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Internal Semidirect with Factors Isomorphic to “Outside” Groups

2

1

$begingroup$

Here is a conjecture of mine:

If $G$ is the internal semidirect product of $N unlhd G$ and $Q le G$, and $phi_1 : N' to N$ and $phi_2 : Q' to Q$ are isomorphisms, then there is some $theta : Q' to textAut N'$ such that $G cong N' rtimes_theta Q'$

My thought was to take $theta (x) = i_x$ with $i_x(y) = phi_1( phi_2(x) phi_1(y) phi_2(x)^-1)$; and then show that $phi : G to N' rtimes_theta Q'$ given by $phi(nq) = (phi_1^-1(n),phi_2^-1(q))$ . Assuming that $theta$ is a homomorphism, I was able to show that $phi$ is an isomorphism. However, when I went back to verify that $theta$ is in fact a homomorphism, I ran into seemingly insuperable difficulties. Is $theta$ as I have defined it a homomorphism? Is it the "right" homomorphism?

group-theory normal-subgroups group-isomorphism semidirect-product automorphism-group

share|cite|improve this question

asked Mar 20 at 11:33

user193319user193319

2,4482927

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean $i_x(y) = phi_1^-1( phi_2(x) phi_1(y) phi_2(x)^-1)$?
  $endgroup$
  – Arnaud D.
  Mar 20 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArnaudD. Dang it! That was my problem. Without that inverse I wasn't getting the right cancellation.
  $endgroup$
  – user193319
  Mar 20 at 11:50
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Here is a conjecture of mine:

If $G$ is the internal semidirect product of $N unlhd G$ and $Q le G$, and $phi_1 : N' to N$ and $phi_2 : Q' to Q$ are isomorphisms, then there is some $theta : Q' to textAut N'$ such that $G cong N' rtimes_theta Q'$

My thought was to take $theta (x) = i_x$ with $i_x(y) = phi_1( phi_2(x) phi_1(y) phi_2(x)^-1)$; and then show that $phi : G to N' rtimes_theta Q'$ given by $phi(nq) = (phi_1^-1(n),phi_2^-1(q))$ . Assuming that $theta$ is a homomorphism, I was able to show that $phi$ is an isomorphism. However, when I went back to verify that $theta$ is in fact a homomorphism, I ran into seemingly insuperable difficulties. Is $theta$ as I have defined it a homomorphism? Is it the "right" homomorphism?

group-theory normal-subgroups group-isomorphism semidirect-product automorphism-group

share|cite|improve this question

asked Mar 20 at 11:33

user193319user193319

2,4482927

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean $i_x(y) = phi_1^-1( phi_2(x) phi_1(y) phi_2(x)^-1)$?
  $endgroup$
  – Arnaud D.
  Mar 20 at 11:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ArnaudD. Dang it! That was my problem. Without that inverse I wasn't getting the right cancellation.
  $endgroup$
  – user193319
  Mar 20 at 11:50
  

  
  
  
  

add a comment | 

$begingroup$

Here is a conjecture of mine:

If $G$ is the internal semidirect product of $N unlhd G$ and $Q le G$, and $phi_1 : N' to N$ and $phi_2 : Q' to Q$ are isomorphisms, then there is some $theta : Q' to textAut N'$ such that $G cong N' rtimes_theta Q'$

My thought was to take $theta (x) = i_x$ with $i_x(y) = phi_1( phi_2(x) phi_1(y) phi_2(x)^-1)$; and then show that $phi : G to N' rtimes_theta Q'$ given by $phi(nq) = (phi_1^-1(n),phi_2^-1(q))$ . Assuming that $theta$ is a homomorphism, I was able to show that $phi$ is an isomorphism. However, when I went back to verify that $theta$ is in fact a homomorphism, I ran into seemingly insuperable difficulties. Is $theta$ as I have defined it a homomorphism? Is it the "right" homomorphism?

group-theory normal-subgroups group-isomorphism semidirect-product automorphism-group

share|cite|improve this question

asked Mar 20 at 11:33

user193319user193319

2,4482927

$endgroup$

share|cite|improve this question

asked Mar 20 at 11:33

user193319user193319

2,4482927

share|cite|improve this question

asked Mar 20 at 11:33

user193319user193319

2,4482927

asked Mar 20 at 11:33

user193319user193319

2,4482927

asked Mar 20 at 11:33

user193319user193319

2,4482927