Invariance under a normal subgroup
$begingroup$
Let $G$ be a group acting on a set $A$. Let $N$ be a non-trivial normal subgroup of $G$. Suppose that $S$ is an $N$-invariant set, i.e. $n cdot s in S$ for all $s in S$, $n in N$. Must $S$ be $G$-invariant?
I suspect the answer is yes (for example, it is true in the situation of a Galois group acting on a Galois field extension). I've been playing around with it for a while and haven't been able to get anything.
group-theory group-actions
asked Dec 6 '18 at 23:33
vukovvukov
788414
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group acting on a set $A$. Let $N$ be a non-trivial normal subgroup of $G$. Suppose that $S$ is an $N$-invariant set, i.e. $n cdot s in S$ for all $s in S$, $n in N$. Must $S$ be $G$-invariant?
I suspect the answer is yes (for example, it is true in the situation of a Galois group acting on a Galois field extension). I've been playing around with it for a while and haven't been able to get anything.
group-theory group-actions
asked Dec 6 '18 at 23:33
vukovvukov
788414
$endgroup$
1
$begingroup$
Sorry N-invariant means N(S)=S?
$endgroup$
– nafhgood
Dec 6 '18 at 23:36
$begingroup$
@mathnoob Yes, I edited it for clarity.
$endgroup$
– vukov
Dec 6 '18 at 23:37
$begingroup$
What if $N$ is the trivial normal subgroup ${e}$?
$endgroup$
– Catalin Zara
Dec 6 '18 at 23:39
$begingroup$
@CatalinZara You're right, I meant to require non-trivial.
$endgroup$
– vukov
Dec 6 '18 at 23:42
add a comment |
$begingroup$
Let $G$ be a group acting on a set $A$. Let $N$ be a non-trivial normal subgroup of $G$. Suppose that $S$ is an $N$-invariant set, i.e. $n cdot s in S$ for all $s in S$, $n in N$. Must $S$ be $G$-invariant?
I suspect the answer is yes (for example, it is true in the situation of a Galois group acting on a Galois field extension). I've been playing around with it for a while and haven't been able to get anything.
group-theory group-actions
asked Dec 6 '18 at 23:33
vukovvukov
788414
$endgroup$
Let $G$ be a group acting on a set $A$. Let $N$ be a non-trivial normal subgroup of $G$. Suppose that $S$ is an $N$-invariant set, i.e. $n cdot s in S$ for all $s in S$, $n in N$. Must $S$ be $G$-invariant?
I suspect the answer is yes (for example, it is true in the situation of a Galois group acting on a Galois field extension). I've been playing around with it for a while and haven't been able to get anything.
group-theory group-actions
group-theory group-actions
asked Dec 6 '18 at 23:33
vukovvukov
788414
asked Dec 6 '18 at 23:33
vukovvukov
788414
edited Dec 6 '18 at 23:41
vukov
asked Dec 6 '18 at 23:33
vukovvukov
788414
asked Dec 6 '18 at 23:33
vukovvukov
788414
asked Dec 6 '18 at 23:33
vukovvukov
788414
788414
1
$begingroup$
Sorry N-invariant means N(S)=S?
$endgroup$
– nafhgood
Dec 6 '18 at 23:36
$begingroup$
@mathnoob Yes, I edited it for clarity.
$endgroup$
– vukov
Dec 6 '18 at 23:37
$begingroup$
What if $N$ is the trivial normal subgroup ${e}$?
$endgroup$
– Catalin Zara
Dec 6 '18 at 23:39
$begingroup$
@CatalinZara You're right, I meant to require non-trivial.
$endgroup$
– vukov
Dec 6 '18 at 23:42
add a comment |
1
$begingroup$
Sorry N-invariant means N(S)=S?
$endgroup$
– nafhgood
Dec 6 '18 at 23:36
$begingroup$
@mathnoob Yes, I edited it for clarity.
$endgroup$
– vukov
Dec 6 '18 at 23:37
$begingroup$
What if $N$ is the trivial normal subgroup ${e}$?
$endgroup$
– Catalin Zara
Dec 6 '18 at 23:39
$begingroup$
@CatalinZara You're right, I meant to require non-trivial.
$endgroup$
– vukov
Dec 6 '18 at 23:42
1
1
$begingroup$
Sorry N-invariant means N(S)=S?
$endgroup$
– nafhgood
Dec 6 '18 at 23:36
$begingroup$
Sorry N-invariant means N(S)=S?
$endgroup$
– nafhgood
Dec 6 '18 at 23:36
$begingroup$
@mathnoob Yes, I edited it for clarity.
$endgroup$
– vukov
Dec 6 '18 at 23:37
$begingroup$
@mathnoob Yes, I edited it for clarity.
$endgroup$
– vukov
Dec 6 '18 at 23:37
$begingroup$
What if $N$ is the trivial normal subgroup ${e}$?
$endgroup$
– Catalin Zara
Dec 6 '18 at 23:39
$begingroup$
What if $N$ is the trivial normal subgroup ${e}$?
$endgroup$
– Catalin Zara
Dec 6 '18 at 23:39
$begingroup$
@CatalinZara You're right, I meant to require non-trivial.
$endgroup$
– vukov
Dec 6 '18 at 23:42
$begingroup$
@CatalinZara You're right, I meant to require non-trivial.
$endgroup$
– vukov
Dec 6 '18 at 23:42
add a comment |
1 Answer
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$begingroup$
Let $G$ be a group and $N$ a normal subgroup of $G$. Let $A=G/N$ be the collection of left cosets with the left multiplication action, and let $S={eN}$. Then letting $Gcdot T={gcdot tmid gin G, tin T}$ for any $Tsubseteq A$, we have in particular $Ncdot S=S$ but $Gcdot S=G/Nneq S$.
answered Dec 6 '18 at 23:48
Sean ClarkSean Clark
1,883813
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$begingroup$
Let $G$ be a group and $N$ a normal subgroup of $G$. Let $A=G/N$ be the collection of left cosets with the left multiplication action, and let $S={eN}$. Then letting $Gcdot T={gcdot tmid gin G, tin T}$ for any $Tsubseteq A$, we have in particular $Ncdot S=S$ but $Gcdot S=G/Nneq S$.
answered Dec 6 '18 at 23:48
Sean ClarkSean Clark
1,883813
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group and $N$ a normal subgroup of $G$. Let $A=G/N$ be the collection of left cosets with the left multiplication action, and let $S={eN}$. Then letting $Gcdot T={gcdot tmid gin G, tin T}$ for any $Tsubseteq A$, we have in particular $Ncdot S=S$ but $Gcdot S=G/Nneq S$.
answered Dec 6 '18 at 23:48
Sean ClarkSean Clark
1,883813
$endgroup$
add a comment |
$begingroup$
Let $G$ be a group and $N$ a normal subgroup of $G$. Let $A=G/N$ be the collection of left cosets with the left multiplication action, and let $S={eN}$. Then letting $Gcdot T={gcdot tmid gin G, tin T}$ for any $Tsubseteq A$, we have in particular $Ncdot S=S$ but $Gcdot S=G/Nneq S$.
answered Dec 6 '18 at 23:48
Sean ClarkSean Clark
1,883813
$endgroup$
Let $G$ be a group and $N$ a normal subgroup of $G$. Let $A=G/N$ be the collection of left cosets with the left multiplication action, and let $S={eN}$. Then letting $Gcdot T={gcdot tmid gin G, tin T}$ for any $Tsubseteq A$, we have in particular $Ncdot S=S$ but $Gcdot S=G/Nneq S$.
answered Dec 6 '18 at 23:48
Sean ClarkSean Clark
1,883813
answered Dec 6 '18 at 23:48
Sean ClarkSean Clark
1,883813
answered Dec 6 '18 at 23:48
Sean ClarkSean Clark
1,883813
answered Dec 6 '18 at 23:48
Sean ClarkSean Clark
1,883813
1,883813
add a comment |
add a comment |
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1
$begingroup$
Sorry N-invariant means N(S)=S?
$endgroup$
– nafhgood
Dec 6 '18 at 23:36
$begingroup$
@mathnoob Yes, I edited it for clarity.
$endgroup$
– vukov
Dec 6 '18 at 23:37
$begingroup$
What if $N$ is the trivial normal subgroup ${e}$?
$endgroup$
– Catalin Zara
Dec 6 '18 at 23:39
$begingroup$
@CatalinZara You're right, I meant to require non-trivial.
$endgroup$
– vukov
Dec 6 '18 at 23:42