Construction of an isomorphism between certain subgroups of $GL_2(mathbb{C})$ and $S_4.$
$begingroup$
Consider the following matrices $A:=$$ begin{pmatrix}
0 & 1 \
-1 & 0
end{pmatrix} , B: =begin{pmatrix}
0 & 1 \
1 & 0
end{pmatrix}$ of multiplicative group $GL_2(mathbb{C})$ and $sigma = (1234), rho = (24) in S_4.$ Let $G_1 = left langle {A,B}right rangle$ and $G_2= left langle {sigma,rho}right rangle.$ Is $G_1 cong G_2?$
Here is my proof. But I have trouble showing $G_1 cong G_2.$ Could someone advise please? Thank you.
Proof:
By direct computation, $o(A)=4$ and $o(B)=2.$ Since $A^i neq B (i=0,1,2,3),$ $G_1 = {A^{i}B^{j}| i=0,1,2,3 wedge j=0,1},$ whence $|G_1|=8.$
$sigma^2=(13)(24), sigma^3=(1432), sigma^4= (1).$ So $o(sigma)=4.$
Similarly, $o(rho)=2.$ By similar reasoning above, $G_2= {sigma^irho^j| i =0,1,2,3 wedge j=0,1},$ whence $|G_2|=8.$
Given $A^iB^j in G_1, $ let $A^iB^j mapsto sigma^irho^j.$ Clearly, this mapping is surjective. But how do I show this mapping is injective homomorphism?
finite-groups group-isomorphism
edited Dec 22 '18 at 5:21
Shaun
9,380113684
asked Nov 26 '13 at 14:09
Alexy VincenzoAlexy Vincenzo
2,1803926
$endgroup$
add a comment |
$begingroup$
Consider the following matrices $A:=$$ begin{pmatrix}
0 & 1 \
-1 & 0
end{pmatrix} , B: =begin{pmatrix}
0 & 1 \
1 & 0
end{pmatrix}$ of multiplicative group $GL_2(mathbb{C})$ and $sigma = (1234), rho = (24) in S_4.$ Let $G_1 = left langle {A,B}right rangle$ and $G_2= left langle {sigma,rho}right rangle.$ Is $G_1 cong G_2?$
Here is my proof. But I have trouble showing $G_1 cong G_2.$ Could someone advise please? Thank you.
Proof:
By direct computation, $o(A)=4$ and $o(B)=2.$ Since $A^i neq B (i=0,1,2,3),$ $G_1 = {A^{i}B^{j}| i=0,1,2,3 wedge j=0,1},$ whence $|G_1|=8.$
$sigma^2=(13)(24), sigma^3=(1432), sigma^4= (1).$ So $o(sigma)=4.$
Similarly, $o(rho)=2.$ By similar reasoning above, $G_2= {sigma^irho^j| i =0,1,2,3 wedge j=0,1},$ whence $|G_2|=8.$
Given $A^iB^j in G_1, $ let $A^iB^j mapsto sigma^irho^j.$ Clearly, this mapping is surjective. But how do I show this mapping is injective homomorphism?
finite-groups group-isomorphism
edited Dec 22 '18 at 5:21
Shaun
9,380113684
asked Nov 26 '13 at 14:09
Alexy VincenzoAlexy Vincenzo
2,1803926
$endgroup$
2
$begingroup$
You have not shown that all elements in the group generated by $A$ and $B$ can be written in the form you give. What about for example $BA$?
$endgroup$
– Tobias Kildetoft
Nov 26 '13 at 14:12
$begingroup$
Thanks for correction. $BAB= A^3.$ Hence, $BA=A^3B^{-1}=A^3B , BA^2=A^3B^{-1}A=A^6B=A^2B, BA^3= A^2BA=A^5B=AB$
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:34
$begingroup$
Also, $rhosigmarho= sigma^{-1}=sigma^3.$ Hence, $rhosigma=sigma^3rho^{-1} = sigma^3rho, rhosigma^2 = sigma^3sigma^{-1}rho= sigma^2rho, rhosigma^3=sigma^2rhosigma=sigma^5rho=sigmarho. $
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:45
add a comment |
$begingroup$
Consider the following matrices $A:=$$ begin{pmatrix}
0 & 1 \
-1 & 0
end{pmatrix} , B: =begin{pmatrix}
0 & 1 \
1 & 0
end{pmatrix}$ of multiplicative group $GL_2(mathbb{C})$ and $sigma = (1234), rho = (24) in S_4.$ Let $G_1 = left langle {A,B}right rangle$ and $G_2= left langle {sigma,rho}right rangle.$ Is $G_1 cong G_2?$
Here is my proof. But I have trouble showing $G_1 cong G_2.$ Could someone advise please? Thank you.
Proof:
By direct computation, $o(A)=4$ and $o(B)=2.$ Since $A^i neq B (i=0,1,2,3),$ $G_1 = {A^{i}B^{j}| i=0,1,2,3 wedge j=0,1},$ whence $|G_1|=8.$
$sigma^2=(13)(24), sigma^3=(1432), sigma^4= (1).$ So $o(sigma)=4.$
Similarly, $o(rho)=2.$ By similar reasoning above, $G_2= {sigma^irho^j| i =0,1,2,3 wedge j=0,1},$ whence $|G_2|=8.$
Given $A^iB^j in G_1, $ let $A^iB^j mapsto sigma^irho^j.$ Clearly, this mapping is surjective. But how do I show this mapping is injective homomorphism?
finite-groups group-isomorphism
edited Dec 22 '18 at 5:21
Shaun
9,380113684
asked Nov 26 '13 at 14:09
Alexy VincenzoAlexy Vincenzo
2,1803926
$endgroup$
Consider the following matrices $A:=$$ begin{pmatrix}
0 & 1 \
-1 & 0
end{pmatrix} , B: =begin{pmatrix}
0 & 1 \
1 & 0
end{pmatrix}$ of multiplicative group $GL_2(mathbb{C})$ and $sigma = (1234), rho = (24) in S_4.$ Let $G_1 = left langle {A,B}right rangle$ and $G_2= left langle {sigma,rho}right rangle.$ Is $G_1 cong G_2?$
Here is my proof. But I have trouble showing $G_1 cong G_2.$ Could someone advise please? Thank you.
Proof:
By direct computation, $o(A)=4$ and $o(B)=2.$ Since $A^i neq B (i=0,1,2,3),$ $G_1 = {A^{i}B^{j}| i=0,1,2,3 wedge j=0,1},$ whence $|G_1|=8.$
$sigma^2=(13)(24), sigma^3=(1432), sigma^4= (1).$ So $o(sigma)=4.$
Similarly, $o(rho)=2.$ By similar reasoning above, $G_2= {sigma^irho^j| i =0,1,2,3 wedge j=0,1},$ whence $|G_2|=8.$
Given $A^iB^j in G_1, $ let $A^iB^j mapsto sigma^irho^j.$ Clearly, this mapping is surjective. But how do I show this mapping is injective homomorphism?
finite-groups group-isomorphism
finite-groups group-isomorphism
edited Dec 22 '18 at 5:21
Shaun
9,380113684
asked Nov 26 '13 at 14:09
Alexy VincenzoAlexy Vincenzo
2,1803926
edited Dec 22 '18 at 5:21
Shaun
9,380113684
asked Nov 26 '13 at 14:09
Alexy VincenzoAlexy Vincenzo
2,1803926
edited Dec 22 '18 at 5:21
Shaun
9,380113684
edited Dec 22 '18 at 5:21
Shaun
9,380113684
edited Dec 22 '18 at 5:21
Shaun
9,380113684
9,380113684
asked Nov 26 '13 at 14:09
Alexy VincenzoAlexy Vincenzo
2,1803926
asked Nov 26 '13 at 14:09
Alexy VincenzoAlexy Vincenzo
2,1803926
asked Nov 26 '13 at 14:09
Alexy VincenzoAlexy Vincenzo
2,1803926
2,1803926
2
$begingroup$
You have not shown that all elements in the group generated by $A$ and $B$ can be written in the form you give. What about for example $BA$?
$endgroup$
– Tobias Kildetoft
Nov 26 '13 at 14:12
$begingroup$
Thanks for correction. $BAB= A^3.$ Hence, $BA=A^3B^{-1}=A^3B , BA^2=A^3B^{-1}A=A^6B=A^2B, BA^3= A^2BA=A^5B=AB$
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:34
$begingroup$
Also, $rhosigmarho= sigma^{-1}=sigma^3.$ Hence, $rhosigma=sigma^3rho^{-1} = sigma^3rho, rhosigma^2 = sigma^3sigma^{-1}rho= sigma^2rho, rhosigma^3=sigma^2rhosigma=sigma^5rho=sigmarho. $
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:45
add a comment |
2
$begingroup$
You have not shown that all elements in the group generated by $A$ and $B$ can be written in the form you give. What about for example $BA$?
$endgroup$
– Tobias Kildetoft
Nov 26 '13 at 14:12
$begingroup$
Thanks for correction. $BAB= A^3.$ Hence, $BA=A^3B^{-1}=A^3B , BA^2=A^3B^{-1}A=A^6B=A^2B, BA^3= A^2BA=A^5B=AB$
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:34
$begingroup$
Also, $rhosigmarho= sigma^{-1}=sigma^3.$ Hence, $rhosigma=sigma^3rho^{-1} = sigma^3rho, rhosigma^2 = sigma^3sigma^{-1}rho= sigma^2rho, rhosigma^3=sigma^2rhosigma=sigma^5rho=sigmarho. $
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:45
2
2
$begingroup$
You have not shown that all elements in the group generated by $A$ and $B$ can be written in the form you give. What about for example $BA$?
$endgroup$
– Tobias Kildetoft
Nov 26 '13 at 14:12
$begingroup$
You have not shown that all elements in the group generated by $A$ and $B$ can be written in the form you give. What about for example $BA$?
$endgroup$
– Tobias Kildetoft
Nov 26 '13 at 14:12
$begingroup$
Thanks for correction. $BAB= A^3.$ Hence, $BA=A^3B^{-1}=A^3B , BA^2=A^3B^{-1}A=A^6B=A^2B, BA^3= A^2BA=A^5B=AB$
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:34
$begingroup$
Thanks for correction. $BAB= A^3.$ Hence, $BA=A^3B^{-1}=A^3B , BA^2=A^3B^{-1}A=A^6B=A^2B, BA^3= A^2BA=A^5B=AB$
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:34
$begingroup$
Also, $rhosigmarho= sigma^{-1}=sigma^3.$ Hence, $rhosigma=sigma^3rho^{-1} = sigma^3rho, rhosigma^2 = sigma^3sigma^{-1}rho= sigma^2rho, rhosigma^3=sigma^2rhosigma=sigma^5rho=sigmarho. $
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:45
$begingroup$
Also, $rhosigmarho= sigma^{-1}=sigma^3.$ Hence, $rhosigma=sigma^3rho^{-1} = sigma^3rho, rhosigma^2 = sigma^3sigma^{-1}rho= sigma^2rho, rhosigma^3=sigma^2rhosigma=sigma^5rho=sigmarho. $
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:45
add a comment |
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$begingroup$
You have not shown that all elements in the group generated by $A$ and $B$ can be written in the form you give. What about for example $BA$?
$endgroup$
– Tobias Kildetoft
Nov 26 '13 at 14:12
$begingroup$
Thanks for correction. $BAB= A^3.$ Hence, $BA=A^3B^{-1}=A^3B , BA^2=A^3B^{-1}A=A^6B=A^2B, BA^3= A^2BA=A^5B=AB$
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:34
$begingroup$
Also, $rhosigmarho= sigma^{-1}=sigma^3.$ Hence, $rhosigma=sigma^3rho^{-1} = sigma^3rho, rhosigma^2 = sigma^3sigma^{-1}rho= sigma^2rho, rhosigma^3=sigma^2rhosigma=sigma^5rho=sigmarho. $
$endgroup$
– Alexy Vincenzo
Nov 26 '13 at 14:45