Internal Semidirect with Factors Isomorphic to “Outside” Groups The Next CEO of Stack Overflowhow 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 $H to operatornameAut(K)$ that coincide at the level of $operatornameOut(K)$ 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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Internal Semidirect with Factors Isomorphic to “Outside” Groups
The Next CEO of Stack Overflowhow 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 $H to operatornameAut(K)$ that coincide at the level of $operatornameOut(K)$ 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 $Artimes_phi_1 B cong Artimes_phi_2 B$?
$begingroup$
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
asked Mar 20 at 11:33
user193319user193319
2,4482927
$endgroup$
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
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
asked Mar 20 at 11:33
user193319user193319
2,4482927
$endgroup$
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
group-theory normal-subgroups group-isomorphism semidirect-product automorphism-group
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
asked Mar 20 at 11:33
user193319user193319
2,4482927
asked Mar 20 at 11:33
user193319user193319
2,4482927
2,4482927
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 |
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
1
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
$begingroup$
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
$begingroup$
@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 |
0
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$begingroup$
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