Finding $m,n,k$ for which $|G|=mn$ and $G$ is nonabelian, where $G$ is given by $langle x,ymid...
$begingroup$
$G$ is defined by the relations $x^m=y^n=1,xy=yx^k$. For which $m,n,k$ does this give a nonabelian group order $mn$?
I started playing around with this using GAP, starting with the case where $m=13$.
For $n=3$ only $k=3,9$ worked, and they generated isomorphic groups, similarly for $n=9$. For $n=5,7,11$ no value of $k$ worked. For $n=2$, $k=12$ worked. For $n=4$, $k=5,8$ worked and generated isomorphic groups. $k=12$ also worked but generated a non-isomorphic group. Similarly for $n=8$. For $n=6$, $k=3,4,9,10,12$ worked. The groups for $k=3,9$ were isomorphic. and those for $k=4,10$ were isomorphic.
For $n=12$, $k=2,dots,12$ all worked. Those for $k=2,6,7,11$ were isomorphic, those for $k=3,9$ were isomorphic, those for $k=4,10$ were isomorphic, and those for $k=5,8$ were isomorphic.
I am having difficulty figuring out what is going on. Is all this well-known material? Is there a complete or partial answer to the opening question (for which $m,n,k$ do we get a non-abelian group order $mn$)? If so, where can I find it?
Also how does one do this stuff by hand? I tried a few cases and found it quite tricky to prove that $x=1$, as often happened for the $k$ that failed.
gap group-presentation finitely-generated
edited Dec 14 '18 at 3:44
Shaun
9,083113683
asked Feb 21 '18 at 19:22
almagestalmagest
12.1k1329
$endgroup$
|
show 4 more comments
$begingroup$
$G$ is defined by the relations $x^m=y^n=1,xy=yx^k$. For which $m,n,k$ does this give a nonabelian group order $mn$?
I started playing around with this using GAP, starting with the case where $m=13$.
For $n=3$ only $k=3,9$ worked, and they generated isomorphic groups, similarly for $n=9$. For $n=5,7,11$ no value of $k$ worked. For $n=2$, $k=12$ worked. For $n=4$, $k=5,8$ worked and generated isomorphic groups. $k=12$ also worked but generated a non-isomorphic group. Similarly for $n=8$. For $n=6$, $k=3,4,9,10,12$ worked. The groups for $k=3,9$ were isomorphic. and those for $k=4,10$ were isomorphic.
For $n=12$, $k=2,dots,12$ all worked. Those for $k=2,6,7,11$ were isomorphic, those for $k=3,9$ were isomorphic, those for $k=4,10$ were isomorphic, and those for $k=5,8$ were isomorphic.
I am having difficulty figuring out what is going on. Is all this well-known material? Is there a complete or partial answer to the opening question (for which $m,n,k$ do we get a non-abelian group order $mn$)? If so, where can I find it?
Also how does one do this stuff by hand? I tried a few cases and found it quite tricky to prove that $x=1$, as often happened for the $k$ that failed.
gap group-presentation finitely-generated
edited Dec 14 '18 at 3:44
Shaun
9,083113683
asked Feb 21 '18 at 19:22
almagestalmagest
12.1k1329
$endgroup$
$begingroup$
Note that if $n=2$ and $k=m-1$, you get a dihedral group. Not very complete, I know :)
$endgroup$
– Patrick Stevens
Feb 21 '18 at 19:25
$begingroup$
Why not have a look at $y^{-n}xy^n$?
$endgroup$
– ancientmathematician
Feb 21 '18 at 20:58
1
$begingroup$
The first thing to note is that the relation can be rewritten as $y^{-1}xy = x^k$. So the requirement is that conjugation by $y$ should give an automorphism of $langle xrangle$ of order dividing $k$. Since the automorphism group of $langle xrangle$ has order $varphi(m)$, this immediately gives some restrictions. On the other hand, once these restrictions are satisfied, one can construct the desired group as a semidirect product.
$endgroup$
– Tobias Kildetoft
Feb 22 '18 at 7:01
1
$begingroup$
@ancientmathematician Yes, that is an excellent point. So the condition is $k^n=1bmod m$. Many thanks.
$endgroup$
– almagest
Feb 23 '18 at 10:53
1
$begingroup$
I suppose to be complete one also needs to see that with this necessary condition one can actually find a non-abelian group meeting this spec. I usually look in the group of all $tmapsto at+b$, $binmathbb{Z}_m$, $ainmathbb{Z}^{*}_m$ for the appropriate subgroup.
$endgroup$
– ancientmathematician
Feb 23 '18 at 11:50
|
show 4 more comments
$begingroup$
$G$ is defined by the relations $x^m=y^n=1,xy=yx^k$. For which $m,n,k$ does this give a nonabelian group order $mn$?
I started playing around with this using GAP, starting with the case where $m=13$.
For $n=3$ only $k=3,9$ worked, and they generated isomorphic groups, similarly for $n=9$. For $n=5,7,11$ no value of $k$ worked. For $n=2$, $k=12$ worked. For $n=4$, $k=5,8$ worked and generated isomorphic groups. $k=12$ also worked but generated a non-isomorphic group. Similarly for $n=8$. For $n=6$, $k=3,4,9,10,12$ worked. The groups for $k=3,9$ were isomorphic. and those for $k=4,10$ were isomorphic.
For $n=12$, $k=2,dots,12$ all worked. Those for $k=2,6,7,11$ were isomorphic, those for $k=3,9$ were isomorphic, those for $k=4,10$ were isomorphic, and those for $k=5,8$ were isomorphic.
I am having difficulty figuring out what is going on. Is all this well-known material? Is there a complete or partial answer to the opening question (for which $m,n,k$ do we get a non-abelian group order $mn$)? If so, where can I find it?
Also how does one do this stuff by hand? I tried a few cases and found it quite tricky to prove that $x=1$, as often happened for the $k$ that failed.
gap group-presentation finitely-generated
edited Dec 14 '18 at 3:44
Shaun
9,083113683
asked Feb 21 '18 at 19:22
almagestalmagest
12.1k1329
$endgroup$
$G$ is defined by the relations $x^m=y^n=1,xy=yx^k$. For which $m,n,k$ does this give a nonabelian group order $mn$?
I started playing around with this using GAP, starting with the case where $m=13$.
For $n=3$ only $k=3,9$ worked, and they generated isomorphic groups, similarly for $n=9$. For $n=5,7,11$ no value of $k$ worked. For $n=2$, $k=12$ worked. For $n=4$, $k=5,8$ worked and generated isomorphic groups. $k=12$ also worked but generated a non-isomorphic group. Similarly for $n=8$. For $n=6$, $k=3,4,9,10,12$ worked. The groups for $k=3,9$ were isomorphic. and those for $k=4,10$ were isomorphic.
For $n=12$, $k=2,dots,12$ all worked. Those for $k=2,6,7,11$ were isomorphic, those for $k=3,9$ were isomorphic, those for $k=4,10$ were isomorphic, and those for $k=5,8$ were isomorphic.
I am having difficulty figuring out what is going on. Is all this well-known material? Is there a complete or partial answer to the opening question (for which $m,n,k$ do we get a non-abelian group order $mn$)? If so, where can I find it?
Also how does one do this stuff by hand? I tried a few cases and found it quite tricky to prove that $x=1$, as often happened for the $k$ that failed.
gap group-presentation finitely-generated
gap group-presentation finitely-generated
edited Dec 14 '18 at 3:44
Shaun
9,083113683
asked Feb 21 '18 at 19:22
almagestalmagest
12.1k1329
edited Dec 14 '18 at 3:44
Shaun
9,083113683
asked Feb 21 '18 at 19:22
almagestalmagest
12.1k1329
edited Dec 14 '18 at 3:44
Shaun
9,083113683
edited Dec 14 '18 at 3:44
Shaun
9,083113683
edited Dec 14 '18 at 3:44
Shaun
9,083113683
9,083113683
asked Feb 21 '18 at 19:22
almagestalmagest
12.1k1329
asked Feb 21 '18 at 19:22
almagestalmagest
12.1k1329
asked Feb 21 '18 at 19:22
almagestalmagest
12.1k1329
12.1k1329
$begingroup$
Note that if $n=2$ and $k=m-1$, you get a dihedral group. Not very complete, I know :)
$endgroup$
– Patrick Stevens
Feb 21 '18 at 19:25
$begingroup$
Why not have a look at $y^{-n}xy^n$?
$endgroup$
– ancientmathematician
Feb 21 '18 at 20:58
1
$begingroup$
The first thing to note is that the relation can be rewritten as $y^{-1}xy = x^k$. So the requirement is that conjugation by $y$ should give an automorphism of $langle xrangle$ of order dividing $k$. Since the automorphism group of $langle xrangle$ has order $varphi(m)$, this immediately gives some restrictions. On the other hand, once these restrictions are satisfied, one can construct the desired group as a semidirect product.
$endgroup$
– Tobias Kildetoft
Feb 22 '18 at 7:01
1
$begingroup$
@ancientmathematician Yes, that is an excellent point. So the condition is $k^n=1bmod m$. Many thanks.
$endgroup$
– almagest
Feb 23 '18 at 10:53
1
$begingroup$
I suppose to be complete one also needs to see that with this necessary condition one can actually find a non-abelian group meeting this spec. I usually look in the group of all $tmapsto at+b$, $binmathbb{Z}_m$, $ainmathbb{Z}^{*}_m$ for the appropriate subgroup.
$endgroup$
– ancientmathematician
Feb 23 '18 at 11:50
|
show 4 more comments
$begingroup$
Note that if $n=2$ and $k=m-1$, you get a dihedral group. Not very complete, I know :)
$endgroup$
– Patrick Stevens
Feb 21 '18 at 19:25
$begingroup$
Why not have a look at $y^{-n}xy^n$?
$endgroup$
– ancientmathematician
Feb 21 '18 at 20:58
1
$begingroup$
The first thing to note is that the relation can be rewritten as $y^{-1}xy = x^k$. So the requirement is that conjugation by $y$ should give an automorphism of $langle xrangle$ of order dividing $k$. Since the automorphism group of $langle xrangle$ has order $varphi(m)$, this immediately gives some restrictions. On the other hand, once these restrictions are satisfied, one can construct the desired group as a semidirect product.
$endgroup$
– Tobias Kildetoft
Feb 22 '18 at 7:01
1
$begingroup$
@ancientmathematician Yes, that is an excellent point. So the condition is $k^n=1bmod m$. Many thanks.
$endgroup$
– almagest
Feb 23 '18 at 10:53
1
$begingroup$
I suppose to be complete one also needs to see that with this necessary condition one can actually find a non-abelian group meeting this spec. I usually look in the group of all $tmapsto at+b$, $binmathbb{Z}_m$, $ainmathbb{Z}^{*}_m$ for the appropriate subgroup.
$endgroup$
– ancientmathematician
Feb 23 '18 at 11:50
$begingroup$
Note that if $n=2$ and $k=m-1$, you get a dihedral group. Not very complete, I know :)
$endgroup$
– Patrick Stevens
Feb 21 '18 at 19:25
$begingroup$
Note that if $n=2$ and $k=m-1$, you get a dihedral group. Not very complete, I know :)
$endgroup$
– Patrick Stevens
Feb 21 '18 at 19:25
$begingroup$
Why not have a look at $y^{-n}xy^n$?
$endgroup$
– ancientmathematician
Feb 21 '18 at 20:58
$begingroup$
Why not have a look at $y^{-n}xy^n$?
$endgroup$
– ancientmathematician
Feb 21 '18 at 20:58
1
1
$begingroup$
The first thing to note is that the relation can be rewritten as $y^{-1}xy = x^k$. So the requirement is that conjugation by $y$ should give an automorphism of $langle xrangle$ of order dividing $k$. Since the automorphism group of $langle xrangle$ has order $varphi(m)$, this immediately gives some restrictions. On the other hand, once these restrictions are satisfied, one can construct the desired group as a semidirect product.
$endgroup$
– Tobias Kildetoft
Feb 22 '18 at 7:01
$begingroup$
The first thing to note is that the relation can be rewritten as $y^{-1}xy = x^k$. So the requirement is that conjugation by $y$ should give an automorphism of $langle xrangle$ of order dividing $k$. Since the automorphism group of $langle xrangle$ has order $varphi(m)$, this immediately gives some restrictions. On the other hand, once these restrictions are satisfied, one can construct the desired group as a semidirect product.
$endgroup$
– Tobias Kildetoft
Feb 22 '18 at 7:01
1
1
$begingroup$
@ancientmathematician Yes, that is an excellent point. So the condition is $k^n=1bmod m$. Many thanks.
$endgroup$
– almagest
Feb 23 '18 at 10:53
$begingroup$
@ancientmathematician Yes, that is an excellent point. So the condition is $k^n=1bmod m$. Many thanks.
$endgroup$
– almagest
Feb 23 '18 at 10:53
1
1
$begingroup$
I suppose to be complete one also needs to see that with this necessary condition one can actually find a non-abelian group meeting this spec. I usually look in the group of all $tmapsto at+b$, $binmathbb{Z}_m$, $ainmathbb{Z}^{*}_m$ for the appropriate subgroup.
$endgroup$
– ancientmathematician
Feb 23 '18 at 11:50
$begingroup$
I suppose to be complete one also needs to see that with this necessary condition one can actually find a non-abelian group meeting this spec. I usually look in the group of all $tmapsto at+b$, $binmathbb{Z}_m$, $ainmathbb{Z}^{*}_m$ for the appropriate subgroup.
$endgroup$
– ancientmathematician
Feb 23 '18 at 11:50
|
show 4 more comments
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$begingroup$
Note that if $n=2$ and $k=m-1$, you get a dihedral group. Not very complete, I know :)
$endgroup$
– Patrick Stevens
Feb 21 '18 at 19:25
$begingroup$
Why not have a look at $y^{-n}xy^n$?
$endgroup$
– ancientmathematician
Feb 21 '18 at 20:58
1
$begingroup$
The first thing to note is that the relation can be rewritten as $y^{-1}xy = x^k$. So the requirement is that conjugation by $y$ should give an automorphism of $langle xrangle$ of order dividing $k$. Since the automorphism group of $langle xrangle$ has order $varphi(m)$, this immediately gives some restrictions. On the other hand, once these restrictions are satisfied, one can construct the desired group as a semidirect product.
$endgroup$
– Tobias Kildetoft
Feb 22 '18 at 7:01
1
$begingroup$
@ancientmathematician Yes, that is an excellent point. So the condition is $k^n=1bmod m$. Many thanks.
$endgroup$
– almagest
Feb 23 '18 at 10:53
1
$begingroup$
I suppose to be complete one also needs to see that with this necessary condition one can actually find a non-abelian group meeting this spec. I usually look in the group of all $tmapsto at+b$, $binmathbb{Z}_m$, $ainmathbb{Z}^{*}_m$ for the appropriate subgroup.
$endgroup$
– ancientmathematician
Feb 23 '18 at 11:50