If $n=3k$, then $langle x,ymid x^n=y^2=1,xy=yx^2rangle$ has the same presentation as $D_6$ after $xmapsto r$...
$begingroup$
I am solving problems in Dummit and Foote; however, this problem I am not able to do.
Show that if $n=3k$, then $X_{2n}$ has order $6$, and it has same generators and relations as $D_6$ when $x$ is replaced by $r$ and $y$ is replaced by $s$, where $$X_{2n}=langle x,ymid x^n=y^2=1, xy=yx^2rangle.$$
So here is my argument
$$begin{align}
xy = yx^2&rightarrow xy^2 = yx^2y\
&rightarrow x = yxxy = yxyx^2 = yy x^2x^2 = x^4\
&rightarrow x^3 = e,
end{align}$$
so ord(x) | 3k and ord(x) | 3 so ord(x) = 1 or ord(x) = 3; however, I tried to show that ord(x) = 1 will lead to a contradiction but I can't get that contradiction.
Ok so thanks to Paul Hammer I was able to prove it. All I need to show is that I derive the relations of $D_6$ from $X_{6k}$ and vice versa.
So first I will go from relation of $X_{6k}$ to relations of $D_6$.
We want to know if $xy * (xy^{-1}) = e$, so $xy * (xy^{-1}) = yx^2xy^{-1} = yx^3y^{-1} = e$. The reason we know $x^3 = e$ is because we know ord(x) = 3 or ord(x) = 1, so both cases it will satisfy $x^3 = e$.
Now going the other way we want to know if $xy = yx^{-1}$ $rightarrow$ $xy = yx^2$.
Well this is same as checking whether $xy*(yx^2)^{-1} = e$.
So we have
$$begin{align}
xy*(yx^2)^{-1}& = xy*x^{-2}y^{-1}\
& = yx^{-1}x^{-1}x^{-1}y^{-1} \
&= yx^{-3}y^{-1} \
&= y*(x^3)^{-1}y^-1\
& = e,
end{align}$$ and we are done!
group-theory proof-verification group-presentation
edited Dec 14 '18 at 2:28
Shaun
9,332113684
$endgroup$
add a comment |
$begingroup$
I am solving problems in Dummit and Foote; however, this problem I am not able to do.
Show that if $n=3k$, then $X_{2n}$ has order $6$, and it has same generators and relations as $D_6$ when $x$ is replaced by $r$ and $y$ is replaced by $s$, where $$X_{2n}=langle x,ymid x^n=y^2=1, xy=yx^2rangle.$$
So here is my argument
$$begin{align}
xy = yx^2&rightarrow xy^2 = yx^2y\
&rightarrow x = yxxy = yxyx^2 = yy x^2x^2 = x^4\
&rightarrow x^3 = e,
end{align}$$
so ord(x) | 3k and ord(x) | 3 so ord(x) = 1 or ord(x) = 3; however, I tried to show that ord(x) = 1 will lead to a contradiction but I can't get that contradiction.
Ok so thanks to Paul Hammer I was able to prove it. All I need to show is that I derive the relations of $D_6$ from $X_{6k}$ and vice versa.
So first I will go from relation of $X_{6k}$ to relations of $D_6$.
We want to know if $xy * (xy^{-1}) = e$, so $xy * (xy^{-1}) = yx^2xy^{-1} = yx^3y^{-1} = e$. The reason we know $x^3 = e$ is because we know ord(x) = 3 or ord(x) = 1, so both cases it will satisfy $x^3 = e$.
Now going the other way we want to know if $xy = yx^{-1}$ $rightarrow$ $xy = yx^2$.
Well this is same as checking whether $xy*(yx^2)^{-1} = e$.
So we have
$$begin{align}
xy*(yx^2)^{-1}& = xy*x^{-2}y^{-1}\
& = yx^{-1}x^{-1}x^{-1}y^{-1} \
&= yx^{-3}y^{-1} \
&= y*(x^3)^{-1}y^-1\
& = e,
end{align}$$ and we are done!
group-theory proof-verification group-presentation
edited Dec 14 '18 at 2:28
Shaun
9,332113684
$endgroup$
$begingroup$
well I know how to do the derive the relations of $D_6$ and vice versa so I just need to show that order of $X_{2n} = 6$ and that would prove it so that is what I am asking.
$endgroup$
– user111750
May 15 '15 at 22:53
1
$begingroup$
You can exhibit a group with two elements satisfying those relations in which $xne e$. Namely, $D_6$. That proves that $xne e$ in $X_{2n}$.
$endgroup$
– anon
May 16 '15 at 0:30
add a comment |
$begingroup$
I am solving problems in Dummit and Foote; however, this problem I am not able to do.
Show that if $n=3k$, then $X_{2n}$ has order $6$, and it has same generators and relations as $D_6$ when $x$ is replaced by $r$ and $y$ is replaced by $s$, where $$X_{2n}=langle x,ymid x^n=y^2=1, xy=yx^2rangle.$$
So here is my argument
$$begin{align}
xy = yx^2&rightarrow xy^2 = yx^2y\
&rightarrow x = yxxy = yxyx^2 = yy x^2x^2 = x^4\
&rightarrow x^3 = e,
end{align}$$
so ord(x) | 3k and ord(x) | 3 so ord(x) = 1 or ord(x) = 3; however, I tried to show that ord(x) = 1 will lead to a contradiction but I can't get that contradiction.
Ok so thanks to Paul Hammer I was able to prove it. All I need to show is that I derive the relations of $D_6$ from $X_{6k}$ and vice versa.
So first I will go from relation of $X_{6k}$ to relations of $D_6$.
We want to know if $xy * (xy^{-1}) = e$, so $xy * (xy^{-1}) = yx^2xy^{-1} = yx^3y^{-1} = e$. The reason we know $x^3 = e$ is because we know ord(x) = 3 or ord(x) = 1, so both cases it will satisfy $x^3 = e$.
Now going the other way we want to know if $xy = yx^{-1}$ $rightarrow$ $xy = yx^2$.
Well this is same as checking whether $xy*(yx^2)^{-1} = e$.
So we have
$$begin{align}
xy*(yx^2)^{-1}& = xy*x^{-2}y^{-1}\
& = yx^{-1}x^{-1}x^{-1}y^{-1} \
&= yx^{-3}y^{-1} \
&= y*(x^3)^{-1}y^-1\
& = e,
end{align}$$ and we are done!
group-theory proof-verification group-presentation
edited Dec 14 '18 at 2:28
Shaun
9,332113684
$endgroup$
I am solving problems in Dummit and Foote; however, this problem I am not able to do.
Show that if $n=3k$, then $X_{2n}$ has order $6$, and it has same generators and relations as $D_6$ when $x$ is replaced by $r$ and $y$ is replaced by $s$, where $$X_{2n}=langle x,ymid x^n=y^2=1, xy=yx^2rangle.$$
So here is my argument
$$begin{align}
xy = yx^2&rightarrow xy^2 = yx^2y\
&rightarrow x = yxxy = yxyx^2 = yy x^2x^2 = x^4\
&rightarrow x^3 = e,
end{align}$$
so ord(x) | 3k and ord(x) | 3 so ord(x) = 1 or ord(x) = 3; however, I tried to show that ord(x) = 1 will lead to a contradiction but I can't get that contradiction.
Ok so thanks to Paul Hammer I was able to prove it. All I need to show is that I derive the relations of $D_6$ from $X_{6k}$ and vice versa.
So first I will go from relation of $X_{6k}$ to relations of $D_6$.
We want to know if $xy * (xy^{-1}) = e$, so $xy * (xy^{-1}) = yx^2xy^{-1} = yx^3y^{-1} = e$. The reason we know $x^3 = e$ is because we know ord(x) = 3 or ord(x) = 1, so both cases it will satisfy $x^3 = e$.
Now going the other way we want to know if $xy = yx^{-1}$ $rightarrow$ $xy = yx^2$.
Well this is same as checking whether $xy*(yx^2)^{-1} = e$.
So we have
$$begin{align}
xy*(yx^2)^{-1}& = xy*x^{-2}y^{-1}\
& = yx^{-1}x^{-1}x^{-1}y^{-1} \
&= yx^{-3}y^{-1} \
&= y*(x^3)^{-1}y^-1\
& = e,
end{align}$$ and we are done!
group-theory proof-verification group-presentation
group-theory proof-verification group-presentation
edited Dec 14 '18 at 2:28
Shaun
9,332113684
edited Dec 14 '18 at 2:28
Shaun
9,332113684
edited Dec 14 '18 at 2:28
Shaun
9,332113684
edited Dec 14 '18 at 2:28
Shaun
9,332113684
edited Dec 14 '18 at 2:28
Shaun
9,332113684
9,332113684
asked May 15 '15 at 22:41
user111750
$begingroup$
well I know how to do the derive the relations of $D_6$ and vice versa so I just need to show that order of $X_{2n} = 6$ and that would prove it so that is what I am asking.
$endgroup$
– user111750
May 15 '15 at 22:53
1
$begingroup$
You can exhibit a group with two elements satisfying those relations in which $xne e$. Namely, $D_6$. That proves that $xne e$ in $X_{2n}$.
$endgroup$
– anon
May 16 '15 at 0:30
add a comment |
$begingroup$
well I know how to do the derive the relations of $D_6$ and vice versa so I just need to show that order of $X_{2n} = 6$ and that would prove it so that is what I am asking.
$endgroup$
– user111750
May 15 '15 at 22:53
1
$begingroup$
You can exhibit a group with two elements satisfying those relations in which $xne e$. Namely, $D_6$. That proves that $xne e$ in $X_{2n}$.
$endgroup$
– anon
May 16 '15 at 0:30
$begingroup$
well I know how to do the derive the relations of $D_6$ and vice versa so I just need to show that order of $X_{2n} = 6$ and that would prove it so that is what I am asking.
$endgroup$
– user111750
May 15 '15 at 22:53
$begingroup$
well I know how to do the derive the relations of $D_6$ and vice versa so I just need to show that order of $X_{2n} = 6$ and that would prove it so that is what I am asking.
$endgroup$
– user111750
May 15 '15 at 22:53
1
1
$begingroup$
You can exhibit a group with two elements satisfying those relations in which $xne e$. Namely, $D_6$. That proves that $xne e$ in $X_{2n}$.
$endgroup$
– anon
May 16 '15 at 0:30
$begingroup$
You can exhibit a group with two elements satisfying those relations in which $xne e$. Namely, $D_6$. That proves that $xne e$ in $X_{2n}$.
$endgroup$
– anon
May 16 '15 at 0:30
add a comment |
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$begingroup$
well I know how to do the derive the relations of $D_6$ and vice versa so I just need to show that order of $X_{2n} = 6$ and that would prove it so that is what I am asking.
$endgroup$
– user111750
May 15 '15 at 22:53
1
$begingroup$
You can exhibit a group with two elements satisfying those relations in which $xne e$. Namely, $D_6$. That proves that $xne e$ in $X_{2n}$.
$endgroup$
– anon
May 16 '15 at 0:30