Let $G=langle Xmid Rrangle$ and $H=langle Xmid Srangle$. If $Rsubseteq S$, the $H$ is isomorphic to a...
1
$begingroup$
Assume that a group $G$ has a presentation $langle X mid R rangle$ and a group $H$ has a presentation $langle X mid S rangle$. If $R subseteq S$, the $H$ is isomorphic to a quotient of $G$.
In fact, if $F_X$ is the free group and $varphi: F_X rightarrow G$ and $psi:F_X rightarrow H$ are the homomorphism, then $ker varphi = R^{F_X} leq S^{F_X} = ker psi$. So, $H cong F_X/kerpsi$ is isomorphic to a quotient of $F_X/kervarphi cong G$.
group-presentation quotient-group
edited Dec 14 '18 at 2:12
Shaun
9,083113683
asked Apr 8 '16 at 5:57
math91math91
62
$endgroup$
add a comment |
1
$begingroup$
Assume that a group $G$ has a presentation $langle X mid R rangle$ and a group $H$ has a presentation $langle X mid S rangle$. If $R subseteq S$, the $H$ is isomorphic to a quotient of $G$.
In fact, if $F_X$ is the free group and $varphi: F_X rightarrow G$ and $psi:F_X rightarrow H$ are the homomorphism, then $ker varphi = R^{F_X} leq S^{F_X} = ker psi$. So, $H cong F_X/kerpsi$ is isomorphic to a quotient of $F_X/kervarphi cong G$.
group-presentation quotient-group
edited Dec 14 '18 at 2:12
Shaun
9,083113683
asked Apr 8 '16 at 5:57
math91math91
62
$endgroup$
$begingroup$
There might be a typo in your group presentation of $G$. If not, then $G$ isn't generated by a single element. To see this note that it has torsion (since $(aba)^4 = 1$) and that $a$ is of infinte order in $G$.
$endgroup$
– Stefan Mesken
Apr 8 '16 at 6:46
$begingroup$
Okay. What are your thoughts? Do you see that $aba = bab$ in $H/ < (aba)^4 >$?
$endgroup$
– Stefan Mesken
Apr 8 '16 at 7:09
$begingroup$
I thought of showing that $ker varphi / ker psi cong <(aba)^4>$ (Third isomorphism theorem), so that the quotient of the normal closures is $<(aba)^4>$
$endgroup$
– math91
Apr 8 '16 at 7:15
2
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
1
$begingroup$
Also mid ($mid$) gives you better spacing as a separator than the pipe character | does.
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
add a comment |
1
1
1
$begingroup$
Assume that a group $G$ has a presentation $langle X mid R rangle$ and a group $H$ has a presentation $langle X mid S rangle$. If $R subseteq S$, the $H$ is isomorphic to a quotient of $G$.
In fact, if $F_X$ is the free group and $varphi: F_X rightarrow G$ and $psi:F_X rightarrow H$ are the homomorphism, then $ker varphi = R^{F_X} leq S^{F_X} = ker psi$. So, $H cong F_X/kerpsi$ is isomorphic to a quotient of $F_X/kervarphi cong G$.
group-presentation quotient-group
edited Dec 14 '18 at 2:12
Shaun
9,083113683
asked Apr 8 '16 at 5:57
math91math91
62
$endgroup$
Assume that a group $G$ has a presentation $langle X mid R rangle$ and a group $H$ has a presentation $langle X mid S rangle$. If $R subseteq S$, the $H$ is isomorphic to a quotient of $G$.
In fact, if $F_X$ is the free group and $varphi: F_X rightarrow G$ and $psi:F_X rightarrow H$ are the homomorphism, then $ker varphi = R^{F_X} leq S^{F_X} = ker psi$. So, $H cong F_X/kerpsi$ is isomorphic to a quotient of $F_X/kervarphi cong G$.
group-presentation quotient-group
group-presentation quotient-group
edited Dec 14 '18 at 2:12
Shaun
9,083113683
asked Apr 8 '16 at 5:57
math91math91
62
edited Dec 14 '18 at 2:12
Shaun
9,083113683
asked Apr 8 '16 at 5:57
math91math91
62
edited Dec 14 '18 at 2:12
Shaun
9,083113683
edited Dec 14 '18 at 2:12
Shaun
9,083113683
edited Dec 14 '18 at 2:12
Shaun
9,083113683
9,083113683
asked Apr 8 '16 at 5:57
math91math91
62
asked Apr 8 '16 at 5:57
math91math91
62
asked Apr 8 '16 at 5:57
math91math91
62
62
$begingroup$
There might be a typo in your group presentation of $G$. If not, then $G$ isn't generated by a single element. To see this note that it has torsion (since $(aba)^4 = 1$) and that $a$ is of infinte order in $G$.
$endgroup$
– Stefan Mesken
Apr 8 '16 at 6:46
$begingroup$
Okay. What are your thoughts? Do you see that $aba = bab$ in $H/ < (aba)^4 >$?
$endgroup$
– Stefan Mesken
Apr 8 '16 at 7:09
$begingroup$
I thought of showing that $ker varphi / ker psi cong <(aba)^4>$ (Third isomorphism theorem), so that the quotient of the normal closures is $<(aba)^4>$
$endgroup$
– math91
Apr 8 '16 at 7:15
2
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
1
$begingroup$
Also mid ($mid$) gives you better spacing as a separator than the pipe character | does.
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
add a comment |
$begingroup$
There might be a typo in your group presentation of $G$. If not, then $G$ isn't generated by a single element. To see this note that it has torsion (since $(aba)^4 = 1$) and that $a$ is of infinte order in $G$.
$endgroup$
– Stefan Mesken
Apr 8 '16 at 6:46
$begingroup$
Okay. What are your thoughts? Do you see that $aba = bab$ in $H/ < (aba)^4 >$?
$endgroup$
– Stefan Mesken
Apr 8 '16 at 7:09
$begingroup$
I thought of showing that $ker varphi / ker psi cong <(aba)^4>$ (Third isomorphism theorem), so that the quotient of the normal closures is $<(aba)^4>$
$endgroup$
– math91
Apr 8 '16 at 7:15
2
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
1
$begingroup$
Also mid ($mid$) gives you better spacing as a separator than the pipe character | does.
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
$begingroup$
There might be a typo in your group presentation of $G$. If not, then $G$ isn't generated by a single element. To see this note that it has torsion (since $(aba)^4 = 1$) and that $a$ is of infinte order in $G$.
$endgroup$
– Stefan Mesken
Apr 8 '16 at 6:46
$begingroup$
There might be a typo in your group presentation of $G$. If not, then $G$ isn't generated by a single element. To see this note that it has torsion (since $(aba)^4 = 1$) and that $a$ is of infinte order in $G$.
$endgroup$
– Stefan Mesken
Apr 8 '16 at 6:46
$begingroup$
Okay. What are your thoughts? Do you see that $aba = bab$ in $H/ < (aba)^4 >$?
$endgroup$
– Stefan Mesken
Apr 8 '16 at 7:09
$begingroup$
Okay. What are your thoughts? Do you see that $aba = bab$ in $H/ < (aba)^4 >$?
$endgroup$
– Stefan Mesken
Apr 8 '16 at 7:09
$begingroup$
I thought of showing that $ker varphi / ker psi cong <(aba)^4>$ (Third isomorphism theorem), so that the quotient of the normal closures is $<(aba)^4>$
$endgroup$
– math91
Apr 8 '16 at 7:15
$begingroup$
I thought of showing that $ker varphi / ker psi cong <(aba)^4>$ (Third isomorphism theorem), so that the quotient of the normal closures is $<(aba)^4>$
$endgroup$
– math91
Apr 8 '16 at 7:15
2
2
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
1
1
$begingroup$
Also mid ($mid$) gives you better spacing as a separator than the pipe character | does.
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
$begingroup$
Also mid ($mid$) gives you better spacing as a separator than the pipe character | does.
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
add a comment |
0
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$begingroup$
There might be a typo in your group presentation of $G$. If not, then $G$ isn't generated by a single element. To see this note that it has torsion (since $(aba)^4 = 1$) and that $a$ is of infinte order in $G$.
$endgroup$
– Stefan Mesken
Apr 8 '16 at 6:46
$begingroup$
Okay. What are your thoughts? Do you see that $aba = bab$ in $H/ < (aba)^4 >$?
$endgroup$
– Stefan Mesken
Apr 8 '16 at 7:09
$begingroup$
I thought of showing that $ker varphi / ker psi cong <(aba)^4>$ (Third isomorphism theorem), so that the quotient of the normal closures is $<(aba)^4>$
$endgroup$
– math91
Apr 8 '16 at 7:15
2
$begingroup$
I'm the langle rangle fairy, here to let you know that $langle, rangle$ plays nicer with TeX than <, > does :)
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17
1
$begingroup$
Also mid ($mid$) gives you better spacing as a separator than the pipe character | does.
$endgroup$
– Patrick Stevens
Apr 8 '16 at 8:17