Proving $H_1 rtimes_{theta_2}K simeq H_1 rtimes_{theta_1}K$
$begingroup$
Let $K=C_p$ be a cyclic group of order $p$ (prime). Let $H_1 = C_p times C_p$, and $theta_1,theta_2 : K to Aut(H_1)$ two homomorphisms. Denote $G_1 = H_1 rtimes_{theta_1}K$ and $G_2 = H_1 rtimes_{theta_2}K$.
Assume $G_1, G_2$ are nonabelian. Prove: $G_1 simeq G_2$
I tried to check what are the (non trivial) homomorphisms $theta :C_p to Aut(C_p times C_p)$, but I can't find any useful property. Any idea how can I characterize those homomorphisms?
abstract-algebra group-theory semidirect-product
asked Dec 22 '18 at 9:56
user401516user401516
1,036311
$endgroup$
add a comment |
$begingroup$
Let $K=C_p$ be a cyclic group of order $p$ (prime). Let $H_1 = C_p times C_p$, and $theta_1,theta_2 : K to Aut(H_1)$ two homomorphisms. Denote $G_1 = H_1 rtimes_{theta_1}K$ and $G_2 = H_1 rtimes_{theta_2}K$.
Assume $G_1, G_2$ are nonabelian. Prove: $G_1 simeq G_2$
I tried to check what are the (non trivial) homomorphisms $theta :C_p to Aut(C_p times C_p)$, but I can't find any useful property. Any idea how can I characterize those homomorphisms?
abstract-algebra group-theory semidirect-product
asked Dec 22 '18 at 9:56
user401516user401516
1,036311
$endgroup$
1
$begingroup$
As stated this is not true, because $theta_1$ could be trivial and $theta_2$ nontrivial. It is true if you assume that $theta_1$ and $theta_2$ are both nontrivial, in which case $G_1$ and $G_2$ are nonabelian groups of order $p^3$.
$endgroup$
– Derek Holt
Dec 22 '18 at 10:39
$begingroup$
Then if $G_1$ and $G_2$ are nonabelian, then $G_1 = G_2 $? does that mean it is true for every $H_1,K$, not only for $K=C_p$, $H_1=C_p times C_p$ ?
$endgroup$
– user401516
Dec 22 '18 at 10:41
$begingroup$
No, why should it mean that?
$endgroup$
– Derek Holt
Dec 22 '18 at 10:54
$begingroup$
First, I edited my question so it is true now ($G_1$ and $G_2$ are nonabelian). As I wrote in my question, I don't think it is true, but I'm rather confused, since it seems like $G_1 = H' rtimes K' = G_2$. What am I missing?
$endgroup$
– user401516
Dec 22 '18 at 11:01
$begingroup$
Perhaps using presentations might help. Here's what a presentation of a semidirect product looks like.
$endgroup$
– Shaun
Dec 22 '18 at 14:38
add a comment |
$begingroup$
Let $K=C_p$ be a cyclic group of order $p$ (prime). Let $H_1 = C_p times C_p$, and $theta_1,theta_2 : K to Aut(H_1)$ two homomorphisms. Denote $G_1 = H_1 rtimes_{theta_1}K$ and $G_2 = H_1 rtimes_{theta_2}K$.
Assume $G_1, G_2$ are nonabelian. Prove: $G_1 simeq G_2$
I tried to check what are the (non trivial) homomorphisms $theta :C_p to Aut(C_p times C_p)$, but I can't find any useful property. Any idea how can I characterize those homomorphisms?
abstract-algebra group-theory semidirect-product
asked Dec 22 '18 at 9:56
user401516user401516
1,036311
$endgroup$
Let $K=C_p$ be a cyclic group of order $p$ (prime). Let $H_1 = C_p times C_p$, and $theta_1,theta_2 : K to Aut(H_1)$ two homomorphisms. Denote $G_1 = H_1 rtimes_{theta_1}K$ and $G_2 = H_1 rtimes_{theta_2}K$.
Assume $G_1, G_2$ are nonabelian. Prove: $G_1 simeq G_2$
I tried to check what are the (non trivial) homomorphisms $theta :C_p to Aut(C_p times C_p)$, but I can't find any useful property. Any idea how can I characterize those homomorphisms?
abstract-algebra group-theory semidirect-product
abstract-algebra group-theory semidirect-product
asked Dec 22 '18 at 9:56
user401516user401516
1,036311
asked Dec 22 '18 at 9:56
user401516user401516
1,036311
edited Dec 23 '18 at 20:34
user401516
asked Dec 22 '18 at 9:56
user401516user401516
1,036311
asked Dec 22 '18 at 9:56
user401516user401516
1,036311
asked Dec 22 '18 at 9:56
user401516user401516
1,036311
1,036311
1
$begingroup$
As stated this is not true, because $theta_1$ could be trivial and $theta_2$ nontrivial. It is true if you assume that $theta_1$ and $theta_2$ are both nontrivial, in which case $G_1$ and $G_2$ are nonabelian groups of order $p^3$.
$endgroup$
– Derek Holt
Dec 22 '18 at 10:39
$begingroup$
Then if $G_1$ and $G_2$ are nonabelian, then $G_1 = G_2 $? does that mean it is true for every $H_1,K$, not only for $K=C_p$, $H_1=C_p times C_p$ ?
$endgroup$
– user401516
Dec 22 '18 at 10:41
$begingroup$
No, why should it mean that?
$endgroup$
– Derek Holt
Dec 22 '18 at 10:54
$begingroup$
First, I edited my question so it is true now ($G_1$ and $G_2$ are nonabelian). As I wrote in my question, I don't think it is true, but I'm rather confused, since it seems like $G_1 = H' rtimes K' = G_2$. What am I missing?
$endgroup$
– user401516
Dec 22 '18 at 11:01
$begingroup$
Perhaps using presentations might help. Here's what a presentation of a semidirect product looks like.
$endgroup$
– Shaun
Dec 22 '18 at 14:38
add a comment |
1
$begingroup$
As stated this is not true, because $theta_1$ could be trivial and $theta_2$ nontrivial. It is true if you assume that $theta_1$ and $theta_2$ are both nontrivial, in which case $G_1$ and $G_2$ are nonabelian groups of order $p^3$.
$endgroup$
– Derek Holt
Dec 22 '18 at 10:39
$begingroup$
Then if $G_1$ and $G_2$ are nonabelian, then $G_1 = G_2 $? does that mean it is true for every $H_1,K$, not only for $K=C_p$, $H_1=C_p times C_p$ ?
$endgroup$
– user401516
Dec 22 '18 at 10:41
$begingroup$
No, why should it mean that?
$endgroup$
– Derek Holt
Dec 22 '18 at 10:54
$begingroup$
First, I edited my question so it is true now ($G_1$ and $G_2$ are nonabelian). As I wrote in my question, I don't think it is true, but I'm rather confused, since it seems like $G_1 = H' rtimes K' = G_2$. What am I missing?
$endgroup$
– user401516
Dec 22 '18 at 11:01
$begingroup$
Perhaps using presentations might help. Here's what a presentation of a semidirect product looks like.
$endgroup$
– Shaun
Dec 22 '18 at 14:38
1
1
$begingroup$
As stated this is not true, because $theta_1$ could be trivial and $theta_2$ nontrivial. It is true if you assume that $theta_1$ and $theta_2$ are both nontrivial, in which case $G_1$ and $G_2$ are nonabelian groups of order $p^3$.
$endgroup$
– Derek Holt
Dec 22 '18 at 10:39
$begingroup$
As stated this is not true, because $theta_1$ could be trivial and $theta_2$ nontrivial. It is true if you assume that $theta_1$ and $theta_2$ are both nontrivial, in which case $G_1$ and $G_2$ are nonabelian groups of order $p^3$.
$endgroup$
– Derek Holt
Dec 22 '18 at 10:39
$begingroup$
Then if $G_1$ and $G_2$ are nonabelian, then $G_1 = G_2 $? does that mean it is true for every $H_1,K$, not only for $K=C_p$, $H_1=C_p times C_p$ ?
$endgroup$
– user401516
Dec 22 '18 at 10:41
$begingroup$
Then if $G_1$ and $G_2$ are nonabelian, then $G_1 = G_2 $? does that mean it is true for every $H_1,K$, not only for $K=C_p$, $H_1=C_p times C_p$ ?
$endgroup$
– user401516
Dec 22 '18 at 10:41
$begingroup$
No, why should it mean that?
$endgroup$
– Derek Holt
Dec 22 '18 at 10:54
$begingroup$
No, why should it mean that?
$endgroup$
– Derek Holt
Dec 22 '18 at 10:54
$begingroup$
First, I edited my question so it is true now ($G_1$ and $G_2$ are nonabelian). As I wrote in my question, I don't think it is true, but I'm rather confused, since it seems like $G_1 = H' rtimes K' = G_2$. What am I missing?
$endgroup$
– user401516
Dec 22 '18 at 11:01
$begingroup$
First, I edited my question so it is true now ($G_1$ and $G_2$ are nonabelian). As I wrote in my question, I don't think it is true, but I'm rather confused, since it seems like $G_1 = H' rtimes K' = G_2$. What am I missing?
$endgroup$
– user401516
Dec 22 '18 at 11:01
$begingroup$
Perhaps using presentations might help. Here's what a presentation of a semidirect product looks like.
$endgroup$
– Shaun
Dec 22 '18 at 14:38
$begingroup$
Perhaps using presentations might help. Here's what a presentation of a semidirect product looks like.
$endgroup$
– Shaun
Dec 22 '18 at 14:38
add a comment |
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$begingroup$
As stated this is not true, because $theta_1$ could be trivial and $theta_2$ nontrivial. It is true if you assume that $theta_1$ and $theta_2$ are both nontrivial, in which case $G_1$ and $G_2$ are nonabelian groups of order $p^3$.
$endgroup$
– Derek Holt
Dec 22 '18 at 10:39
$begingroup$
Then if $G_1$ and $G_2$ are nonabelian, then $G_1 = G_2 $? does that mean it is true for every $H_1,K$, not only for $K=C_p$, $H_1=C_p times C_p$ ?
$endgroup$
– user401516
Dec 22 '18 at 10:41
$begingroup$
No, why should it mean that?
$endgroup$
– Derek Holt
Dec 22 '18 at 10:54
$begingroup$
First, I edited my question so it is true now ($G_1$ and $G_2$ are nonabelian). As I wrote in my question, I don't think it is true, but I'm rather confused, since it seems like $G_1 = H' rtimes K' = G_2$. What am I missing?
$endgroup$
– user401516
Dec 22 '18 at 11:01
$begingroup$
Perhaps using presentations might help. Here's what a presentation of a semidirect product looks like.
$endgroup$
– Shaun
Dec 22 '18 at 14:38