Let $G$ be transitive on $S$. Show that the action is primitive if and only if every...
$begingroup$
I am self-studying "Classical Groups and Geometric Algebra" by Larry C. Grove. This is the 2nd question of the exercises of the 0th Chapter.
Let $G$ be transitive on $S$. Show that the action is primitive if and only if every $operatorname{Stab}_G(a), ain S$, is a maximal subgroup of $G$.
Unfortunately, I can't even start.
$G$ acts transitively on $S$ $iff$ $operatorname{Orb}_G(a)=S, forall ain S$.
So, we can reach any element of $S$ applying $G$ to any element of $S$.- If the action is primitive there is no block like $Bsubseteq S$,
with $|B|ge 2, Bneq S$, such that for each $x in G$ either $xB=B$
or $xBcap B = emptyset$.
$operatorname{Stab}_G(a)$ is a maximal subgroup if there is no subgroup containing
$operatorname{Stab}_G(a)$ other than $operatorname{Stab}_G(a)$ itself and $G$.
group-theory group-actions maximal-subgroup
edited Dec 28 '18 at 19:53
Shaun
10.5k113687
asked Dec 28 '18 at 19:11
waterwater
215
$endgroup$
add a comment |
$begingroup$
I am self-studying "Classical Groups and Geometric Algebra" by Larry C. Grove. This is the 2nd question of the exercises of the 0th Chapter.
Let $G$ be transitive on $S$. Show that the action is primitive if and only if every $operatorname{Stab}_G(a), ain S$, is a maximal subgroup of $G$.
Unfortunately, I can't even start.
$G$ acts transitively on $S$ $iff$ $operatorname{Orb}_G(a)=S, forall ain S$.
So, we can reach any element of $S$ applying $G$ to any element of $S$.- If the action is primitive there is no block like $Bsubseteq S$,
with $|B|ge 2, Bneq S$, such that for each $x in G$ either $xB=B$
or $xBcap B = emptyset$.
$operatorname{Stab}_G(a)$ is a maximal subgroup if there is no subgroup containing
$operatorname{Stab}_G(a)$ other than $operatorname{Stab}_G(a)$ itself and $G$.
group-theory group-actions maximal-subgroup
edited Dec 28 '18 at 19:53
Shaun
10.5k113687
asked Dec 28 '18 at 19:11
waterwater
215
$endgroup$
4
$begingroup$
Outline of proof: if the stabilizer is not maximal, then it is contained in a larger subgroups $H$, and the orbit of $a$ under $H$ is a block. Conversely, if the the action is imprimitive, and $B$ is a block with $a in B$, then the stabilizer of $B$ is a subgroup properly containing ${rm Stab}_G(a)$.
$endgroup$
– Derek Holt
Dec 28 '18 at 19:42
$begingroup$
@DerekHolt thank you very much. I will try to answer it.
$endgroup$
– water
Dec 30 '18 at 6:19
$begingroup$
@DerekHolt May I ask your comment about my answer?
$endgroup$
– water
Dec 30 '18 at 17:23
add a comment |
$begingroup$
I am self-studying "Classical Groups and Geometric Algebra" by Larry C. Grove. This is the 2nd question of the exercises of the 0th Chapter.
Let $G$ be transitive on $S$. Show that the action is primitive if and only if every $operatorname{Stab}_G(a), ain S$, is a maximal subgroup of $G$.
Unfortunately, I can't even start.
$G$ acts transitively on $S$ $iff$ $operatorname{Orb}_G(a)=S, forall ain S$.
So, we can reach any element of $S$ applying $G$ to any element of $S$.- If the action is primitive there is no block like $Bsubseteq S$,
with $|B|ge 2, Bneq S$, such that for each $x in G$ either $xB=B$
or $xBcap B = emptyset$.
$operatorname{Stab}_G(a)$ is a maximal subgroup if there is no subgroup containing
$operatorname{Stab}_G(a)$ other than $operatorname{Stab}_G(a)$ itself and $G$.
group-theory group-actions maximal-subgroup
edited Dec 28 '18 at 19:53
Shaun
10.5k113687
asked Dec 28 '18 at 19:11
waterwater
215
$endgroup$
I am self-studying "Classical Groups and Geometric Algebra" by Larry C. Grove. This is the 2nd question of the exercises of the 0th Chapter.
Let $G$ be transitive on $S$. Show that the action is primitive if and only if every $operatorname{Stab}_G(a), ain S$, is a maximal subgroup of $G$.
Unfortunately, I can't even start.
$G$ acts transitively on $S$ $iff$ $operatorname{Orb}_G(a)=S, forall ain S$.
So, we can reach any element of $S$ applying $G$ to any element of $S$.- If the action is primitive there is no block like $Bsubseteq S$,
with $|B|ge 2, Bneq S$, such that for each $x in G$ either $xB=B$
or $xBcap B = emptyset$.
$operatorname{Stab}_G(a)$ is a maximal subgroup if there is no subgroup containing
$operatorname{Stab}_G(a)$ other than $operatorname{Stab}_G(a)$ itself and $G$.
group-theory group-actions maximal-subgroup
group-theory group-actions maximal-subgroup
edited Dec 28 '18 at 19:53
Shaun
10.5k113687
asked Dec 28 '18 at 19:11
waterwater
215
edited Dec 28 '18 at 19:53
Shaun
10.5k113687
asked Dec 28 '18 at 19:11
waterwater
215
edited Dec 28 '18 at 19:53
Shaun
10.5k113687
edited Dec 28 '18 at 19:53
Shaun
10.5k113687
edited Dec 28 '18 at 19:53
Shaun
10.5k113687
10.5k113687
asked Dec 28 '18 at 19:11
waterwater
215
asked Dec 28 '18 at 19:11
waterwater
215
asked Dec 28 '18 at 19:11
waterwater
215
215
4
$begingroup$
Outline of proof: if the stabilizer is not maximal, then it is contained in a larger subgroups $H$, and the orbit of $a$ under $H$ is a block. Conversely, if the the action is imprimitive, and $B$ is a block with $a in B$, then the stabilizer of $B$ is a subgroup properly containing ${rm Stab}_G(a)$.
$endgroup$
– Derek Holt
Dec 28 '18 at 19:42
$begingroup$
@DerekHolt thank you very much. I will try to answer it.
$endgroup$
– water
Dec 30 '18 at 6:19
$begingroup$
@DerekHolt May I ask your comment about my answer?
$endgroup$
– water
Dec 30 '18 at 17:23
add a comment |
4
$begingroup$
Outline of proof: if the stabilizer is not maximal, then it is contained in a larger subgroups $H$, and the orbit of $a$ under $H$ is a block. Conversely, if the the action is imprimitive, and $B$ is a block with $a in B$, then the stabilizer of $B$ is a subgroup properly containing ${rm Stab}_G(a)$.
$endgroup$
– Derek Holt
Dec 28 '18 at 19:42
$begingroup$
@DerekHolt thank you very much. I will try to answer it.
$endgroup$
– water
Dec 30 '18 at 6:19
$begingroup$
@DerekHolt May I ask your comment about my answer?
$endgroup$
– water
Dec 30 '18 at 17:23
4
4
$begingroup$
Outline of proof: if the stabilizer is not maximal, then it is contained in a larger subgroups $H$, and the orbit of $a$ under $H$ is a block. Conversely, if the the action is imprimitive, and $B$ is a block with $a in B$, then the stabilizer of $B$ is a subgroup properly containing ${rm Stab}_G(a)$.
$endgroup$
– Derek Holt
Dec 28 '18 at 19:42
$begingroup$
Outline of proof: if the stabilizer is not maximal, then it is contained in a larger subgroups $H$, and the orbit of $a$ under $H$ is a block. Conversely, if the the action is imprimitive, and $B$ is a block with $a in B$, then the stabilizer of $B$ is a subgroup properly containing ${rm Stab}_G(a)$.
$endgroup$
– Derek Holt
Dec 28 '18 at 19:42
$begingroup$
@DerekHolt thank you very much. I will try to answer it.
$endgroup$
– water
Dec 30 '18 at 6:19
$begingroup$
@DerekHolt thank you very much. I will try to answer it.
$endgroup$
– water
Dec 30 '18 at 6:19
$begingroup$
@DerekHolt May I ask your comment about my answer?
$endgroup$
– water
Dec 30 '18 at 17:23
$begingroup$
@DerekHolt May I ask your comment about my answer?
$endgroup$
– water
Dec 30 '18 at 17:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
($implies$)
Let $H$ is a proper subgroup of $G$ containing $Stab_G(a)$ and $B={ha|hin H}$ is a subset of $S$.
Assume that $xBcap B neq emptyset$ for some $x in G$. There exist $h,h'in H$ such that $xha=h'a$. Then, $h'^{-1}xha=aimplies h'^{-1}xhin Stab_G(a)implies h'^{-1}xhin Himplies h'(h'^{-1}xh)h^{-1}=xin Himplies xB=Bimplies Btext{ is a block.}$
$B$ is a block means $B={a}$ or $B=S$ because $G$ acts primitively on $S$.
$B={ha|hin H}={a} implies ha=a, forall hin H implies H=Stab_G(a)$.
$B={ha|hin H}=S implies Htext{ acts transitively on }B.$ We also have $G$ acts transitively on $B$. So there exist $gin G$ and $bar hin H$ such that $ga=bar ha$. Then, $bar{h}^{-1}ga=aimplies bar{h}^{-1}gin Himplies Gsubseteq Himplies H=G$.
($impliedby$)
Let $G$ acts on $S$ imprimitively such that there exists some block $Bsubseteq S$ with $ain B$. Then, $Stab_G(B)={g:gb=b, gin G, and forall bin B}supset{g:ga=a, gin G }=Stab_G(a)$. Let $alpha, betain Stab_G(B)$, and $bin B$. Then $(alphabeta)cdot b=alphacdot(betacdot b)=alphacdot b=bimplies alphabetain Stab_G(B).$ Moreover, $alpha^{-1}cdot b=alpha^{-1}cdot(alphacdot b)=(alpha^{-1}alpha)cdot b=bimplies alpha^{-1}in Stab_G(B).$ Thus, $Stab_G(B)$ is a proper subgroup containing $Stab_G(a)$.
answered Dec 30 '18 at 6:20
waterwater
215
$endgroup$
1
$begingroup$
Your proof of $Rightarrow$ is OK, but the proof of $Leftarrow$ is confused. You have confused the block $B$, which is a subset of $S$ with its stabilizer, which is a subgroup of $G$.
$endgroup$
– Derek Holt
Dec 30 '18 at 18:21
$begingroup$
@DerekHolt Is the definition of the stabilizer of the set B either $Stab_G(B)={g:gb=b, gin G, and forall bin B}$ or $Stab_G(B) = {gin G:gB=B}$?
$endgroup$
– water
Dec 30 '18 at 19:18
1
$begingroup$
Only one of those two subgroups could strictly contain ${rm Stab}_G(a)$. I think I have answered enough questions!
$endgroup$
– Derek Holt
Dec 30 '18 at 19:56
$begingroup$
@DerekHolt I see, thank you very much.
$endgroup$
– water
Dec 30 '18 at 22:49
add a comment |
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1 Answer
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$begingroup$
($implies$)
Let $H$ is a proper subgroup of $G$ containing $Stab_G(a)$ and $B={ha|hin H}$ is a subset of $S$.
Assume that $xBcap B neq emptyset$ for some $x in G$. There exist $h,h'in H$ such that $xha=h'a$. Then, $h'^{-1}xha=aimplies h'^{-1}xhin Stab_G(a)implies h'^{-1}xhin Himplies h'(h'^{-1}xh)h^{-1}=xin Himplies xB=Bimplies Btext{ is a block.}$
$B$ is a block means $B={a}$ or $B=S$ because $G$ acts primitively on $S$.
$B={ha|hin H}={a} implies ha=a, forall hin H implies H=Stab_G(a)$.
$B={ha|hin H}=S implies Htext{ acts transitively on }B.$ We also have $G$ acts transitively on $B$. So there exist $gin G$ and $bar hin H$ such that $ga=bar ha$. Then, $bar{h}^{-1}ga=aimplies bar{h}^{-1}gin Himplies Gsubseteq Himplies H=G$.
($impliedby$)
Let $G$ acts on $S$ imprimitively such that there exists some block $Bsubseteq S$ with $ain B$. Then, $Stab_G(B)={g:gb=b, gin G, and forall bin B}supset{g:ga=a, gin G }=Stab_G(a)$. Let $alpha, betain Stab_G(B)$, and $bin B$. Then $(alphabeta)cdot b=alphacdot(betacdot b)=alphacdot b=bimplies alphabetain Stab_G(B).$ Moreover, $alpha^{-1}cdot b=alpha^{-1}cdot(alphacdot b)=(alpha^{-1}alpha)cdot b=bimplies alpha^{-1}in Stab_G(B).$ Thus, $Stab_G(B)$ is a proper subgroup containing $Stab_G(a)$.
answered Dec 30 '18 at 6:20
waterwater
215
$endgroup$
1
$begingroup$
Your proof of $Rightarrow$ is OK, but the proof of $Leftarrow$ is confused. You have confused the block $B$, which is a subset of $S$ with its stabilizer, which is a subgroup of $G$.
$endgroup$
– Derek Holt
Dec 30 '18 at 18:21
$begingroup$
@DerekHolt Is the definition of the stabilizer of the set B either $Stab_G(B)={g:gb=b, gin G, and forall bin B}$ or $Stab_G(B) = {gin G:gB=B}$?
$endgroup$
– water
Dec 30 '18 at 19:18
1
$begingroup$
Only one of those two subgroups could strictly contain ${rm Stab}_G(a)$. I think I have answered enough questions!
$endgroup$
– Derek Holt
Dec 30 '18 at 19:56
$begingroup$
@DerekHolt I see, thank you very much.
$endgroup$
– water
Dec 30 '18 at 22:49
add a comment |
$begingroup$
($implies$)
Let $H$ is a proper subgroup of $G$ containing $Stab_G(a)$ and $B={ha|hin H}$ is a subset of $S$.
Assume that $xBcap B neq emptyset$ for some $x in G$. There exist $h,h'in H$ such that $xha=h'a$. Then, $h'^{-1}xha=aimplies h'^{-1}xhin Stab_G(a)implies h'^{-1}xhin Himplies h'(h'^{-1}xh)h^{-1}=xin Himplies xB=Bimplies Btext{ is a block.}$
$B$ is a block means $B={a}$ or $B=S$ because $G$ acts primitively on $S$.
$B={ha|hin H}={a} implies ha=a, forall hin H implies H=Stab_G(a)$.
$B={ha|hin H}=S implies Htext{ acts transitively on }B.$ We also have $G$ acts transitively on $B$. So there exist $gin G$ and $bar hin H$ such that $ga=bar ha$. Then, $bar{h}^{-1}ga=aimplies bar{h}^{-1}gin Himplies Gsubseteq Himplies H=G$.
($impliedby$)
Let $G$ acts on $S$ imprimitively such that there exists some block $Bsubseteq S$ with $ain B$. Then, $Stab_G(B)={g:gb=b, gin G, and forall bin B}supset{g:ga=a, gin G }=Stab_G(a)$. Let $alpha, betain Stab_G(B)$, and $bin B$. Then $(alphabeta)cdot b=alphacdot(betacdot b)=alphacdot b=bimplies alphabetain Stab_G(B).$ Moreover, $alpha^{-1}cdot b=alpha^{-1}cdot(alphacdot b)=(alpha^{-1}alpha)cdot b=bimplies alpha^{-1}in Stab_G(B).$ Thus, $Stab_G(B)$ is a proper subgroup containing $Stab_G(a)$.
answered Dec 30 '18 at 6:20
waterwater
215
$endgroup$
1
$begingroup$
Your proof of $Rightarrow$ is OK, but the proof of $Leftarrow$ is confused. You have confused the block $B$, which is a subset of $S$ with its stabilizer, which is a subgroup of $G$.
$endgroup$
– Derek Holt
Dec 30 '18 at 18:21
$begingroup$
@DerekHolt Is the definition of the stabilizer of the set B either $Stab_G(B)={g:gb=b, gin G, and forall bin B}$ or $Stab_G(B) = {gin G:gB=B}$?
$endgroup$
– water
Dec 30 '18 at 19:18
1
$begingroup$
Only one of those two subgroups could strictly contain ${rm Stab}_G(a)$. I think I have answered enough questions!
$endgroup$
– Derek Holt
Dec 30 '18 at 19:56
$begingroup$
@DerekHolt I see, thank you very much.
$endgroup$
– water
Dec 30 '18 at 22:49
add a comment |
$begingroup$
($implies$)
Let $H$ is a proper subgroup of $G$ containing $Stab_G(a)$ and $B={ha|hin H}$ is a subset of $S$.
Assume that $xBcap B neq emptyset$ for some $x in G$. There exist $h,h'in H$ such that $xha=h'a$. Then, $h'^{-1}xha=aimplies h'^{-1}xhin Stab_G(a)implies h'^{-1}xhin Himplies h'(h'^{-1}xh)h^{-1}=xin Himplies xB=Bimplies Btext{ is a block.}$
$B$ is a block means $B={a}$ or $B=S$ because $G$ acts primitively on $S$.
$B={ha|hin H}={a} implies ha=a, forall hin H implies H=Stab_G(a)$.
$B={ha|hin H}=S implies Htext{ acts transitively on }B.$ We also have $G$ acts transitively on $B$. So there exist $gin G$ and $bar hin H$ such that $ga=bar ha$. Then, $bar{h}^{-1}ga=aimplies bar{h}^{-1}gin Himplies Gsubseteq Himplies H=G$.
($impliedby$)
Let $G$ acts on $S$ imprimitively such that there exists some block $Bsubseteq S$ with $ain B$. Then, $Stab_G(B)={g:gb=b, gin G, and forall bin B}supset{g:ga=a, gin G }=Stab_G(a)$. Let $alpha, betain Stab_G(B)$, and $bin B$. Then $(alphabeta)cdot b=alphacdot(betacdot b)=alphacdot b=bimplies alphabetain Stab_G(B).$ Moreover, $alpha^{-1}cdot b=alpha^{-1}cdot(alphacdot b)=(alpha^{-1}alpha)cdot b=bimplies alpha^{-1}in Stab_G(B).$ Thus, $Stab_G(B)$ is a proper subgroup containing $Stab_G(a)$.
answered Dec 30 '18 at 6:20
waterwater
215
$endgroup$
($implies$)
Let $H$ is a proper subgroup of $G$ containing $Stab_G(a)$ and $B={ha|hin H}$ is a subset of $S$.
Assume that $xBcap B neq emptyset$ for some $x in G$. There exist $h,h'in H$ such that $xha=h'a$. Then, $h'^{-1}xha=aimplies h'^{-1}xhin Stab_G(a)implies h'^{-1}xhin Himplies h'(h'^{-1}xh)h^{-1}=xin Himplies xB=Bimplies Btext{ is a block.}$
$B$ is a block means $B={a}$ or $B=S$ because $G$ acts primitively on $S$.
$B={ha|hin H}={a} implies ha=a, forall hin H implies H=Stab_G(a)$.
$B={ha|hin H}=S implies Htext{ acts transitively on }B.$ We also have $G$ acts transitively on $B$. So there exist $gin G$ and $bar hin H$ such that $ga=bar ha$. Then, $bar{h}^{-1}ga=aimplies bar{h}^{-1}gin Himplies Gsubseteq Himplies H=G$.
($impliedby$)
Let $G$ acts on $S$ imprimitively such that there exists some block $Bsubseteq S$ with $ain B$. Then, $Stab_G(B)={g:gb=b, gin G, and forall bin B}supset{g:ga=a, gin G }=Stab_G(a)$. Let $alpha, betain Stab_G(B)$, and $bin B$. Then $(alphabeta)cdot b=alphacdot(betacdot b)=alphacdot b=bimplies alphabetain Stab_G(B).$ Moreover, $alpha^{-1}cdot b=alpha^{-1}cdot(alphacdot b)=(alpha^{-1}alpha)cdot b=bimplies alpha^{-1}in Stab_G(B).$ Thus, $Stab_G(B)$ is a proper subgroup containing $Stab_G(a)$.
answered Dec 30 '18 at 6:20
waterwater
215
edited Dec 30 '18 at 22:48
answered Dec 30 '18 at 6:20
waterwater
215
answered Dec 30 '18 at 6:20
waterwater
215
answered Dec 30 '18 at 6:20
waterwater
215
215
1
$begingroup$
Your proof of $Rightarrow$ is OK, but the proof of $Leftarrow$ is confused. You have confused the block $B$, which is a subset of $S$ with its stabilizer, which is a subgroup of $G$.
$endgroup$
– Derek Holt
Dec 30 '18 at 18:21
$begingroup$
@DerekHolt Is the definition of the stabilizer of the set B either $Stab_G(B)={g:gb=b, gin G, and forall bin B}$ or $Stab_G(B) = {gin G:gB=B}$?
$endgroup$
– water
Dec 30 '18 at 19:18
1
$begingroup$
Only one of those two subgroups could strictly contain ${rm Stab}_G(a)$. I think I have answered enough questions!
$endgroup$
– Derek Holt
Dec 30 '18 at 19:56
$begingroup$
@DerekHolt I see, thank you very much.
$endgroup$
– water
Dec 30 '18 at 22:49
add a comment |
1
$begingroup$
Your proof of $Rightarrow$ is OK, but the proof of $Leftarrow$ is confused. You have confused the block $B$, which is a subset of $S$ with its stabilizer, which is a subgroup of $G$.
$endgroup$
– Derek Holt
Dec 30 '18 at 18:21
$begingroup$
@DerekHolt Is the definition of the stabilizer of the set B either $Stab_G(B)={g:gb=b, gin G, and forall bin B}$ or $Stab_G(B) = {gin G:gB=B}$?
$endgroup$
– water
Dec 30 '18 at 19:18
1
$begingroup$
Only one of those two subgroups could strictly contain ${rm Stab}_G(a)$. I think I have answered enough questions!
$endgroup$
– Derek Holt
Dec 30 '18 at 19:56
$begingroup$
@DerekHolt I see, thank you very much.
$endgroup$
– water
Dec 30 '18 at 22:49
1
1
$begingroup$
Your proof of $Rightarrow$ is OK, but the proof of $Leftarrow$ is confused. You have confused the block $B$, which is a subset of $S$ with its stabilizer, which is a subgroup of $G$.
$endgroup$
– Derek Holt
Dec 30 '18 at 18:21
$begingroup$
Your proof of $Rightarrow$ is OK, but the proof of $Leftarrow$ is confused. You have confused the block $B$, which is a subset of $S$ with its stabilizer, which is a subgroup of $G$.
$endgroup$
– Derek Holt
Dec 30 '18 at 18:21
$begingroup$
@DerekHolt Is the definition of the stabilizer of the set B either $Stab_G(B)={g:gb=b, gin G, and forall bin B}$ or $Stab_G(B) = {gin G:gB=B}$?
$endgroup$
– water
Dec 30 '18 at 19:18
$begingroup$
@DerekHolt Is the definition of the stabilizer of the set B either $Stab_G(B)={g:gb=b, gin G, and forall bin B}$ or $Stab_G(B) = {gin G:gB=B}$?
$endgroup$
– water
Dec 30 '18 at 19:18
1
1
$begingroup$
Only one of those two subgroups could strictly contain ${rm Stab}_G(a)$. I think I have answered enough questions!
$endgroup$
– Derek Holt
Dec 30 '18 at 19:56
$begingroup$
Only one of those two subgroups could strictly contain ${rm Stab}_G(a)$. I think I have answered enough questions!
$endgroup$
– Derek Holt
Dec 30 '18 at 19:56
$begingroup$
@DerekHolt I see, thank you very much.
$endgroup$
– water
Dec 30 '18 at 22:49
$begingroup$
@DerekHolt I see, thank you very much.
$endgroup$
– water
Dec 30 '18 at 22:49
add a comment |
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$begingroup$
Outline of proof: if the stabilizer is not maximal, then it is contained in a larger subgroups $H$, and the orbit of $a$ under $H$ is a block. Conversely, if the the action is imprimitive, and $B$ is a block with $a in B$, then the stabilizer of $B$ is a subgroup properly containing ${rm Stab}_G(a)$.
$endgroup$
– Derek Holt
Dec 28 '18 at 19:42
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@DerekHolt thank you very much. I will try to answer it.
$endgroup$
– water
Dec 30 '18 at 6:19
$begingroup$
@DerekHolt May I ask your comment about my answer?
$endgroup$
– water
Dec 30 '18 at 17:23