Mixing
multiplying two normal subgroups is still normal?
$begingroup$
Let $G_i triangleleft G_{i+1}$ both subgroups of $G$. Let $N$ be a normal group.
Does $G_iN triangleleft G_{i+1}N$?
Does $(G_iN/N) triangleleft (G_{i+1}N/N)$?
I know that $q:Glongrightarrow G/N$ preserves normality. Hence if $G_iN triangleleft G_{i+1}N$ then their quotients would be normal.
I have tried by considering an element in $gin G_iN$ and $hin G_{i+1}N$ and trying to find out if $ghg^{-1}in G_iN$. But I am a bit confused of those multilpication groups. And that is why I can not play with this expression.
group-theory normal-subgroups
edited Jan 22 at 20:59
Ethan Bolker
45k553120
asked Jan 22 at 20:50
idriskameniidriskameni
759321
$endgroup$
add a comment |
$begingroup$
Let $G_i triangleleft G_{i+1}$ both subgroups of $G$. Let $N$ be a normal group.
Does $G_iN triangleleft G_{i+1}N$?
Does $(G_iN/N) triangleleft (G_{i+1}N/N)$?
I know that $q:Glongrightarrow G/N$ preserves normality. Hence if $G_iN triangleleft G_{i+1}N$ then their quotients would be normal.
I have tried by considering an element in $gin G_iN$ and $hin G_{i+1}N$ and trying to find out if $ghg^{-1}in G_iN$. But I am a bit confused of those multilpication groups. And that is why I can not play with this expression.
group-theory normal-subgroups
edited Jan 22 at 20:59
Ethan Bolker
45k553120
asked Jan 22 at 20:50
idriskameniidriskameni
759321
$endgroup$
$begingroup$
What do you know? What have you tried?
$endgroup$
– verret
Jan 22 at 20:55
$begingroup$
Edited. @verret
$endgroup$
– idriskameni
Jan 22 at 20:58
$begingroup$
Have you heard of the second/third isomorphism theorem?
$endgroup$
– DonAntonio
Jan 22 at 21:00
1
$begingroup$
Yes I do. I have been playing with that expression too. $G_i/(G_icap H)triangleleft G_{i+1}/(G_{i+1}cap H)$. But still nothing.
$endgroup$
– idriskameni
Jan 22 at 21:02
1
$begingroup$
I have tried a lot of things. Is not that I haven't worked on it. Don't you think so. I mean, it is easy to come and coment these things.
$endgroup$
– idriskameni
Jan 22 at 21:02
add a comment |
$begingroup$
Let $G_i triangleleft G_{i+1}$ both subgroups of $G$. Let $N$ be a normal group.
Does $G_iN triangleleft G_{i+1}N$?
Does $(G_iN/N) triangleleft (G_{i+1}N/N)$?
I know that $q:Glongrightarrow G/N$ preserves normality. Hence if $G_iN triangleleft G_{i+1}N$ then their quotients would be normal.
I have tried by considering an element in $gin G_iN$ and $hin G_{i+1}N$ and trying to find out if $ghg^{-1}in G_iN$. But I am a bit confused of those multilpication groups. And that is why I can not play with this expression.
group-theory normal-subgroups
edited Jan 22 at 20:59
Ethan Bolker
45k553120
asked Jan 22 at 20:50
idriskameniidriskameni
759321
$endgroup$
Let $G_i triangleleft G_{i+1}$ both subgroups of $G$. Let $N$ be a normal group.
Does $G_iN triangleleft G_{i+1}N$?
Does $(G_iN/N) triangleleft (G_{i+1}N/N)$?
I know that $q:Glongrightarrow G/N$ preserves normality. Hence if $G_iN triangleleft G_{i+1}N$ then their quotients would be normal.
I have tried by considering an element in $gin G_iN$ and $hin G_{i+1}N$ and trying to find out if $ghg^{-1}in G_iN$. But I am a bit confused of those multilpication groups. And that is why I can not play with this expression.
group-theory normal-subgroups
group-theory normal-subgroups
edited Jan 22 at 20:59
Ethan Bolker
45k553120
asked Jan 22 at 20:50
idriskameniidriskameni
759321
edited Jan 22 at 20:59
Ethan Bolker
45k553120
asked Jan 22 at 20:50
idriskameniidriskameni
759321
edited Jan 22 at 20:59
Ethan Bolker
45k553120
edited Jan 22 at 20:59
Ethan Bolker
45k553120
edited Jan 22 at 20:59
Ethan Bolker
45k553120
45k553120
asked Jan 22 at 20:50
idriskameniidriskameni
759321
asked Jan 22 at 20:50
idriskameniidriskameni
759321
asked Jan 22 at 20:50
idriskameniidriskameni
759321
759321
$begingroup$
What do you know? What have you tried?
$endgroup$
– verret
Jan 22 at 20:55
$begingroup$
Edited. @verret
$endgroup$
– idriskameni
Jan 22 at 20:58
$begingroup$
Have you heard of the second/third isomorphism theorem?
$endgroup$
– DonAntonio
Jan 22 at 21:00
1
$begingroup$
Yes I do. I have been playing with that expression too. $G_i/(G_icap H)triangleleft G_{i+1}/(G_{i+1}cap H)$. But still nothing.
$endgroup$
– idriskameni
Jan 22 at 21:02
1
$begingroup$
I have tried a lot of things. Is not that I haven't worked on it. Don't you think so. I mean, it is easy to come and coment these things.
$endgroup$
– idriskameni
Jan 22 at 21:02
add a comment |
$begingroup$
What do you know? What have you tried?
$endgroup$
– verret
Jan 22 at 20:55
$begingroup$
Edited. @verret
$endgroup$
– idriskameni
Jan 22 at 20:58
$begingroup$
Have you heard of the second/third isomorphism theorem?
$endgroup$
– DonAntonio
Jan 22 at 21:00
1
$begingroup$
Yes I do. I have been playing with that expression too. $G_i/(G_icap H)triangleleft G_{i+1}/(G_{i+1}cap H)$. But still nothing.
$endgroup$
– idriskameni
Jan 22 at 21:02
1
$begingroup$
I have tried a lot of things. Is not that I haven't worked on it. Don't you think so. I mean, it is easy to come and coment these things.
$endgroup$
– idriskameni
Jan 22 at 21:02
$begingroup$
What do you know? What have you tried?
$endgroup$
– verret
Jan 22 at 20:55
$begingroup$
What do you know? What have you tried?
$endgroup$
– verret
Jan 22 at 20:55
$begingroup$
Edited. @verret
$endgroup$
– idriskameni
Jan 22 at 20:58
$begingroup$
Edited. @verret
$endgroup$
– idriskameni
Jan 22 at 20:58
$begingroup$
Have you heard of the second/third isomorphism theorem?
$endgroup$
– DonAntonio
Jan 22 at 21:00
$begingroup$
Have you heard of the second/third isomorphism theorem?
$endgroup$
– DonAntonio
Jan 22 at 21:00
1
1
$begingroup$
Yes I do. I have been playing with that expression too. $G_i/(G_icap H)triangleleft G_{i+1}/(G_{i+1}cap H)$. But still nothing.
$endgroup$
– idriskameni
Jan 22 at 21:02
$begingroup$
Yes I do. I have been playing with that expression too. $G_i/(G_icap H)triangleleft G_{i+1}/(G_{i+1}cap H)$. But still nothing.
$endgroup$
– idriskameni
Jan 22 at 21:02
1
1
$begingroup$
I have tried a lot of things. Is not that I haven't worked on it. Don't you think so. I mean, it is easy to come and coment these things.
$endgroup$
– idriskameni
Jan 22 at 21:02
$begingroup$
I have tried a lot of things. Is not that I haven't worked on it. Don't you think so. I mean, it is easy to come and coment these things.
$endgroup$
– idriskameni
Jan 22 at 21:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I want to change the notation to make stuff clearer: Suppose $Hlhd K< G$, and let $Nlhd G$ (so really $H=G_i$ and $K=G_{i+1}$). Then a proof that $HNlhd KN$ by considering a conjugator is:
Suppose $kin K$, $hin H$, and $n, n'in N$. Firstly, note that
$$n^{-1}k^{-1}(hn')kn=(n^{-1}k^{-1}hkn)cdot (n^{-1}k^{-1}n'kn)$$
Clearly $(n^{-1}k^{-1}n'kn)in N$, while $(n^{-1}k^{-1}hkn)in HN$ by the following:
$$begin{align*}
k^{-1}hk&in H\
Rightarrow k^{-1}hkcdot (k^{-1}hk)^{-1}n^{-1}(k^{-1}hk)&in HN\
n^{-1}(k^{-1}hk)&in HN\
n^{-1}(k^{-1}hk)n&in HN
end{align*}
$$
The result follows.
answered Jan 23 at 8:49
user1729user1729
17.4k64193
$endgroup$
add a comment |
$begingroup$
Hint:
The first one should be almost immediate, and for the second one:
$$G_iN/Ncong G_i/(G_icap N);,;;; G_{i+1}/(G_{i+1}cap N)cong G_{i+1}N/N$$
answered Jan 23 at 8:30
DonAntonioDonAntonio
179k1494233
$endgroup$
2
$begingroup$
I am afraid that I cannot see how this helps solve the problem. The second one follows directly from the first which, as you say. is straightforward.
$endgroup$
– Derek Holt
Jan 23 at 8:47
$begingroup$
@DerekHolt That isomorphism helps, imo. to see that (1) those quotient groups are well-defined and (2) one is normal in the other. I also think both are more or less straightforward (though this usually depends on the beholder...). Did I miss anything?
$endgroup$
– DonAntonio
Jan 23 at 8:55
$begingroup$
I don't really see how the isomorphisms help prove that one is normal in the other. (The downvote is not from me.)
$endgroup$
– Derek Holt
Jan 23 at 9:11
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I want to change the notation to make stuff clearer: Suppose $Hlhd K< G$, and let $Nlhd G$ (so really $H=G_i$ and $K=G_{i+1}$). Then a proof that $HNlhd KN$ by considering a conjugator is:
Suppose $kin K$, $hin H$, and $n, n'in N$. Firstly, note that
$$n^{-1}k^{-1}(hn')kn=(n^{-1}k^{-1}hkn)cdot (n^{-1}k^{-1}n'kn)$$
Clearly $(n^{-1}k^{-1}n'kn)in N$, while $(n^{-1}k^{-1}hkn)in HN$ by the following:
$$begin{align*}
k^{-1}hk&in H\
Rightarrow k^{-1}hkcdot (k^{-1}hk)^{-1}n^{-1}(k^{-1}hk)&in HN\
n^{-1}(k^{-1}hk)&in HN\
n^{-1}(k^{-1}hk)n&in HN
end{align*}
$$
The result follows.
answered Jan 23 at 8:49
user1729user1729
17.4k64193
$endgroup$
add a comment |
$begingroup$
I want to change the notation to make stuff clearer: Suppose $Hlhd K< G$, and let $Nlhd G$ (so really $H=G_i$ and $K=G_{i+1}$). Then a proof that $HNlhd KN$ by considering a conjugator is:
Suppose $kin K$, $hin H$, and $n, n'in N$. Firstly, note that
$$n^{-1}k^{-1}(hn')kn=(n^{-1}k^{-1}hkn)cdot (n^{-1}k^{-1}n'kn)$$
Clearly $(n^{-1}k^{-1}n'kn)in N$, while $(n^{-1}k^{-1}hkn)in HN$ by the following:
$$begin{align*}
k^{-1}hk&in H\
Rightarrow k^{-1}hkcdot (k^{-1}hk)^{-1}n^{-1}(k^{-1}hk)&in HN\
n^{-1}(k^{-1}hk)&in HN\
n^{-1}(k^{-1}hk)n&in HN
end{align*}
$$
The result follows.
answered Jan 23 at 8:49
user1729user1729
17.4k64193
$endgroup$
add a comment |
$begingroup$
I want to change the notation to make stuff clearer: Suppose $Hlhd K< G$, and let $Nlhd G$ (so really $H=G_i$ and $K=G_{i+1}$). Then a proof that $HNlhd KN$ by considering a conjugator is:
Suppose $kin K$, $hin H$, and $n, n'in N$. Firstly, note that
$$n^{-1}k^{-1}(hn')kn=(n^{-1}k^{-1}hkn)cdot (n^{-1}k^{-1}n'kn)$$
Clearly $(n^{-1}k^{-1}n'kn)in N$, while $(n^{-1}k^{-1}hkn)in HN$ by the following:
$$begin{align*}
k^{-1}hk&in H\
Rightarrow k^{-1}hkcdot (k^{-1}hk)^{-1}n^{-1}(k^{-1}hk)&in HN\
n^{-1}(k^{-1}hk)&in HN\
n^{-1}(k^{-1}hk)n&in HN
end{align*}
$$
The result follows.
answered Jan 23 at 8:49
user1729user1729
17.4k64193
$endgroup$
I want to change the notation to make stuff clearer: Suppose $Hlhd K< G$, and let $Nlhd G$ (so really $H=G_i$ and $K=G_{i+1}$). Then a proof that $HNlhd KN$ by considering a conjugator is:
Suppose $kin K$, $hin H$, and $n, n'in N$. Firstly, note that
$$n^{-1}k^{-1}(hn')kn=(n^{-1}k^{-1}hkn)cdot (n^{-1}k^{-1}n'kn)$$
Clearly $(n^{-1}k^{-1}n'kn)in N$, while $(n^{-1}k^{-1}hkn)in HN$ by the following:
$$begin{align*}
k^{-1}hk&in H\
Rightarrow k^{-1}hkcdot (k^{-1}hk)^{-1}n^{-1}(k^{-1}hk)&in HN\
n^{-1}(k^{-1}hk)&in HN\
n^{-1}(k^{-1}hk)n&in HN
end{align*}
$$
The result follows.
answered Jan 23 at 8:49
user1729user1729
17.4k64193
edited Jan 23 at 9:06
answered Jan 23 at 8:49
user1729user1729
17.4k64193
answered Jan 23 at 8:49
user1729user1729
17.4k64193
answered Jan 23 at 8:49
user1729user1729
17.4k64193
17.4k64193
add a comment |
add a comment |
$begingroup$
Hint:
The first one should be almost immediate, and for the second one:
$$G_iN/Ncong G_i/(G_icap N);,;;; G_{i+1}/(G_{i+1}cap N)cong G_{i+1}N/N$$
answered Jan 23 at 8:30
DonAntonioDonAntonio
179k1494233
$endgroup$
2
$begingroup$
I am afraid that I cannot see how this helps solve the problem. The second one follows directly from the first which, as you say. is straightforward.
$endgroup$
– Derek Holt
Jan 23 at 8:47
$begingroup$
@DerekHolt That isomorphism helps, imo. to see that (1) those quotient groups are well-defined and (2) one is normal in the other. I also think both are more or less straightforward (though this usually depends on the beholder...). Did I miss anything?
$endgroup$
– DonAntonio
Jan 23 at 8:55
$begingroup$
I don't really see how the isomorphisms help prove that one is normal in the other. (The downvote is not from me.)
$endgroup$
– Derek Holt
Jan 23 at 9:11
add a comment |
$begingroup$
Hint:
The first one should be almost immediate, and for the second one:
$$G_iN/Ncong G_i/(G_icap N);,;;; G_{i+1}/(G_{i+1}cap N)cong G_{i+1}N/N$$
answered Jan 23 at 8:30
DonAntonioDonAntonio
179k1494233
$endgroup$
2
$begingroup$
I am afraid that I cannot see how this helps solve the problem. The second one follows directly from the first which, as you say. is straightforward.
$endgroup$
– Derek Holt
Jan 23 at 8:47
$begingroup$
@DerekHolt That isomorphism helps, imo. to see that (1) those quotient groups are well-defined and (2) one is normal in the other. I also think both are more or less straightforward (though this usually depends on the beholder...). Did I miss anything?
$endgroup$
– DonAntonio
Jan 23 at 8:55
$begingroup$
I don't really see how the isomorphisms help prove that one is normal in the other. (The downvote is not from me.)
$endgroup$
– Derek Holt
Jan 23 at 9:11
add a comment |
$begingroup$
Hint:
The first one should be almost immediate, and for the second one:
$$G_iN/Ncong G_i/(G_icap N);,;;; G_{i+1}/(G_{i+1}cap N)cong G_{i+1}N/N$$
answered Jan 23 at 8:30
DonAntonioDonAntonio
179k1494233
$endgroup$
Hint:
The first one should be almost immediate, and for the second one:
$$G_iN/Ncong G_i/(G_icap N);,;;; G_{i+1}/(G_{i+1}cap N)cong G_{i+1}N/N$$
answered Jan 23 at 8:30
DonAntonioDonAntonio
179k1494233
answered Jan 23 at 8:30
DonAntonioDonAntonio
179k1494233
answered Jan 23 at 8:30
DonAntonioDonAntonio
179k1494233
answered Jan 23 at 8:30
DonAntonioDonAntonio
179k1494233
179k1494233
2
$begingroup$
I am afraid that I cannot see how this helps solve the problem. The second one follows directly from the first which, as you say. is straightforward.
$endgroup$
– Derek Holt
Jan 23 at 8:47
$begingroup$
@DerekHolt That isomorphism helps, imo. to see that (1) those quotient groups are well-defined and (2) one is normal in the other. I also think both are more or less straightforward (though this usually depends on the beholder...). Did I miss anything?
$endgroup$
– DonAntonio
Jan 23 at 8:55
$begingroup$
I don't really see how the isomorphisms help prove that one is normal in the other. (The downvote is not from me.)
$endgroup$
– Derek Holt
Jan 23 at 9:11
add a comment |
2
$begingroup$
I am afraid that I cannot see how this helps solve the problem. The second one follows directly from the first which, as you say. is straightforward.
$endgroup$
– Derek Holt
Jan 23 at 8:47
$begingroup$
@DerekHolt That isomorphism helps, imo. to see that (1) those quotient groups are well-defined and (2) one is normal in the other. I also think both are more or less straightforward (though this usually depends on the beholder...). Did I miss anything?
$endgroup$
– DonAntonio
Jan 23 at 8:55
$begingroup$
I don't really see how the isomorphisms help prove that one is normal in the other. (The downvote is not from me.)
$endgroup$
– Derek Holt
Jan 23 at 9:11
2
2
$begingroup$
I am afraid that I cannot see how this helps solve the problem. The second one follows directly from the first which, as you say. is straightforward.
$endgroup$
– Derek Holt
Jan 23 at 8:47
$begingroup$
I am afraid that I cannot see how this helps solve the problem. The second one follows directly from the first which, as you say. is straightforward.
$endgroup$
– Derek Holt
Jan 23 at 8:47
$begingroup$
@DerekHolt That isomorphism helps, imo. to see that (1) those quotient groups are well-defined and (2) one is normal in the other. I also think both are more or less straightforward (though this usually depends on the beholder...). Did I miss anything?
$endgroup$
– DonAntonio
Jan 23 at 8:55
$begingroup$
@DerekHolt That isomorphism helps, imo. to see that (1) those quotient groups are well-defined and (2) one is normal in the other. I also think both are more or less straightforward (though this usually depends on the beholder...). Did I miss anything?
$endgroup$
– DonAntonio
Jan 23 at 8:55
$begingroup$
I don't really see how the isomorphisms help prove that one is normal in the other. (The downvote is not from me.)
$endgroup$
– Derek Holt
Jan 23 at 9:11
$begingroup$
I don't really see how the isomorphisms help prove that one is normal in the other. (The downvote is not from me.)
$endgroup$
– Derek Holt
Jan 23 at 9:11
add a comment |
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$begingroup$
What do you know? What have you tried?
$endgroup$
– verret
Jan 22 at 20:55
$begingroup$
Edited. @verret
$endgroup$
– idriskameni
Jan 22 at 20:58
$begingroup$
Have you heard of the second/third isomorphism theorem?
$endgroup$
– DonAntonio
Jan 22 at 21:00
1
$begingroup$
Yes I do. I have been playing with that expression too. $G_i/(G_icap H)triangleleft G_{i+1}/(G_{i+1}cap H)$. But still nothing.
$endgroup$
– idriskameni
Jan 22 at 21:02
1
$begingroup$
I have tried a lot of things. Is not that I haven't worked on it. Don't you think so. I mean, it is easy to come and coment these things.
$endgroup$
– idriskameni
Jan 22 at 21:02