Does $X_{2n}=langle x,y|x^n=y^2=1,xy=yx^2rangle$ really have order $6$?
Here is the entire question:
Let $X_{2n}=langle x,ymid x^n=y^2=1,xy=yx^2rangle $ and $n=3k$ then show that $X_{2n}$ has order $6$.
Here is my question:
How do we know that $X_{2n}$ has order $6$ and not order $2$?
While tackling this question I assumed that the statement
$x^n=1$
would mean that
$x^kneq 1$ for $1leq kleq n-1$
But as you go through this problem you can prove that $x^3=1$.
Well this lead to a deeper question about the uniqueness presentations.
Sure we could have the group $G={1,x,x^2,y,yx,yx^2}$ satisfy the relation, but why doesn't $G={1,y}$ not also satify the relation?
group-theory finite-groups group-presentation
edited Dec 3 '18 at 1:58
Shaun
8,750113680
asked Jun 17 '17 at 19:43
user160110
1,314716
|
show 5 more comments
Here is the entire question:
Let $X_{2n}=langle x,ymid x^n=y^2=1,xy=yx^2rangle $ and $n=3k$ then show that $X_{2n}$ has order $6$.
Here is my question:
How do we know that $X_{2n}$ has order $6$ and not order $2$?
While tackling this question I assumed that the statement
$x^n=1$
would mean that
$x^kneq 1$ for $1leq kleq n-1$
But as you go through this problem you can prove that $x^3=1$.
Well this lead to a deeper question about the uniqueness presentations.
Sure we could have the group $G={1,x,x^2,y,yx,yx^2}$ satisfy the relation, but why doesn't $G={1,y}$ not also satify the relation?
group-theory finite-groups group-presentation
edited Dec 3 '18 at 1:58
Shaun
8,750113680
asked Jun 17 '17 at 19:43
user160110
1,314716
Adding the relation $x=e$ changes the group. The trivial group also satisfies the relations.
– lulu
Jun 17 '17 at 19:44
@lulu So presentations aren't unique?
– user160110
Jun 17 '17 at 19:49
Not following. Are you asking if one group can have more than one presentation? If so, the answer is "of course". The Word Problem famously tells you that there is no algorithm for deciding if a given presentation leads to the trivial group, for example.
– lulu
Jun 17 '17 at 19:51
@lulu I am asking if a presentation can be satisfied by nonisomorpic groups.
– user160110
Jun 17 '17 at 19:52
Then the answer is "of course not". In your case you added a new relation (namely $x=e$) so the presentations are not the same.
– lulu
Jun 17 '17 at 19:53
|
show 5 more comments
Here is the entire question:
Let $X_{2n}=langle x,ymid x^n=y^2=1,xy=yx^2rangle $ and $n=3k$ then show that $X_{2n}$ has order $6$.
Here is my question:
How do we know that $X_{2n}$ has order $6$ and not order $2$?
While tackling this question I assumed that the statement
$x^n=1$
would mean that
$x^kneq 1$ for $1leq kleq n-1$
But as you go through this problem you can prove that $x^3=1$.
Well this lead to a deeper question about the uniqueness presentations.
Sure we could have the group $G={1,x,x^2,y,yx,yx^2}$ satisfy the relation, but why doesn't $G={1,y}$ not also satify the relation?
group-theory finite-groups group-presentation
edited Dec 3 '18 at 1:58
Shaun
8,750113680
asked Jun 17 '17 at 19:43
user160110
1,314716
Here is the entire question:
Let $X_{2n}=langle x,ymid x^n=y^2=1,xy=yx^2rangle $ and $n=3k$ then show that $X_{2n}$ has order $6$.
Here is my question:
How do we know that $X_{2n}$ has order $6$ and not order $2$?
While tackling this question I assumed that the statement
$x^n=1$
would mean that
$x^kneq 1$ for $1leq kleq n-1$
But as you go through this problem you can prove that $x^3=1$.
Well this lead to a deeper question about the uniqueness presentations.
Sure we could have the group $G={1,x,x^2,y,yx,yx^2}$ satisfy the relation, but why doesn't $G={1,y}$ not also satify the relation?
group-theory finite-groups group-presentation
group-theory finite-groups group-presentation
edited Dec 3 '18 at 1:58
Shaun
8,750113680
asked Jun 17 '17 at 19:43
user160110
1,314716
edited Dec 3 '18 at 1:58
Shaun
8,750113680
asked Jun 17 '17 at 19:43
user160110
1,314716
edited Dec 3 '18 at 1:58
Shaun
8,750113680
edited Dec 3 '18 at 1:58
Shaun
8,750113680
edited Dec 3 '18 at 1:58
Shaun
8,750113680
8,750113680
asked Jun 17 '17 at 19:43
user160110
1,314716
asked Jun 17 '17 at 19:43
user160110
1,314716
asked Jun 17 '17 at 19:43
user160110
1,314716
1,314716
Adding the relation $x=e$ changes the group. The trivial group also satisfies the relations.
– lulu
Jun 17 '17 at 19:44
@lulu So presentations aren't unique?
– user160110
Jun 17 '17 at 19:49
Not following. Are you asking if one group can have more than one presentation? If so, the answer is "of course". The Word Problem famously tells you that there is no algorithm for deciding if a given presentation leads to the trivial group, for example.
– lulu
Jun 17 '17 at 19:51
@lulu I am asking if a presentation can be satisfied by nonisomorpic groups.
– user160110
Jun 17 '17 at 19:52
Then the answer is "of course not". In your case you added a new relation (namely $x=e$) so the presentations are not the same.
– lulu
Jun 17 '17 at 19:53
|
show 5 more comments
Adding the relation $x=e$ changes the group. The trivial group also satisfies the relations.
– lulu
Jun 17 '17 at 19:44
@lulu So presentations aren't unique?
– user160110
Jun 17 '17 at 19:49
Not following. Are you asking if one group can have more than one presentation? If so, the answer is "of course". The Word Problem famously tells you that there is no algorithm for deciding if a given presentation leads to the trivial group, for example.
– lulu
Jun 17 '17 at 19:51
@lulu I am asking if a presentation can be satisfied by nonisomorpic groups.
– user160110
Jun 17 '17 at 19:52
Then the answer is "of course not". In your case you added a new relation (namely $x=e$) so the presentations are not the same.
– lulu
Jun 17 '17 at 19:53
Adding the relation $x=e$ changes the group. The trivial group also satisfies the relations.
– lulu
Jun 17 '17 at 19:44
Adding the relation $x=e$ changes the group. The trivial group also satisfies the relations.
– lulu
Jun 17 '17 at 19:44
@lulu So presentations aren't unique?
– user160110
Jun 17 '17 at 19:49
@lulu So presentations aren't unique?
– user160110
Jun 17 '17 at 19:49
Not following. Are you asking if one group can have more than one presentation? If so, the answer is "of course". The Word Problem famously tells you that there is no algorithm for deciding if a given presentation leads to the trivial group, for example.
– lulu
Jun 17 '17 at 19:51
Not following. Are you asking if one group can have more than one presentation? If so, the answer is "of course". The Word Problem famously tells you that there is no algorithm for deciding if a given presentation leads to the trivial group, for example.
– lulu
Jun 17 '17 at 19:51
@lulu I am asking if a presentation can be satisfied by nonisomorpic groups.
– user160110
Jun 17 '17 at 19:52
@lulu I am asking if a presentation can be satisfied by nonisomorpic groups.
– user160110
Jun 17 '17 at 19:52
Then the answer is "of course not". In your case you added a new relation (namely $x=e$) so the presentations are not the same.
– lulu
Jun 17 '17 at 19:53
Then the answer is "of course not". In your case you added a new relation (namely $x=e$) so the presentations are not the same.
– lulu
Jun 17 '17 at 19:53
|
show 5 more comments
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Adding the relation $x=e$ changes the group. The trivial group also satisfies the relations.
– lulu
Jun 17 '17 at 19:44
@lulu So presentations aren't unique?
– user160110
Jun 17 '17 at 19:49
Not following. Are you asking if one group can have more than one presentation? If so, the answer is "of course". The Word Problem famously tells you that there is no algorithm for deciding if a given presentation leads to the trivial group, for example.
– lulu
Jun 17 '17 at 19:51
@lulu I am asking if a presentation can be satisfied by nonisomorpic groups.
– user160110
Jun 17 '17 at 19:52
Then the answer is "of course not". In your case you added a new relation (namely $x=e$) so the presentations are not the same.
– lulu
Jun 17 '17 at 19:53