Mixing
Normal subgroups are not transitive, but $H subseteq N lhd G, H lhd G implies H lhd N$?
Let $H,N,G$ be groups where $H subseteq N subseteq G$. For the 3 statements below:
$H$ is a normal subgroup of $G$: $ghg^{-1} in H forall h in H, forall g in G$
$N$ is a normal subgroup of $G$: $gng^{-1} in N forall n in n, forall g in G$
$H$ is a normal subgroup of $N$: $nhn^{-1} in H forall h in H, forall n in N$
Is 1 enough to imply 3?
abstract-algebra group-theory normal-subgroups
add a comment |
Let $H,N,G$ be groups where $H subseteq N subseteq G$. For the 3 statements below:
$H$ is a normal subgroup of $G$: $ghg^{-1} in H forall h in H, forall g in G$
$N$ is a normal subgroup of $G$: $gng^{-1} in N forall n in n, forall g in G$
$H$ is a normal subgroup of $N$: $nhn^{-1} in H forall h in H, forall n in N$
Is 1 enough to imply 3?
abstract-algebra group-theory normal-subgroups
1
yes, you are correct
– user10354138
Nov 10 '18 at 3:17
@user10354138 Thank you!
– user198044
Nov 10 '18 at 3:33
1
Yes infact you did not even need N to be normal in G. Just H being normal in G was enough.
– Mustang
Nov 10 '18 at 6:31
@Mustang Thank you!
– user198044
Nov 10 '18 at 9:00
add a comment |
Let $H,N,G$ be groups where $H subseteq N subseteq G$. For the 3 statements below:
$H$ is a normal subgroup of $G$: $ghg^{-1} in H forall h in H, forall g in G$
$N$ is a normal subgroup of $G$: $gng^{-1} in N forall n in n, forall g in G$
$H$ is a normal subgroup of $N$: $nhn^{-1} in H forall h in H, forall n in N$
Is 1 enough to imply 3?
abstract-algebra group-theory normal-subgroups
Let $H,N,G$ be groups where $H subseteq N subseteq G$. For the 3 statements below:
$H$ is a normal subgroup of $G$: $ghg^{-1} in H forall h in H, forall g in G$
$N$ is a normal subgroup of $G$: $gng^{-1} in N forall n in n, forall g in G$
$H$ is a normal subgroup of $N$: $nhn^{-1} in H forall h in H, forall n in N$
Is 1 enough to imply 3?
abstract-algebra group-theory normal-subgroups
abstract-algebra group-theory normal-subgroups
edited Nov 20 '18 at 13:44
asked Nov 10 '18 at 3:14
user198044
1
yes, you are correct
– user10354138
Nov 10 '18 at 3:17
@user10354138 Thank you!
– user198044
Nov 10 '18 at 3:33
1
Yes infact you did not even need N to be normal in G. Just H being normal in G was enough.
– Mustang
Nov 10 '18 at 6:31
@Mustang Thank you!
– user198044
Nov 10 '18 at 9:00
add a comment |
1
yes, you are correct
– user10354138
Nov 10 '18 at 3:17
@user10354138 Thank you!
– user198044
Nov 10 '18 at 3:33
1
Yes infact you did not even need N to be normal in G. Just H being normal in G was enough.
– Mustang
Nov 10 '18 at 6:31
@Mustang Thank you!
– user198044
Nov 10 '18 at 9:00
1
1
yes, you are correct
– user10354138
Nov 10 '18 at 3:17
yes, you are correct
– user10354138
Nov 10 '18 at 3:17
@user10354138 Thank you!
– user198044
Nov 10 '18 at 3:33
@user10354138 Thank you!
– user198044
Nov 10 '18 at 3:33
1
1
Yes infact you did not even need N to be normal in G. Just H being normal in G was enough.
– Mustang
Nov 10 '18 at 6:31
Yes infact you did not even need N to be normal in G. Just H being normal in G was enough.
– Mustang
Nov 10 '18 at 6:31
@Mustang Thank you!
– user198044
Nov 10 '18 at 9:00
@Mustang Thank you!
– user198044
Nov 10 '18 at 9:00
add a comment |
1 Answer
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I think we just let $g=n$:
If $H$ is a normal subgroup of $G$, then $ghg^{-1} in H forall h in H, forall g in G$ including $n in N subseteq G$
Therefore, not only
$H subseteq N lhd G, H lhd G implies H lhd N$
but also
$H subseteq N subset G, H lhd G implies H lhd N$?
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
active
oldest
votes
I think we just let $g=n$:
If $H$ is a normal subgroup of $G$, then $ghg^{-1} in H forall h in H, forall g in G$ including $n in N subseteq G$
Therefore, not only
$H subseteq N lhd G, H lhd G implies H lhd N$
but also
$H subseteq N subset G, H lhd G implies H lhd N$?
add a comment |
I think we just let $g=n$:
If $H$ is a normal subgroup of $G$, then $ghg^{-1} in H forall h in H, forall g in G$ including $n in N subseteq G$
Therefore, not only
$H subseteq N lhd G, H lhd G implies H lhd N$
but also
$H subseteq N subset G, H lhd G implies H lhd N$?
add a comment |
I think we just let $g=n$:
If $H$ is a normal subgroup of $G$, then $ghg^{-1} in H forall h in H, forall g in G$ including $n in N subseteq G$
Therefore, not only
$H subseteq N lhd G, H lhd G implies H lhd N$
but also
$H subseteq N subset G, H lhd G implies H lhd N$?
I think we just let $g=n$:
If $H$ is a normal subgroup of $G$, then $ghg^{-1} in H forall h in H, forall g in G$ including $n in N subseteq G$
Therefore, not only
$H subseteq N lhd G, H lhd G implies H lhd N$
but also
$H subseteq N subset G, H lhd G implies H lhd N$?
answered Nov 20 '18 at 13:44
user198044
add a comment |
add a comment |
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1
yes, you are correct
– user10354138
Nov 10 '18 at 3:17
@user10354138 Thank you!
– user198044
Nov 10 '18 at 3:33
1
Yes infact you did not even need N to be normal in G. Just H being normal in G was enough.
– Mustang
Nov 10 '18 at 6:31
@Mustang Thank you!
– user198044
Nov 10 '18 at 9:00