13 at 19:11
2
$begingroup$
I do not understand the votes to close - is it merely because this is an old, yet unanswered question? Personally, I think it should be left open and that someone should, you know, answer it...
$endgroup$
– user1729
Jun 24 '13 at 19:36
1
$begingroup$
@user1729 I completely agree.
$endgroup$
– Potato
Jun 25 '13 at 18:12
1
$begingroup$
Look at this
$endgroup$
– Bumblebee
Apr 5 '18 at 23:29
$begingroup$
Actually you want a proof for this: PSL$(2,mathbb C)$ is a simple group. Try google!
$endgroup$
– user26857
Mar 3 '13 at 12:38
$begingroup$
Actually you want a proof for this: PSL$(2,mathbb C)$ is a simple group. Try google!
$endgroup$
– user26857
Mar 3 '13 at 12:38
$begingroup$
What is $phi$?
$endgroup$
– Julian Kuelshammer
Jun 24 '13 at 19:11
$begingroup$
What is $phi$?
$endgroup$
– Julian Kuelshammer
Jun 24 '13 at 19:11
2
2
$begingroup$
I do not understand the votes to close - is it merely because this is an old, yet unanswered question? Personally, I think it should be left open and that someone should, you know, answer it...
$endgroup$
– user1729
Jun 24 '13 at 19:36
$begingroup$
I do not understand the votes to close - is it merely because this is an old, yet unanswered question? Personally, I think it should be left open and that someone should, you know, answer it...
$endgroup$
– user1729
Jun 24 '13 at 19:36
1
1
$begingroup$
@user1729 I completely agree.
$endgroup$
– Potato
Jun 25 '13 at 18:12
$begingroup$
@user1729 I completely agree.
$endgroup$
– Potato
Jun 25 '13 at 18:12
1
1
$begingroup$
Look at this
$endgroup$
– Bumblebee
Apr 5 '18 at 23:29
$begingroup$
Look at this
$endgroup$
– Bumblebee
Apr 5 '18 at 23:29
add a comment |
1 Answer
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votes
$begingroup$
You can prove this without having to look at $PSL_2 (mathbb{C})$. First, consider that any Möbius transformation can be expressed as the composition of dilations, translations and the inversion. In particular, if $T(z) =dfrac{az+b}{cz+d}$, then $T = S_4 circ S_3 circ S_2 circ S_1$, where $S_1$ and $S_4$ are translations, $S_3$ a dilation and $S_2$ the inversion. Now, let $N trianglelefteq M$ be non-trivial and suppose $T in N$ is a translation, i.e., $T(z) = z+b$, with $b in mathbb{C^{*}}$ (we don't have to consider $b=0$ because the identity is always there). Let $S(z)=az$, with $a in mathbb{C}^{*}$. We have that $S^{-1}circ T circ S = z+dfrac{b}{a} in N$, so, for any $c in mathbb{C^{*}}$, we may choose $a=b/c$ and obtain every translation. In particular, $R(z) = z-1 in mathbb{C^{*}}$ and if we take $I(z) = 1/z$ to be the inversion, we can obtain that
$$Icirc R circ I^{-1} = dfrac{z}{1-z}$$
$$left(dfrac{z}{1-z}right)circ R = dfrac{z-1}{-z}$$
$$(-I)circ left(dfrac{z-1}{-z}right) circ (-I) = dfrac{1}{z+1}$$
$$left(dfrac{1}{z+1}right) circ R = dfrac{1}{z} in N$$
Then, if we take $I circ S circ I circ S^{-1} = a^2 z$, we can obtain every dilation and so $N=M$. Now, suppose we have $S(z)=az in N$, $ain mathbb{C^{*}-{1}}$. If we take $T(z) = z+b$, $b in mathbb{C^{*}}$, then $ S^{-1} circ T circ S circ T^{-1} = z+ab-b in N$ $Rightarrow z+c in N, forall c in mathbb{C^{*}}$. We have proved the following implications, for $N trianglelefteq M$:
$(text{i) }z+b in N$, for some $b in mathbb{C^{*}} Rightarrow$ $z+c in N, forall c in mathbb{C^{*}} Rightarrow 1/z in N Rightarrow az in N, forall a in mathbb{C^{*}} Rightarrow N = M;$
$(text{ii) }az in N$, for some $a in mathbb{C^{*}-{1}} Rightarrow$ $z+ab-b in N Rightarrow$ $N=M;$
$(text{iii) }1/z in NRightarrow$ $az in N, forall a in mathbb{C^{*}} Rightarrow$ $N=M;$
Finally, it suffices to prove that if we have a general Mobius transformation $T(z) =dfrac{az+b}{cz+d} in N$, then we can obtain a dilation, a translation or a inversion. We have that $T = S_4 circ S_3 circ S_2 circ S_1$. This implies that $ S_1 circ S_4 circ S_3 circ S_2 in N$ and $ S_2 circ S_1 circ S_3 circ S_3 in N$, so $L = (S_1 circ S_3 circ S_3)^2 in N$. Since $L$ is the composition of translations and dilations, it has the form $L(z)=xz+y $, with $x,y in mathbb{C^{*}}$. If $x=1$ and $y=0$, this would imply that $S_1 circ S_4 circ S_3 = z Rightarrow S_4 circ S_3 = S_1^{-1}$ $Rightarrow T = S_1^{-1} circ S_2 circ S_1 Rightarrow 1/z in N$
$ Rightarrow N=M$. If $x=1$ or $y=0$, we are done. Finally, if $xneq 1$ and $y neq 0$, then $$L circ (z-y) circ L^{-1} circ (z+y) = z-xy+y in N$$ and, since $-xy+y neq 0$, $N=M$.
answered Feb 2 at 21:15
NowrasNowras
563
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add a comment |
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$begingroup$
You can prove this without having to look at $PSL_2 (mathbb{C})$. First, consider that any Möbius transformation can be expressed as the composition of dilations, translations and the inversion. In particular, if $T(z) =dfrac{az+b}{cz+d}$, then $T = S_4 circ S_3 circ S_2 circ S_1$, where $S_1$ and $S_4$ are translations, $S_3$ a dilation and $S_2$ the inversion. Now, let $N trianglelefteq M$ be non-trivial and suppose $T in N$ is a translation, i.e., $T(z) = z+b$, with $b in mathbb{C^{*}}$ (we don't have to consider $b=0$ because the identity is always there). Let $S(z)=az$, with $a in mathbb{C}^{*}$. We have that $S^{-1}circ T circ S = z+dfrac{b}{a} in N$, so, for any $c in mathbb{C^{*}}$, we may choose $a=b/c$ and obtain every translation. In particular, $R(z) = z-1 in mathbb{C^{*}}$ and if we take $I(z) = 1/z$ to be the inversion, we can obtain that
$$Icirc R circ I^{-1} = dfrac{z}{1-z}$$
$$left(dfrac{z}{1-z}right)circ R = dfrac{z-1}{-z}$$
$$(-I)circ left(dfrac{z-1}{-z}right) circ (-I) = dfrac{1}{z+1}$$
$$left(dfrac{1}{z+1}right) circ R = dfrac{1}{z} in N$$
Then, if we take $I circ S circ I circ S^{-1} = a^2 z$, we can obtain every dilation and so $N=M$. Now, suppose we have $S(z)=az in N$, $ain mathbb{C^{*}-{1}}$. If we take $T(z) = z+b$, $b in mathbb{C^{*}}$, then $ S^{-1} circ T circ S circ T^{-1} = z+ab-b in N$ $Rightarrow z+c in N, forall c in mathbb{C^{*}}$. We have proved the following implications, for $N trianglelefteq M$:
$(text{i) }z+b in N$, for some $b in mathbb{C^{*}} Rightarrow$ $z+c in N, forall c in mathbb{C^{*}} Rightarrow 1/z in N Rightarrow az in N, forall a in mathbb{C^{*}} Rightarrow N = M;$
$(text{ii) }az in N$, for some $a in mathbb{C^{*}-{1}} Rightarrow$ $z+ab-b in N Rightarrow$ $N=M;$
$(text{iii) }1/z in NRightarrow$ $az in N, forall a in mathbb{C^{*}} Rightarrow$ $N=M;$
Finally, it suffices to prove that if we have a general Mobius transformation $T(z) =dfrac{az+b}{cz+d} in N$, then we can obtain a dilation, a translation or a inversion. We have that $T = S_4 circ S_3 circ S_2 circ S_1$. This implies that $ S_1 circ S_4 circ S_3 circ S_2 in N$ and $ S_2 circ S_1 circ S_3 circ S_3 in N$, so $L = (S_1 circ S_3 circ S_3)^2 in N$. Since $L$ is the composition of translations and dilations, it has the form $L(z)=xz+y $, with $x,y in mathbb{C^{*}}$. If $x=1$ and $y=0$, this would imply that $S_1 circ S_4 circ S_3 = z Rightarrow S_4 circ S_3 = S_1^{-1}$ $Rightarrow T = S_1^{-1} circ S_2 circ S_1 Rightarrow 1/z in N$
$ Rightarrow N=M$. If $x=1$ or $y=0$, we are done. Finally, if $xneq 1$ and $y neq 0$, then $$L circ (z-y) circ L^{-1} circ (z+y) = z-xy+y in N$$ and, since $-xy+y neq 0$, $N=M$.
answered Feb 2 at 21:15
NowrasNowras
563
$endgroup$
add a comment |
$begingroup$
You can prove this without having to look at $PSL_2 (mathbb{C})$. First, consider that any Möbius transformation can be expressed as the composition of dilations, translations and the inversion. In particular, if $T(z) =dfrac{az+b}{cz+d}$, then $T = S_4 circ S_3 circ S_2 circ S_1$, where $S_1$ and $S_4$ are translations, $S_3$ a dilation and $S_2$ the inversion. Now, let $N trianglelefteq M$ be non-trivial and suppose $T in N$ is a translation, i.e., $T(z) = z+b$, with $b in mathbb{C^{*}}$ (we don't have to consider $b=0$ because the identity is always there). Let $S(z)=az$, with $a in mathbb{C}^{*}$. We have that $S^{-1}circ T circ S = z+dfrac{b}{a} in N$, so, for any $c in mathbb{C^{*}}$, we may choose $a=b/c$ and obtain every translation. In particular, $R(z) = z-1 in mathbb{C^{*}}$ and if we take $I(z) = 1/z$ to be the inversion, we can obtain that
$$Icirc R circ I^{-1} = dfrac{z}{1-z}$$
$$left(dfrac{z}{1-z}right)circ R = dfrac{z-1}{-z}$$
$$(-I)circ left(dfrac{z-1}{-z}right) circ (-I) = dfrac{1}{z+1}$$
$$left(dfrac{1}{z+1}right) circ R = dfrac{1}{z} in N$$
Then, if we take $I circ S circ I circ S^{-1} = a^2 z$, we can obtain every dilation and so $N=M$. Now, suppose we have $S(z)=az in N$, $ain mathbb{C^{*}-{1}}$. If we take $T(z) = z+b$, $b in mathbb{C^{*}}$, then $ S^{-1} circ T circ S circ T^{-1} = z+ab-b in N$ $Rightarrow z+c in N, forall c in mathbb{C^{*}}$. We have proved the following implications, for $N trianglelefteq M$:
$(text{i) }z+b in N$, for some $b in mathbb{C^{*}} Rightarrow$ $z+c in N, forall c in mathbb{C^{*}} Rightarrow 1/z in N Rightarrow az in N, forall a in mathbb{C^{*}} Rightarrow N = M;$
$(text{ii) }az in N$, for some $a in mathbb{C^{*}-{1}} Rightarrow$ $z+ab-b in N Rightarrow$ $N=M;$
$(text{iii) }1/z in NRightarrow$ $az in N, forall a in mathbb{C^{*}} Rightarrow$ $N=M;$
Finally, it suffices to prove that if we have a general Mobius transformation $T(z) =dfrac{az+b}{cz+d} in N$, then we can obtain a dilation, a translation or a inversion. We have that $T = S_4 circ S_3 circ S_2 circ S_1$. This implies that $ S_1 circ S_4 circ S_3 circ S_2 in N$ and $ S_2 circ S_1 circ S_3 circ S_3 in N$, so $L = (S_1 circ S_3 circ S_3)^2 in N$. Since $L$ is the composition of translations and dilations, it has the form $L(z)=xz+y $, with $x,y in mathbb{C^{*}}$. If $x=1$ and $y=0$, this would imply that $S_1 circ S_4 circ S_3 = z Rightarrow S_4 circ S_3 = S_1^{-1}$ $Rightarrow T = S_1^{-1} circ S_2 circ S_1 Rightarrow 1/z in N$
$ Rightarrow N=M$. If $x=1$ or $y=0$, we are done. Finally, if $xneq 1$ and $y neq 0$, then $$L circ (z-y) circ L^{-1} circ (z+y) = z-xy+y in N$$ and, since $-xy+y neq 0$, $N=M$.
answered Feb 2 at 21:15
NowrasNowras
563
$endgroup$
add a comment |
$begingroup$
You can prove this without having to look at $PSL_2 (mathbb{C})$. First, consider that any Möbius transformation can be expressed as the composition of dilations, translations and the inversion. In particular, if $T(z) =dfrac{az+b}{cz+d}$, then $T = S_4 circ S_3 circ S_2 circ S_1$, where $S_1$ and $S_4$ are translations, $S_3$ a dilation and $S_2$ the inversion. Now, let $N trianglelefteq M$ be non-trivial and suppose $T in N$ is a translation, i.e., $T(z) = z+b$, with $b in mathbb{C^{*}}$ (we don't have to consider $b=0$ because the identity is always there). Let $S(z)=az$, with $a in mathbb{C}^{*}$. We have that $S^{-1}circ T circ S = z+dfrac{b}{a} in N$, so, for any $c in mathbb{C^{*}}$, we may choose $a=b/c$ and obtain every translation. In particular, $R(z) = z-1 in mathbb{C^{*}}$ and if we take $I(z) = 1/z$ to be the inversion, we can obtain that
$$Icirc R circ I^{-1} = dfrac{z}{1-z}$$
$$left(dfrac{z}{1-z}right)circ R = dfrac{z-1}{-z}$$
$$(-I)circ left(dfrac{z-1}{-z}right) circ (-I) = dfrac{1}{z+1}$$
$$left(dfrac{1}{z+1}right) circ R = dfrac{1}{z} in N$$
Then, if we take $I circ S circ I circ S^{-1} = a^2 z$, we can obtain every dilation and so $N=M$. Now, suppose we have $S(z)=az in N$, $ain mathbb{C^{*}-{1}}$. If we take $T(z) = z+b$, $b in mathbb{C^{*}}$, then $ S^{-1} circ T circ S circ T^{-1} = z+ab-b in N$ $Rightarrow z+c in N, forall c in mathbb{C^{*}}$. We have proved the following implications, for $N trianglelefteq M$:
$(text{i) }z+b in N$, for some $b in mathbb{C^{*}} Rightarrow$ $z+c in N, forall c in mathbb{C^{*}} Rightarrow 1/z in N Rightarrow az in N, forall a in mathbb{C^{*}} Rightarrow N = M;$
$(text{ii) }az in N$, for some $a in mathbb{C^{*}-{1}} Rightarrow$ $z+ab-b in N Rightarrow$ $N=M;$
$(text{iii) }1/z in NRightarrow$ $az in N, forall a in mathbb{C^{*}} Rightarrow$ $N=M;$
Finally, it suffices to prove that if we have a general Mobius transformation $T(z) =dfrac{az+b}{cz+d} in N$, then we can obtain a dilation, a translation or a inversion. We have that $T = S_4 circ S_3 circ S_2 circ S_1$. This implies that $ S_1 circ S_4 circ S_3 circ S_2 in N$ and $ S_2 circ S_1 circ S_3 circ S_3 in N$, so $L = (S_1 circ S_3 circ S_3)^2 in N$. Since $L$ is the composition of translations and dilations, it has the form $L(z)=xz+y $, with $x,y in mathbb{C^{*}}$. If $x=1$ and $y=0$, this would imply that $S_1 circ S_4 circ S_3 = z Rightarrow S_4 circ S_3 = S_1^{-1}$ $Rightarrow T = S_1^{-1} circ S_2 circ S_1 Rightarrow 1/z in N$
$ Rightarrow N=M$. If $x=1$ or $y=0$, we are done. Finally, if $xneq 1$ and $y neq 0$, then $$L circ (z-y) circ L^{-1} circ (z+y) = z-xy+y in N$$ and, since $-xy+y neq 0$, $N=M$.
answered Feb 2 at 21:15
NowrasNowras
563
$endgroup$
You can prove this without having to look at $PSL_2 (mathbb{C})$. First, consider that any Möbius transformation can be expressed as the composition of dilations, translations and the inversion. In particular, if $T(z) =dfrac{az+b}{cz+d}$, then $T = S_4 circ S_3 circ S_2 circ S_1$, where $S_1$ and $S_4$ are translations, $S_3$ a dilation and $S_2$ the inversion. Now, let $N trianglelefteq M$ be non-trivial and suppose $T in N$ is a translation, i.e., $T(z) = z+b$, with $b in mathbb{C^{*}}$ (we don't have to consider $b=0$ because the identity is always there). Let $S(z)=az$, with $a in mathbb{C}^{*}$. We have that $S^{-1}circ T circ S = z+dfrac{b}{a} in N$, so, for any $c in mathbb{C^{*}}$, we may choose $a=b/c$ and obtain every translation. In particular, $R(z) = z-1 in mathbb{C^{*}}$ and if we take $I(z) = 1/z$ to be the inversion, we can obtain that
$$Icirc R circ I^{-1} = dfrac{z}{1-z}$$
$$left(dfrac{z}{1-z}right)circ R = dfrac{z-1}{-z}$$
$$(-I)circ left(dfrac{z-1}{-z}right) circ (-I) = dfrac{1}{z+1}$$
$$left(dfrac{1}{z+1}right) circ R = dfrac{1}{z} in N$$
Then, if we take $I circ S circ I circ S^{-1} = a^2 z$, we can obtain every dilation and so $N=M$. Now, suppose we have $S(z)=az in N$, $ain mathbb{C^{*}-{1}}$. If we take $T(z) = z+b$, $b in mathbb{C^{*}}$, then $ S^{-1} circ T circ S circ T^{-1} = z+ab-b in N$ $Rightarrow z+c in N, forall c in mathbb{C^{*}}$. We have proved the following implications, for $N trianglelefteq M$:
$(text{i) }z+b in N$, for some $b in mathbb{C^{*}} Rightarrow$ $z+c in N, forall c in mathbb{C^{*}} Rightarrow 1/z in N Rightarrow az in N, forall a in mathbb{C^{*}} Rightarrow N = M;$
$(text{ii) }az in N$, for some $a in mathbb{C^{*}-{1}} Rightarrow$ $z+ab-b in N Rightarrow$ $N=M;$
$(text{iii) }1/z in NRightarrow$ $az in N, forall a in mathbb{C^{*}} Rightarrow$ $N=M;$
Finally, it suffices to prove that if we have a general Mobius transformation $T(z) =dfrac{az+b}{cz+d} in N$, then we can obtain a dilation, a translation or a inversion. We have that $T = S_4 circ S_3 circ S_2 circ S_1$. This implies that $ S_1 circ S_4 circ S_3 circ S_2 in N$ and $ S_2 circ S_1 circ S_3 circ S_3 in N$, so $L = (S_1 circ S_3 circ S_3)^2 in N$. Since $L$ is the composition of translations and dilations, it has the form $L(z)=xz+y $, with $x,y in mathbb{C^{*}}$. If $x=1$ and $y=0$, this would imply that $S_1 circ S_4 circ S_3 = z Rightarrow S_4 circ S_3 = S_1^{-1}$ $Rightarrow T = S_1^{-1} circ S_2 circ S_1 Rightarrow 1/z in N$
$ Rightarrow N=M$. If $x=1$ or $y=0$, we are done. Finally, if $xneq 1$ and $y neq 0$, then $$L circ (z-y) circ L^{-1} circ (z+y) = z-xy+y in N$$ and, since $-xy+y neq 0$, $N=M$.
answered Feb 2 at 21:15
NowrasNowras
563
answered Feb 2 at 21:15
NowrasNowras
563
answered Feb 2 at 21:15
NowrasNowras
563
answered Feb 2 at 21:15
NowrasNowras
563
563
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$begingroup$
Actually you want a proof for this: PSL$(2,mathbb C)$ is a simple group. Try google!
$endgroup$
– user26857
Mar 3 '13 at 12:38
$begingroup$
What is $phi$?
$endgroup$
– Julian Kuelshammer
Jun 24 '13 at 19:11
2
$begingroup$
I do not understand the votes to close - is it merely because this is an old, yet unanswered question? Personally, I think it should be left open and that someone should, you know, answer it...
$endgroup$
– user1729
Jun 24 '13 at 19:36
1
$begingroup$
@user1729 I completely agree.
$endgroup$
– Potato
Jun 25 '13 at 18:12
1
$begingroup$
Look at this
$endgroup$
– Bumblebee
Apr 5 '18 at 23:29