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how to prove that semidirect products are not isomorphicClassifing groups of order 56: problems with the semidirect productIs there a nontrivial semidirect product of two groups isomorphic to their direct product?$ K rtimes_{phi_1} C cong K rtimes_{phi_2} C$ when $phi_1(C), phi_2(C)$ are conjugated and $C$ is a product of two cyclic groupsDoes an isomorphism of groups that can be written as a direct product induce isomorphisms on the factors?Semidirect product when the action factorsDo homomorphisms $H to operatorname{Aut}(K)$ that coincide at the level of $operatorname{Out}(K)$ induce isomorphic semidirect products?A question on classification of groups of order 30Groups of order 56 with Sylow 2-subgroup isomorphic $Q_8$When is $Artimes_{phi_1} B cong Artimes_{phi_2} B$?

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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 $theta : Q' to text{Aut } N'$ 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

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asked Mar 20 at 11:33

user193319user193319

2,4552927

$endgroup$

• 
  
  
  

  
  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 $theta : Q' to text{Aut } N'$ 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

2,4552927

$endgroup$

• 
  
  
  

  
  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 $theta : Q' to text{Aut } N'$ 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

2,4552927

$endgroup$

share|cite|improve this question

asked Mar 20 at 11:33

user193319user193319

2,4552927

share|cite|improve this question

asked Mar 20 at 11:33

user193319user193319

2,4552927

asked Mar 20 at 11:33

user193319user193319

2,4552927

asked Mar 20 at 11:33

user193319user193319

2,4552927