How can we determine the group with the presentation $G:=leftlangle g,hmid g^4=h^4=1,hg=g^{-1}h rightrangle?$
5
$begingroup$
Please how can we determine the finite group whose presentation is given as:
$$G:=leftlangle g,hmid g^4=h^4=1,hg=g^{-1}h rightrangle?$$
See this comment for some context from the person asking this question.
finite-groups group-presentation
edited Dec 22 '18 at 13:55
Shaun
9,332113684
asked Mar 2 '15 at 12:39
ChuksChuks
915518
$endgroup$
|
show 3 more comments
5
$begingroup$
Please how can we determine the finite group whose presentation is given as:
$$G:=leftlangle g,hmid g^4=h^4=1,hg=g^{-1}h rightrangle?$$
See this comment for some context from the person asking this question.
finite-groups group-presentation
edited Dec 22 '18 at 13:55
Shaun
9,332113684
asked Mar 2 '15 at 12:39
ChuksChuks
915518
$endgroup$
$begingroup$
As a first step you could try to prove that all elements of $G$ can be written as $g^ih^j$ with $0le ile 3$ and $0le jle 3$.
$endgroup$
– j.p.
Mar 2 '15 at 14:19
$begingroup$
Thanks J.P. Your hint implies that $|G|leq 16$. But I am pretty sure that $|G|geq 16$ (as no group of order less than 16 has this nature). So, going by your suggestion, $|G|$ must be 16. However, I am still unable to identify the exact nonabelian group of order 16 that satisfies this.
$endgroup$
– Chuks
Mar 2 '15 at 14:37
$begingroup$
Next try to show that both $g^2$ and $h^2$ are in the center of $G$. (I'm not sure yet which group we will get, I just work comment to comment, I hope you don't mind. It might be that the given relations imply additional relations, but our group has at least $8$ elements.)
$endgroup$
– j.p.
Mar 2 '15 at 14:57
$begingroup$
Forget my last comment. Just try to find a semi-direct product of two cyclic groups of order $4$ that is the image of your group $G$. This shows $|G|ge 16$, and $|G|le 16$ you know by my first comment. Hence $G$ is isomorphic to that semi-direct product.
$endgroup$
– j.p.
Mar 2 '15 at 17:22
$begingroup$
Using GAP, tryF:=FreeGroup(2); rels:=[ (F.1)^4, (F.2)^4, (F.1)*(F.2)*(F.1)*(F.2)^(-1) ]; G:=F/rels; StructureDescription(G);.
$endgroup$
– Shaun
Dec 14 '18 at 2:06
|
show 3 more comments
5
5
5
1
$begingroup$
Please how can we determine the finite group whose presentation is given as:
$$G:=leftlangle g,hmid g^4=h^4=1,hg=g^{-1}h rightrangle?$$
See this comment for some context from the person asking this question.
finite-groups group-presentation
edited Dec 22 '18 at 13:55
Shaun
9,332113684
asked Mar 2 '15 at 12:39
ChuksChuks
915518
$endgroup$
Please how can we determine the finite group whose presentation is given as:
$$G:=leftlangle g,hmid g^4=h^4=1,hg=g^{-1}h rightrangle?$$
See this comment for some context from the person asking this question.
finite-groups group-presentation
finite-groups group-presentation
edited Dec 22 '18 at 13:55
Shaun
9,332113684
asked Mar 2 '15 at 12:39
ChuksChuks
915518
edited Dec 22 '18 at 13:55
Shaun
9,332113684
asked Mar 2 '15 at 12:39
ChuksChuks
915518
edited Dec 22 '18 at 13:55
Shaun
9,332113684
edited Dec 22 '18 at 13:55
Shaun
9,332113684
edited Dec 22 '18 at 13:55
Shaun
9,332113684
9,332113684
asked Mar 2 '15 at 12:39
ChuksChuks
915518
asked Mar 2 '15 at 12:39
ChuksChuks
915518
asked Mar 2 '15 at 12:39
ChuksChuks
915518
915518
$begingroup$
As a first step you could try to prove that all elements of $G$ can be written as $g^ih^j$ with $0le ile 3$ and $0le jle 3$.
$endgroup$
– j.p.
Mar 2 '15 at 14:19
$begingroup$
Thanks J.P. Your hint implies that $|G|leq 16$. But I am pretty sure that $|G|geq 16$ (as no group of order less than 16 has this nature). So, going by your suggestion, $|G|$ must be 16. However, I am still unable to identify the exact nonabelian group of order 16 that satisfies this.
$endgroup$
– Chuks
Mar 2 '15 at 14:37
$begingroup$
Next try to show that both $g^2$ and $h^2$ are in the center of $G$. (I'm not sure yet which group we will get, I just work comment to comment, I hope you don't mind. It might be that the given relations imply additional relations, but our group has at least $8$ elements.)
$endgroup$
– j.p.
Mar 2 '15 at 14:57
$begingroup$
Forget my last comment. Just try to find a semi-direct product of two cyclic groups of order $4$ that is the image of your group $G$. This shows $|G|ge 16$, and $|G|le 16$ you know by my first comment. Hence $G$ is isomorphic to that semi-direct product.
$endgroup$
– j.p.
Mar 2 '15 at 17:22
$begingroup$
Using GAP, tryF:=FreeGroup(2); rels:=[ (F.1)^4, (F.2)^4, (F.1)*(F.2)*(F.1)*(F.2)^(-1) ]; G:=F/rels; StructureDescription(G);.
$endgroup$
– Shaun
Dec 14 '18 at 2:06
|
show 3 more comments
$begingroup$
As a first step you could try to prove that all elements of $G$ can be written as $g^ih^j$ with $0le ile 3$ and $0le jle 3$.
$endgroup$
– j.p.
Mar 2 '15 at 14:19
$begingroup$
Thanks J.P. Your hint implies that $|G|leq 16$. But I am pretty sure that $|G|geq 16$ (as no group of order less than 16 has this nature). So, going by your suggestion, $|G|$ must be 16. However, I am still unable to identify the exact nonabelian group of order 16 that satisfies this.
$endgroup$
– Chuks
Mar 2 '15 at 14:37
$begingroup$
Next try to show that both $g^2$ and $h^2$ are in the center of $G$. (I'm not sure yet which group we will get, I just work comment to comment, I hope you don't mind. It might be that the given relations imply additional relations, but our group has at least $8$ elements.)
$endgroup$
– j.p.
Mar 2 '15 at 14:57
$begingroup$
Forget my last comment. Just try to find a semi-direct product of two cyclic groups of order $4$ that is the image of your group $G$. This shows $|G|ge 16$, and $|G|le 16$ you know by my first comment. Hence $G$ is isomorphic to that semi-direct product.
$endgroup$
– j.p.
Mar 2 '15 at 17:22
$begingroup$
Using GAP, tryF:=FreeGroup(2); rels:=[ (F.1)^4, (F.2)^4, (F.1)*(F.2)*(F.1)*(F.2)^(-1) ]; G:=F/rels; StructureDescription(G);.
$endgroup$
– Shaun
Dec 14 '18 at 2:06
$begingroup$
As a first step you could try to prove that all elements of $G$ can be written as $g^ih^j$ with $0le ile 3$ and $0le jle 3$.
$endgroup$
– j.p.
Mar 2 '15 at 14:19
$begingroup$
As a first step you could try to prove that all elements of $G$ can be written as $g^ih^j$ with $0le ile 3$ and $0le jle 3$.
$endgroup$
– j.p.
Mar 2 '15 at 14:19
$begingroup$
Thanks J.P. Your hint implies that $|G|leq 16$. But I am pretty sure that $|G|geq 16$ (as no group of order less than 16 has this nature). So, going by your suggestion, $|G|$ must be 16. However, I am still unable to identify the exact nonabelian group of order 16 that satisfies this.
$endgroup$
– Chuks
Mar 2 '15 at 14:37
$begingroup$
Thanks J.P. Your hint implies that $|G|leq 16$. But I am pretty sure that $|G|geq 16$ (as no group of order less than 16 has this nature). So, going by your suggestion, $|G|$ must be 16. However, I am still unable to identify the exact nonabelian group of order 16 that satisfies this.
$endgroup$
– Chuks
Mar 2 '15 at 14:37
$begingroup$
Next try to show that both $g^2$ and $h^2$ are in the center of $G$. (I'm not sure yet which group we will get, I just work comment to comment, I hope you don't mind. It might be that the given relations imply additional relations, but our group has at least $8$ elements.)
$endgroup$
– j.p.
Mar 2 '15 at 14:57
$begingroup$
Next try to show that both $g^2$ and $h^2$ are in the center of $G$. (I'm not sure yet which group we will get, I just work comment to comment, I hope you don't mind. It might be that the given relations imply additional relations, but our group has at least $8$ elements.)
$endgroup$
– j.p.
Mar 2 '15 at 14:57
$begingroup$
Forget my last comment. Just try to find a semi-direct product of two cyclic groups of order $4$ that is the image of your group $G$. This shows $|G|ge 16$, and $|G|le 16$ you know by my first comment. Hence $G$ is isomorphic to that semi-direct product.
$endgroup$
– j.p.
Mar 2 '15 at 17:22
$begingroup$
Forget my last comment. Just try to find a semi-direct product of two cyclic groups of order $4$ that is the image of your group $G$. This shows $|G|ge 16$, and $|G|le 16$ you know by my first comment. Hence $G$ is isomorphic to that semi-direct product.
$endgroup$
– j.p.
Mar 2 '15 at 17:22
$begingroup$
Using GAP, try
F:=FreeGroup(2); rels:=[ (F.1)^4, (F.2)^4, (F.1)*(F.2)*(F.1)*(F.2)^(-1) ]; G:=F/rels; StructureDescription(G);.$endgroup$
– Shaun
Dec 14 '18 at 2:06
$begingroup$
Using GAP, try
F:=FreeGroup(2); rels:=[ (F.1)^4, (F.2)^4, (F.1)*(F.2)*(F.1)*(F.2)^(-1) ]; G:=F/rels; StructureDescription(G);.$endgroup$
– Shaun
Dec 14 '18 at 2:06
|
show 3 more comments
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$begingroup$
As a first step you could try to prove that all elements of $G$ can be written as $g^ih^j$ with $0le ile 3$ and $0le jle 3$.
$endgroup$
– j.p.
Mar 2 '15 at 14:19
$begingroup$
Thanks J.P. Your hint implies that $|G|leq 16$. But I am pretty sure that $|G|geq 16$ (as no group of order less than 16 has this nature). So, going by your suggestion, $|G|$ must be 16. However, I am still unable to identify the exact nonabelian group of order 16 that satisfies this.
$endgroup$
– Chuks
Mar 2 '15 at 14:37
$begingroup$
Next try to show that both $g^2$ and $h^2$ are in the center of $G$. (I'm not sure yet which group we will get, I just work comment to comment, I hope you don't mind. It might be that the given relations imply additional relations, but our group has at least $8$ elements.)
$endgroup$
– j.p.
Mar 2 '15 at 14:57
$begingroup$
Forget my last comment. Just try to find a semi-direct product of two cyclic groups of order $4$ that is the image of your group $G$. This shows $|G|ge 16$, and $|G|le 16$ you know by my first comment. Hence $G$ is isomorphic to that semi-direct product.
$endgroup$
– j.p.
Mar 2 '15 at 17:22
$begingroup$
Using GAP, try
F:=FreeGroup(2); rels:=[ (F.1)^4, (F.2)^4, (F.1)*(F.2)*(F.1)*(F.2)^(-1) ]; G:=F/rels; StructureDescription(G);.$endgroup$
– Shaun
Dec 14 '18 at 2:06