Let $G=langle X, Y|X^p=Y^q=(XY)^r=1, XY=YXrangle,$ for $p$ prime, $pleq qleq r$. Show that if $pnmid r$ then...
$begingroup$
I have a group $G$ that has presentation $$langle X, Y mid X^p=Y^q=(XY)^r=1, XY=YX rangle,$$ where $p$ is prime such that $pleq q leq r$. I need to show that if $p nmid r$ then $G = C_b$ where $b =gcd(q,pr)$.
I prove it like this.
If $p nmid r$, so $exists , s in mathbb{Z}$ such that $r=sp+d$, where $0<d<p$.
We have $1=(XY)^r=X^rY^r=X^dY^r$. Also $gcd(p,d)=1$, so $exists$ $ u,v in mathbb{Z}$ such that $1=up+vd$. So $X=X^{up+vd}=X^{vd}=Y^{-rv}$. Hence $X$ can be written with some power of $Y$.
We also have $(XY)^r=1$. So $(XY)^{rp}=1$ this implies $Y^{rp}=1$ and we already have$Y^q=1$.
By substituting $X=Y^{-rv}$ in the presentation of a group G,
we see that $X^p=Y^{-rvp}=(Y^{rp})^{-v}=1$ and $(XY)^r=Y^{r(1-rv)}$.
So how do I reduce to the presentation $G=langle Y mid Y^q=Y^{rp}=1 rangle $ to prove that these are the only relations of this group? But we have an other relation $Y^{r(1-rv)}=1$. So how could prove $Y$ has order $b=mathtt{gcd}(q,rp)$?
I know one way of proving is to find the $rm{gcd}(pr,r(1-rv))=pr$, but for this, I need to show that $p|1-rv$ that I am unable to get.
Could anyone please help me in it?
finite-groups group-presentation combinatorial-group-theory
edited Nov 29 '18 at 23:42
Shaun
9,083113683
asked Aug 16 '17 at 16:13
S786S786
526312
$endgroup$
add a comment |
$begingroup$
I have a group $G$ that has presentation $$langle X, Y mid X^p=Y^q=(XY)^r=1, XY=YX rangle,$$ where $p$ is prime such that $pleq q leq r$. I need to show that if $p nmid r$ then $G = C_b$ where $b =gcd(q,pr)$.
I prove it like this.
If $p nmid r$, so $exists , s in mathbb{Z}$ such that $r=sp+d$, where $0<d<p$.
We have $1=(XY)^r=X^rY^r=X^dY^r$. Also $gcd(p,d)=1$, so $exists$ $ u,v in mathbb{Z}$ such that $1=up+vd$. So $X=X^{up+vd}=X^{vd}=Y^{-rv}$. Hence $X$ can be written with some power of $Y$.
We also have $(XY)^r=1$. So $(XY)^{rp}=1$ this implies $Y^{rp}=1$ and we already have$Y^q=1$.
By substituting $X=Y^{-rv}$ in the presentation of a group G,
we see that $X^p=Y^{-rvp}=(Y^{rp})^{-v}=1$ and $(XY)^r=Y^{r(1-rv)}$.
So how do I reduce to the presentation $G=langle Y mid Y^q=Y^{rp}=1 rangle $ to prove that these are the only relations of this group? But we have an other relation $Y^{r(1-rv)}=1$. So how could prove $Y$ has order $b=mathtt{gcd}(q,rp)$?
I know one way of proving is to find the $rm{gcd}(pr,r(1-rv))=pr$, but for this, I need to show that $p|1-rv$ that I am unable to get.
Could anyone please help me in it?
finite-groups group-presentation combinatorial-group-theory
edited Nov 29 '18 at 23:42
Shaun
9,083113683
asked Aug 16 '17 at 16:13
S786S786
526312
$endgroup$
3
$begingroup$
$1 = up+vd = up + v(r-sp)$ so $1-rv = up - vsp$.
$endgroup$
– Derek Holt
Aug 16 '17 at 17:10
$begingroup$
@Derek Thank you very much. I didn't realise it during the proof. Thanks.
$endgroup$
– S786
Aug 16 '17 at 22:37
add a comment |
$begingroup$
I have a group $G$ that has presentation $$langle X, Y mid X^p=Y^q=(XY)^r=1, XY=YX rangle,$$ where $p$ is prime such that $pleq q leq r$. I need to show that if $p nmid r$ then $G = C_b$ where $b =gcd(q,pr)$.
I prove it like this.
If $p nmid r$, so $exists , s in mathbb{Z}$ such that $r=sp+d$, where $0<d<p$.
We have $1=(XY)^r=X^rY^r=X^dY^r$. Also $gcd(p,d)=1$, so $exists$ $ u,v in mathbb{Z}$ such that $1=up+vd$. So $X=X^{up+vd}=X^{vd}=Y^{-rv}$. Hence $X$ can be written with some power of $Y$.
We also have $(XY)^r=1$. So $(XY)^{rp}=1$ this implies $Y^{rp}=1$ and we already have$Y^q=1$.
By substituting $X=Y^{-rv}$ in the presentation of a group G,
we see that $X^p=Y^{-rvp}=(Y^{rp})^{-v}=1$ and $(XY)^r=Y^{r(1-rv)}$.
So how do I reduce to the presentation $G=langle Y mid Y^q=Y^{rp}=1 rangle $ to prove that these are the only relations of this group? But we have an other relation $Y^{r(1-rv)}=1$. So how could prove $Y$ has order $b=mathtt{gcd}(q,rp)$?
I know one way of proving is to find the $rm{gcd}(pr,r(1-rv))=pr$, but for this, I need to show that $p|1-rv$ that I am unable to get.
Could anyone please help me in it?
finite-groups group-presentation combinatorial-group-theory
edited Nov 29 '18 at 23:42
Shaun
9,083113683
asked Aug 16 '17 at 16:13
S786S786
526312
$endgroup$
I have a group $G$ that has presentation $$langle X, Y mid X^p=Y^q=(XY)^r=1, XY=YX rangle,$$ where $p$ is prime such that $pleq q leq r$. I need to show that if $p nmid r$ then $G = C_b$ where $b =gcd(q,pr)$.
I prove it like this.
If $p nmid r$, so $exists , s in mathbb{Z}$ such that $r=sp+d$, where $0<d<p$.
We have $1=(XY)^r=X^rY^r=X^dY^r$. Also $gcd(p,d)=1$, so $exists$ $ u,v in mathbb{Z}$ such that $1=up+vd$. So $X=X^{up+vd}=X^{vd}=Y^{-rv}$. Hence $X$ can be written with some power of $Y$.
We also have $(XY)^r=1$. So $(XY)^{rp}=1$ this implies $Y^{rp}=1$ and we already have$Y^q=1$.
By substituting $X=Y^{-rv}$ in the presentation of a group G,
we see that $X^p=Y^{-rvp}=(Y^{rp})^{-v}=1$ and $(XY)^r=Y^{r(1-rv)}$.
So how do I reduce to the presentation $G=langle Y mid Y^q=Y^{rp}=1 rangle $ to prove that these are the only relations of this group? But we have an other relation $Y^{r(1-rv)}=1$. So how could prove $Y$ has order $b=mathtt{gcd}(q,rp)$?
I know one way of proving is to find the $rm{gcd}(pr,r(1-rv))=pr$, but for this, I need to show that $p|1-rv$ that I am unable to get.
Could anyone please help me in it?
finite-groups group-presentation combinatorial-group-theory
finite-groups group-presentation combinatorial-group-theory
edited Nov 29 '18 at 23:42
Shaun
9,083113683
asked Aug 16 '17 at 16:13
S786S786
526312
edited Nov 29 '18 at 23:42
Shaun
9,083113683
asked Aug 16 '17 at 16:13
S786S786
526312
edited Nov 29 '18 at 23:42
Shaun
9,083113683
edited Nov 29 '18 at 23:42
Shaun
9,083113683
edited Nov 29 '18 at 23:42
Shaun
9,083113683
9,083113683
asked Aug 16 '17 at 16:13
S786S786
526312
asked Aug 16 '17 at 16:13
S786S786
526312
asked Aug 16 '17 at 16:13
S786S786
526312
526312
3
$begingroup$
$1 = up+vd = up + v(r-sp)$ so $1-rv = up - vsp$.
$endgroup$
– Derek Holt
Aug 16 '17 at 17:10
$begingroup$
@Derek Thank you very much. I didn't realise it during the proof. Thanks.
$endgroup$
– S786
Aug 16 '17 at 22:37
add a comment |
3
$begingroup$
$1 = up+vd = up + v(r-sp)$ so $1-rv = up - vsp$.
$endgroup$
– Derek Holt
Aug 16 '17 at 17:10
$begingroup$
@Derek Thank you very much. I didn't realise it during the proof. Thanks.
$endgroup$
– S786
Aug 16 '17 at 22:37
3
3
$begingroup$
$1 = up+vd = up + v(r-sp)$ so $1-rv = up - vsp$.
$endgroup$
– Derek Holt
Aug 16 '17 at 17:10
$begingroup$
$1 = up+vd = up + v(r-sp)$ so $1-rv = up - vsp$.
$endgroup$
– Derek Holt
Aug 16 '17 at 17:10
$begingroup$
@Derek Thank you very much. I didn't realise it during the proof. Thanks.
$endgroup$
– S786
Aug 16 '17 at 22:37
$begingroup$
@Derek Thank you very much. I didn't realise it during the proof. Thanks.
$endgroup$
– S786
Aug 16 '17 at 22:37
add a comment |
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3
$begingroup$
$1 = up+vd = up + v(r-sp)$ so $1-rv = up - vsp$.
$endgroup$
– Derek Holt
Aug 16 '17 at 17:10
$begingroup$
@Derek Thank you very much. I didn't realise it during the proof. Thanks.
$endgroup$
– S786
Aug 16 '17 at 22:37