How to construct the homomorphism in semidirect product of $Z_3$ and $Z_{13}$?
0
$begingroup$
I know that in the semidirect product of $A$ and $B$, the homomorphism $phi:Arightarrow Aut(B)$ should be $phi_y(x) = yxy^{-1}$ but have no idea how to construct one for $phi:Z_3rightarrow Aut(Z_{13})$. Any help is appreciated. The presentation of such a group is given here Finding presentation of group of order 39 but I don't know what the explicit homomorphism would be.
abstract-algebra group-theory finite-groups semidirect-product automorphism-group
edited Dec 3 '18 at 22:12
Arturo Magidin
263k34587915
asked Dec 3 '18 at 21:52
manifoldedmanifolded
1907
$endgroup$
|
show 2 more comments
0
$begingroup$
I know that in the semidirect product of $A$ and $B$, the homomorphism $phi:Arightarrow Aut(B)$ should be $phi_y(x) = yxy^{-1}$ but have no idea how to construct one for $phi:Z_3rightarrow Aut(Z_{13})$. Any help is appreciated. The presentation of such a group is given here Finding presentation of group of order 39 but I don't know what the explicit homomorphism would be.
abstract-algebra group-theory finite-groups semidirect-product automorphism-group
edited Dec 3 '18 at 22:12
Arturo Magidin
263k34587915
asked Dec 3 '18 at 21:52
manifoldedmanifolded
1907
$endgroup$
2
$begingroup$
“Homeomorphism” is a topological term: it means a continuous bijection with continuous inverse. Presumably, you mean homomorphism.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 21:57
$begingroup$
See this question.
$endgroup$
– Dietrich Burde
Dec 3 '18 at 22:01
$begingroup$
@ArturoMagidin Yes, corrected that.
$endgroup$
– manifolded
Dec 3 '18 at 22:02
$begingroup$
... not everywhere... but now it’s fixed.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:12
$begingroup$
@DietrichBurde I wanted to know which homomorphism corresponds to the presentation ${x,y|x^{13}=y^3=1,yxy^{-1} = x^3}$? And that question answers how to find all the homomorphisms.
$endgroup$
– manifolded
Dec 3 '18 at 22:20
|
show 2 more comments
0
0
0
$begingroup$
I know that in the semidirect product of $A$ and $B$, the homomorphism $phi:Arightarrow Aut(B)$ should be $phi_y(x) = yxy^{-1}$ but have no idea how to construct one for $phi:Z_3rightarrow Aut(Z_{13})$. Any help is appreciated. The presentation of such a group is given here Finding presentation of group of order 39 but I don't know what the explicit homomorphism would be.
abstract-algebra group-theory finite-groups semidirect-product automorphism-group
edited Dec 3 '18 at 22:12
Arturo Magidin
263k34587915
asked Dec 3 '18 at 21:52
manifoldedmanifolded
1907
$endgroup$
I know that in the semidirect product of $A$ and $B$, the homomorphism $phi:Arightarrow Aut(B)$ should be $phi_y(x) = yxy^{-1}$ but have no idea how to construct one for $phi:Z_3rightarrow Aut(Z_{13})$. Any help is appreciated. The presentation of such a group is given here Finding presentation of group of order 39 but I don't know what the explicit homomorphism would be.
abstract-algebra group-theory finite-groups semidirect-product automorphism-group
abstract-algebra group-theory finite-groups semidirect-product automorphism-group
edited Dec 3 '18 at 22:12
Arturo Magidin
263k34587915
asked Dec 3 '18 at 21:52
manifoldedmanifolded
1907
edited Dec 3 '18 at 22:12
Arturo Magidin
263k34587915
asked Dec 3 '18 at 21:52
manifoldedmanifolded
1907
edited Dec 3 '18 at 22:12
Arturo Magidin
263k34587915
edited Dec 3 '18 at 22:12
Arturo Magidin
263k34587915
edited Dec 3 '18 at 22:12
Arturo Magidin
263k34587915
263k34587915
asked Dec 3 '18 at 21:52
manifoldedmanifolded
1907
asked Dec 3 '18 at 21:52
manifoldedmanifolded
1907
asked Dec 3 '18 at 21:52
manifoldedmanifolded
1907
1907
2
$begingroup$
“Homeomorphism” is a topological term: it means a continuous bijection with continuous inverse. Presumably, you mean homomorphism.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 21:57
$begingroup$
See this question.
$endgroup$
– Dietrich Burde
Dec 3 '18 at 22:01
$begingroup$
@ArturoMagidin Yes, corrected that.
$endgroup$
– manifolded
Dec 3 '18 at 22:02
$begingroup$
... not everywhere... but now it’s fixed.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:12
$begingroup$
@DietrichBurde I wanted to know which homomorphism corresponds to the presentation ${x,y|x^{13}=y^3=1,yxy^{-1} = x^3}$? And that question answers how to find all the homomorphisms.
$endgroup$
– manifolded
Dec 3 '18 at 22:20
|
show 2 more comments
2
$begingroup$
“Homeomorphism” is a topological term: it means a continuous bijection with continuous inverse. Presumably, you mean homomorphism.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 21:57
$begingroup$
See this question.
$endgroup$
– Dietrich Burde
Dec 3 '18 at 22:01
$begingroup$
@ArturoMagidin Yes, corrected that.
$endgroup$
– manifolded
Dec 3 '18 at 22:02
$begingroup$
... not everywhere... but now it’s fixed.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:12
$begingroup$
@DietrichBurde I wanted to know which homomorphism corresponds to the presentation ${x,y|x^{13}=y^3=1,yxy^{-1} = x^3}$? And that question answers how to find all the homomorphisms.
$endgroup$
– manifolded
Dec 3 '18 at 22:20
2
2
$begingroup$
“Homeomorphism” is a topological term: it means a continuous bijection with continuous inverse. Presumably, you mean homomorphism.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 21:57
$begingroup$
“Homeomorphism” is a topological term: it means a continuous bijection with continuous inverse. Presumably, you mean homomorphism.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 21:57
$begingroup$
See this question.
$endgroup$
– Dietrich Burde
Dec 3 '18 at 22:01
$begingroup$
See this question.
$endgroup$
– Dietrich Burde
Dec 3 '18 at 22:01
$begingroup$
@ArturoMagidin Yes, corrected that.
$endgroup$
– manifolded
Dec 3 '18 at 22:02
$begingroup$
@ArturoMagidin Yes, corrected that.
$endgroup$
– manifolded
Dec 3 '18 at 22:02
$begingroup$
... not everywhere... but now it’s fixed.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:12
$begingroup$
... not everywhere... but now it’s fixed.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:12
$begingroup$
@DietrichBurde I wanted to know which homomorphism corresponds to the presentation ${x,y|x^{13}=y^3=1,yxy^{-1} = x^3}$? And that question answers how to find all the homomorphisms.
$endgroup$
– manifolded
Dec 3 '18 at 22:20
$begingroup$
@DietrichBurde I wanted to know which homomorphism corresponds to the presentation ${x,y|x^{13}=y^3=1,yxy^{-1} = x^3}$? And that question answers how to find all the homomorphisms.
$endgroup$
– manifolded
Dec 3 '18 at 22:20
|
show 2 more comments
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2
$begingroup$
“Homeomorphism” is a topological term: it means a continuous bijection with continuous inverse. Presumably, you mean homomorphism.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 21:57
$begingroup$
See this question.
$endgroup$
– Dietrich Burde
Dec 3 '18 at 22:01
$begingroup$
@ArturoMagidin Yes, corrected that.
$endgroup$
– manifolded
Dec 3 '18 at 22:02
$begingroup$
... not everywhere... but now it’s fixed.
$endgroup$
– Arturo Magidin
Dec 3 '18 at 22:12
$begingroup$
@DietrichBurde I wanted to know which homomorphism corresponds to the presentation ${x,y|x^{13}=y^3=1,yxy^{-1} = x^3}$? And that question answers how to find all the homomorphisms.
$endgroup$
– manifolded
Dec 3 '18 at 22:20