Prove that $f^-1mathbbR^+fsubsetneqq Aut(mathbbI)$!Example where $[E:mathbbQ]<|mathrmAut_mathbbQE|$Find Aut$(G)$, Inn$(G)$ and $dfractextAut(G)textInn(G)$ for $G = mathbbZ_2 times mathbbZ_2$Prove that $psiinoperatornameAut(Q_8)$Showing that $mathbbH^*$ maps onto $mathrmAut(mathbbH)$Prove that Aut$(G) cong mathbbZ_4$Showing $textAut(mathbbQ)$ is the trivial group and calculating $textAut(mathbbQ(sqrt2))$What should be isomorphism class of $Aut(mathbbZ_p^atimes mathbbZ_p^b q^c)$?Exhibiting $textAut(N|L)$Prove that $mathbbQ[sqrt2] subsetneqqmathbbQ[sqrt2+sqrt3] $How to prove that a filter intersects all the dense sets?
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Prove that $f^-1mathbbR^+fsubsetneqq Aut(mathbbI)$!
Example where $[E:mathbbQ]<|mathrmAut_mathbbQE|$Find Aut$(G)$, Inn$(G)$ and $dfractextAut(G)textInn(G)$ for $G = mathbbZ_2 times mathbbZ_2$Prove that $psiinoperatornameAut(Q_8)$Showing that $mathbbH^*$ maps onto $mathrmAut(mathbbH)$Prove that Aut$(G) cong mathbbZ_4$Showing $textAut(mathbbQ)$ is the trivial group and calculating $textAut(mathbbQ(sqrt2))$What should be isomorphism class of $Aut(mathbbZ_p^atimes mathbbZ_p^b q^c)$?Exhibiting $textAut(N|L)$Prove that $mathbbQ[sqrt2] subsetneqqmathbbQ[sqrt2+sqrt3] $How to prove that a filter intersects all the dense sets?
$begingroup$
Let $mathbbI=([0,1],leq)$ and suppose
$Aut(mathbbI)=f$.
For any $fin Aut(mathbbI)$ and $rin mathbbR^+$, define $rf$ by $(rf)(x)=f(x)^r$. I am trying to show that
$f^-1mathbbR^+fsubsetneqq Aut(mathbbI)$ where $f^-1mathbbR^+f=f^-1rf$ and $(f^-1rf)(x)=f^-1(f(x)^r)$ for all $xin [0,1]$. Of course it is obvious that $f^-1mathbbR^+fsubseteq Aut(mathbbI)$. But why $f^-1mathbbR^+fneq Aut(mathbbI)$? Thanks.
abstract-algebra order-theory lattice-orders ordered-fields
asked Mar 28 at 15:37
S Ali MousaviS Ali Mousavi
456411
$endgroup$
|
show 4 more comments
$begingroup$
Let $mathbbI=([0,1],leq)$ and suppose
$Aut(mathbbI)=f$.
For any $fin Aut(mathbbI)$ and $rin mathbbR^+$, define $rf$ by $(rf)(x)=f(x)^r$. I am trying to show that
$f^-1mathbbR^+fsubsetneqq Aut(mathbbI)$ where $f^-1mathbbR^+f=f^-1rf$ and $(f^-1rf)(x)=f^-1(f(x)^r)$ for all $xin [0,1]$. Of course it is obvious that $f^-1mathbbR^+fsubseteq Aut(mathbbI)$. But why $f^-1mathbbR^+fneq Aut(mathbbI)$? Thanks.
abstract-algebra order-theory lattice-orders ordered-fields
asked Mar 28 at 15:37
S Ali MousaviS Ali Mousavi
456411
$endgroup$
$begingroup$
Are you using $mathbbR^+ = (0,+infty)$?
$endgroup$
– Gary Moon
Mar 28 at 15:48
$begingroup$
You seem to be testing whether this statement is true for a fixed $f$. It clearly is false in the case that $f=operatornameId_mathbb I$ because then you would be asserting that every automorphism is of the form $xmapsto x^r$.
$endgroup$
– MPW
Mar 28 at 15:48
$begingroup$
@GaryMoon Yes...
$endgroup$
– S Ali Mousavi
Mar 28 at 15:50
$begingroup$
@MPW $f$ is fix but arbitrary, not necessarily identity.
$endgroup$
– S Ali Mousavi
Mar 28 at 15:52
$begingroup$
Yes, but it fails in that case, so it can't be true for every $f$. Which $f$ are you considering?
$endgroup$
– MPW
Mar 28 at 15:55
|
show 4 more comments
$begingroup$
Let $mathbbI=([0,1],leq)$ and suppose
$Aut(mathbbI)=f$.
For any $fin Aut(mathbbI)$ and $rin mathbbR^+$, define $rf$ by $(rf)(x)=f(x)^r$. I am trying to show that
$f^-1mathbbR^+fsubsetneqq Aut(mathbbI)$ where $f^-1mathbbR^+f=f^-1rf$ and $(f^-1rf)(x)=f^-1(f(x)^r)$ for all $xin [0,1]$. Of course it is obvious that $f^-1mathbbR^+fsubseteq Aut(mathbbI)$. But why $f^-1mathbbR^+fneq Aut(mathbbI)$? Thanks.
abstract-algebra order-theory lattice-orders ordered-fields
asked Mar 28 at 15:37
S Ali MousaviS Ali Mousavi
456411
$endgroup$
Let $mathbbI=([0,1],leq)$ and suppose
$Aut(mathbbI)=f$.
For any $fin Aut(mathbbI)$ and $rin mathbbR^+$, define $rf$ by $(rf)(x)=f(x)^r$. I am trying to show that
$f^-1mathbbR^+fsubsetneqq Aut(mathbbI)$ where $f^-1mathbbR^+f=f^-1rf$ and $(f^-1rf)(x)=f^-1(f(x)^r)$ for all $xin [0,1]$. Of course it is obvious that $f^-1mathbbR^+fsubseteq Aut(mathbbI)$. But why $f^-1mathbbR^+fneq Aut(mathbbI)$? Thanks.
abstract-algebra order-theory lattice-orders ordered-fields
abstract-algebra order-theory lattice-orders ordered-fields
asked Mar 28 at 15:37
S Ali MousaviS Ali Mousavi
456411
asked Mar 28 at 15:37
S Ali MousaviS Ali Mousavi
456411
edited Mar 28 at 15:39
S Ali Mousavi
asked Mar 28 at 15:37
S Ali MousaviS Ali Mousavi
456411
asked Mar 28 at 15:37
S Ali MousaviS Ali Mousavi
456411
asked Mar 28 at 15:37
S Ali MousaviS Ali Mousavi
456411
456411
$begingroup$
Are you using $mathbbR^+ = (0,+infty)$?
$endgroup$
– Gary Moon
Mar 28 at 15:48
$begingroup$
You seem to be testing whether this statement is true for a fixed $f$. It clearly is false in the case that $f=operatornameId_mathbb I$ because then you would be asserting that every automorphism is of the form $xmapsto x^r$.
$endgroup$
– MPW
Mar 28 at 15:48
$begingroup$
@GaryMoon Yes...
$endgroup$
– S Ali Mousavi
Mar 28 at 15:50
$begingroup$
@MPW $f$ is fix but arbitrary, not necessarily identity.
$endgroup$
– S Ali Mousavi
Mar 28 at 15:52
$begingroup$
Yes, but it fails in that case, so it can't be true for every $f$. Which $f$ are you considering?
$endgroup$
– MPW
Mar 28 at 15:55
|
show 4 more comments
$begingroup$
Are you using $mathbbR^+ = (0,+infty)$?
$endgroup$
– Gary Moon
Mar 28 at 15:48
$begingroup$
You seem to be testing whether this statement is true for a fixed $f$. It clearly is false in the case that $f=operatornameId_mathbb I$ because then you would be asserting that every automorphism is of the form $xmapsto x^r$.
$endgroup$
– MPW
Mar 28 at 15:48
$begingroup$
@GaryMoon Yes...
$endgroup$
– S Ali Mousavi
Mar 28 at 15:50
$begingroup$
@MPW $f$ is fix but arbitrary, not necessarily identity.
$endgroup$
– S Ali Mousavi
Mar 28 at 15:52
$begingroup$
Yes, but it fails in that case, so it can't be true for every $f$. Which $f$ are you considering?
$endgroup$
– MPW
Mar 28 at 15:55
$begingroup$
Are you using $mathbbR^+ = (0,+infty)$?
$endgroup$
– Gary Moon
Mar 28 at 15:48
$begingroup$
Are you using $mathbbR^+ = (0,+infty)$?
$endgroup$
– Gary Moon
Mar 28 at 15:48
$begingroup$
You seem to be testing whether this statement is true for a fixed $f$. It clearly is false in the case that $f=operatornameId_mathbb I$ because then you would be asserting that every automorphism is of the form $xmapsto x^r$.
$endgroup$
– MPW
Mar 28 at 15:48
$begingroup$
You seem to be testing whether this statement is true for a fixed $f$. It clearly is false in the case that $f=operatornameId_mathbb I$ because then you would be asserting that every automorphism is of the form $xmapsto x^r$.
$endgroup$
– MPW
Mar 28 at 15:48
$begingroup$
@GaryMoon Yes...
$endgroup$
– S Ali Mousavi
Mar 28 at 15:50
$begingroup$
@GaryMoon Yes...
$endgroup$
– S Ali Mousavi
Mar 28 at 15:50
$begingroup$
@MPW $f$ is fix but arbitrary, not necessarily identity.
$endgroup$
– S Ali Mousavi
Mar 28 at 15:52
$begingroup$
@MPW $f$ is fix but arbitrary, not necessarily identity.
$endgroup$
– S Ali Mousavi
Mar 28 at 15:52
$begingroup$
Yes, but it fails in that case, so it can't be true for every $f$. Which $f$ are you considering?
$endgroup$
– MPW
Mar 28 at 15:55
$begingroup$
Yes, but it fails in that case, so it can't be true for every $f$. Which $f$ are you considering?
$endgroup$
– MPW
Mar 28 at 15:55
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
We consider $r>0$ as the isomorphism $xmapsto x^r$, so $rf$ and $f^-1rf$ are just compositions. So $mathbb R^+$ is a (proper) subgroup of $Aut(mathbb I)$, and $f^-1mathbb R^+f$ are its conjugates, hence also proper subgroups.
answered Mar 28 at 16:08
SMMSMM
2,918510
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We consider $r>0$ as the isomorphism $xmapsto x^r$, so $rf$ and $f^-1rf$ are just compositions. So $mathbb R^+$ is a (proper) subgroup of $Aut(mathbb I)$, and $f^-1mathbb R^+f$ are its conjugates, hence also proper subgroups.
answered Mar 28 at 16:08
SMMSMM
2,918510
$endgroup$
add a comment |
$begingroup$
We consider $r>0$ as the isomorphism $xmapsto x^r$, so $rf$ and $f^-1rf$ are just compositions. So $mathbb R^+$ is a (proper) subgroup of $Aut(mathbb I)$, and $f^-1mathbb R^+f$ are its conjugates, hence also proper subgroups.
answered Mar 28 at 16:08
SMMSMM
2,918510
$endgroup$
add a comment |
$begingroup$
We consider $r>0$ as the isomorphism $xmapsto x^r$, so $rf$ and $f^-1rf$ are just compositions. So $mathbb R^+$ is a (proper) subgroup of $Aut(mathbb I)$, and $f^-1mathbb R^+f$ are its conjugates, hence also proper subgroups.
answered Mar 28 at 16:08
SMMSMM
2,918510
$endgroup$
We consider $r>0$ as the isomorphism $xmapsto x^r$, so $rf$ and $f^-1rf$ are just compositions. So $mathbb R^+$ is a (proper) subgroup of $Aut(mathbb I)$, and $f^-1mathbb R^+f$ are its conjugates, hence also proper subgroups.
answered Mar 28 at 16:08
SMMSMM
2,918510
answered Mar 28 at 16:08
SMMSMM
2,918510
answered Mar 28 at 16:08
SMMSMM
2,918510
answered Mar 28 at 16:08
SMMSMM
2,918510
2,918510
add a comment |
add a comment |
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$begingroup$
Are you using $mathbbR^+ = (0,+infty)$?
$endgroup$
– Gary Moon
Mar 28 at 15:48
$begingroup$
You seem to be testing whether this statement is true for a fixed $f$. It clearly is false in the case that $f=operatornameId_mathbb I$ because then you would be asserting that every automorphism is of the form $xmapsto x^r$.
$endgroup$
– MPW
Mar 28 at 15:48
$begingroup$
@GaryMoon Yes...
$endgroup$
– S Ali Mousavi
Mar 28 at 15:50
$begingroup$
@MPW $f$ is fix but arbitrary, not necessarily identity.
$endgroup$
– S Ali Mousavi
Mar 28 at 15:52
$begingroup$
Yes, but it fails in that case, so it can't be true for every $f$. Which $f$ are you considering?
$endgroup$
– MPW
Mar 28 at 15:55