Question about a statement concerning normal subgroups
I'm asking the following: it is true that if $K$ is a normal subgroup of $G$ and $Kleq Hleq G$ then $K$ is normal in $H$? I tried to prove it but I failed to do so, so I'm starting to suspect that it is not true. Can you provide me a proof or a counterexample of this statement or hint about its proof?
abstract-algebra group-theory
asked Nov 22 '18 at 17:30
user573497user573497
16619
add a comment |
I'm asking the following: it is true that if $K$ is a normal subgroup of $G$ and $Kleq Hleq G$ then $K$ is normal in $H$? I tried to prove it but I failed to do so, so I'm starting to suspect that it is not true. Can you provide me a proof or a counterexample of this statement or hint about its proof?
abstract-algebra group-theory
asked Nov 22 '18 at 17:30
user573497user573497
16619
add a comment |
I'm asking the following: it is true that if $K$ is a normal subgroup of $G$ and $Kleq Hleq G$ then $K$ is normal in $H$? I tried to prove it but I failed to do so, so I'm starting to suspect that it is not true. Can you provide me a proof or a counterexample of this statement or hint about its proof?
abstract-algebra group-theory
asked Nov 22 '18 at 17:30
user573497user573497
16619
I'm asking the following: it is true that if $K$ is a normal subgroup of $G$ and $Kleq Hleq G$ then $K$ is normal in $H$? I tried to prove it but I failed to do so, so I'm starting to suspect that it is not true. Can you provide me a proof or a counterexample of this statement or hint about its proof?
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 22 '18 at 17:30
user573497user573497
16619
asked Nov 22 '18 at 17:30
user573497user573497
16619
asked Nov 22 '18 at 17:30
user573497user573497
16619
asked Nov 22 '18 at 17:30
user573497user573497
16619
asked Nov 22 '18 at 17:30
user573497user573497
16619
16619
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Hint:
You know that
$$;Klhd Gimplies g^{-1}kgin K;,;;text{for all elements};;kin K,,,,gin G;$$
Now, $;Hsubset G; $, so...
answered Nov 22 '18 at 17:39
DonAntonioDonAntonio
177k1492225
add a comment |
It is trivially true because $Ktrianglelefteq G$ is equivalent to $gKg^{-1}=K,,forall gin G$. This condition is met $forall gin H$, as $Hsubset G$. Note we do not need normality of $H$.
answered Nov 22 '18 at 17:44
Chris CusterChris Custer
10.9k3824
1
Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
– user573497
Nov 22 '18 at 20:02
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
You know that
$$;Klhd Gimplies g^{-1}kgin K;,;;text{for all elements};;kin K,,,,gin G;$$
Now, $;Hsubset G; $, so...
answered Nov 22 '18 at 17:39
DonAntonioDonAntonio
177k1492225
add a comment |
Hint:
You know that
$$;Klhd Gimplies g^{-1}kgin K;,;;text{for all elements};;kin K,,,,gin G;$$
Now, $;Hsubset G; $, so...
answered Nov 22 '18 at 17:39
DonAntonioDonAntonio
177k1492225
add a comment |
Hint:
You know that
$$;Klhd Gimplies g^{-1}kgin K;,;;text{for all elements};;kin K,,,,gin G;$$
Now, $;Hsubset G; $, so...
answered Nov 22 '18 at 17:39
DonAntonioDonAntonio
177k1492225
Hint:
You know that
$$;Klhd Gimplies g^{-1}kgin K;,;;text{for all elements};;kin K,,,,gin G;$$
Now, $;Hsubset G; $, so...
answered Nov 22 '18 at 17:39
DonAntonioDonAntonio
177k1492225
answered Nov 22 '18 at 17:39
DonAntonioDonAntonio
177k1492225
answered Nov 22 '18 at 17:39
DonAntonioDonAntonio
177k1492225
answered Nov 22 '18 at 17:39
DonAntonioDonAntonio
177k1492225
177k1492225
add a comment |
add a comment |
It is trivially true because $Ktrianglelefteq G$ is equivalent to $gKg^{-1}=K,,forall gin G$. This condition is met $forall gin H$, as $Hsubset G$. Note we do not need normality of $H$.
answered Nov 22 '18 at 17:44
Chris CusterChris Custer
10.9k3824
1
Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
– user573497
Nov 22 '18 at 20:02
add a comment |
It is trivially true because $Ktrianglelefteq G$ is equivalent to $gKg^{-1}=K,,forall gin G$. This condition is met $forall gin H$, as $Hsubset G$. Note we do not need normality of $H$.
answered Nov 22 '18 at 17:44
Chris CusterChris Custer
10.9k3824
1
Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
– user573497
Nov 22 '18 at 20:02
add a comment |
It is trivially true because $Ktrianglelefteq G$ is equivalent to $gKg^{-1}=K,,forall gin G$. This condition is met $forall gin H$, as $Hsubset G$. Note we do not need normality of $H$.
answered Nov 22 '18 at 17:44
Chris CusterChris Custer
10.9k3824
It is trivially true because $Ktrianglelefteq G$ is equivalent to $gKg^{-1}=K,,forall gin G$. This condition is met $forall gin H$, as $Hsubset G$. Note we do not need normality of $H$.
answered Nov 22 '18 at 17:44
Chris CusterChris Custer
10.9k3824
answered Nov 22 '18 at 17:44
Chris CusterChris Custer
10.9k3824
answered Nov 22 '18 at 17:44
Chris CusterChris Custer
10.9k3824
answered Nov 22 '18 at 17:44
Chris CusterChris Custer
10.9k3824
10.9k3824
1
Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
– user573497
Nov 22 '18 at 20:02
add a comment |
1
Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
– user573497
Nov 22 '18 at 20:02
1
1
Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
– user573497
Nov 22 '18 at 20:02
Oh, now I get it! Since $H$ is a subgroup of $G$ and $gkg^(-1)$ belongs to $K$ for all $g$ in $G$ it also holds for all $h$ in $H$.
– user573497
Nov 22 '18 at 20:02
add a comment |
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