Mixing
Quotient group of the dihedral group by $langle r^2 rangle.$
$begingroup$
Show that $G/H$ is abelian, where $G$ is the dihedral group $$ G={langle r,, f mid r^n=f^2=1,, rf=fr^{-1}rangle}$$
and $H$ is the subgroup $langle r^2 rangle.$
I've tried showing that for $a,b in G$ then $ar^{2i}br^{2j}=br^{2i}ar^{2j}$ but im sure if the steps I used are true.
$$begin{align}
ar^{2i}br^{2j} &=r^{2i}a^{-1}br^{2j}\
&=r^{2i}b^{-1}ar^{2j} \
&=br^{2i}ar^{2j}.
end{align}$$
group-theory finite-groups group-presentation dihedral-groups quotient-group
edited Jan 12 at 12:34
Shaun
9,083113683
$endgroup$
add a comment |
$begingroup$
Show that $G/H$ is abelian, where $G$ is the dihedral group $$ G={langle r,, f mid r^n=f^2=1,, rf=fr^{-1}rangle}$$
and $H$ is the subgroup $langle r^2 rangle.$
I've tried showing that for $a,b in G$ then $ar^{2i}br^{2j}=br^{2i}ar^{2j}$ but im sure if the steps I used are true.
$$begin{align}
ar^{2i}br^{2j} &=r^{2i}a^{-1}br^{2j}\
&=r^{2i}b^{-1}ar^{2j} \
&=br^{2i}ar^{2j}.
end{align}$$
group-theory finite-groups group-presentation dihedral-groups quotient-group
edited Jan 12 at 12:34
Shaun
9,083113683
$endgroup$
add a comment |
$begingroup$
Show that $G/H$ is abelian, where $G$ is the dihedral group $$ G={langle r,, f mid r^n=f^2=1,, rf=fr^{-1}rangle}$$
and $H$ is the subgroup $langle r^2 rangle.$
I've tried showing that for $a,b in G$ then $ar^{2i}br^{2j}=br^{2i}ar^{2j}$ but im sure if the steps I used are true.
$$begin{align}
ar^{2i}br^{2j} &=r^{2i}a^{-1}br^{2j}\
&=r^{2i}b^{-1}ar^{2j} \
&=br^{2i}ar^{2j}.
end{align}$$
group-theory finite-groups group-presentation dihedral-groups quotient-group
edited Jan 12 at 12:34
Shaun
9,083113683
$endgroup$
Show that $G/H$ is abelian, where $G$ is the dihedral group $$ G={langle r,, f mid r^n=f^2=1,, rf=fr^{-1}rangle}$$
and $H$ is the subgroup $langle r^2 rangle.$
I've tried showing that for $a,b in G$ then $ar^{2i}br^{2j}=br^{2i}ar^{2j}$ but im sure if the steps I used are true.
$$begin{align}
ar^{2i}br^{2j} &=r^{2i}a^{-1}br^{2j}\
&=r^{2i}b^{-1}ar^{2j} \
&=br^{2i}ar^{2j}.
end{align}$$
group-theory finite-groups group-presentation dihedral-groups quotient-group
group-theory finite-groups group-presentation dihedral-groups quotient-group
edited Jan 12 at 12:34
Shaun
9,083113683
edited Jan 12 at 12:34
Shaun
9,083113683
edited Jan 12 at 12:34
Shaun
9,083113683
edited Jan 12 at 12:34
Shaun
9,083113683
edited Jan 12 at 12:34
Shaun
9,083113683
9,083113683
asked Jan 11 at 17:51
user633852
add a comment |
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2 Answers
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$begingroup$
Hint If $n$ is odd, $midlangle r^2ranglemid=midlangle rrangle mid=n$. Hence the index is $2$. That is, $D_{2n}/langle r^2rangle=C_2$.
If $n$ is even, $mid langle r^2rangle mid=n/2$. Thus the index is $4$. Thus $D_{2n}/langle r^2rangle $ has order $4$.
answered Jan 11 at 18:33
Chris CusterChris Custer
13.1k3827
$endgroup$
add a comment |
$begingroup$
One way to look at this is to note that
$$begin{align}
G/H &conglangle r, fmid r^2=1, r^n=f^2=1, rf=fr^{-1}rangle \
&conglangle r, fmid r^{gcd(2, n)}=f^2=1, underbrace{rf=fr}_{r^2=1}rangle \
&cong Bbb Z_{gcd(2,n)}timesBbb Z_2,
end{align}$$
where the first isomorphism$^dagger$ holds because $G/H$ is, essentially, just introducing the relation $r^2=1$ to $G$; the second isomorphism is a consequence of manipulating the relations in the first presentation above (and a standard result on the $gcd$ of powers of the same generator of a presentation); and the latter is, again, standard, since one just has to recognise a typical presentation for the direct product cyclic groups.
$dagger$ A presentation is not a group, technically speaking. What I am referring to is the group defined by the presentation.
answered Jan 12 at 12:50
ShaunShaun
9,083113683
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Hint If $n$ is odd, $midlangle r^2ranglemid=midlangle rrangle mid=n$. Hence the index is $2$. That is, $D_{2n}/langle r^2rangle=C_2$.
If $n$ is even, $mid langle r^2rangle mid=n/2$. Thus the index is $4$. Thus $D_{2n}/langle r^2rangle $ has order $4$.
answered Jan 11 at 18:33
Chris CusterChris Custer
13.1k3827
$endgroup$
add a comment |
$begingroup$
Hint If $n$ is odd, $midlangle r^2ranglemid=midlangle rrangle mid=n$. Hence the index is $2$. That is, $D_{2n}/langle r^2rangle=C_2$.
If $n$ is even, $mid langle r^2rangle mid=n/2$. Thus the index is $4$. Thus $D_{2n}/langle r^2rangle $ has order $4$.
answered Jan 11 at 18:33
Chris CusterChris Custer
13.1k3827
$endgroup$
add a comment |
$begingroup$
Hint If $n$ is odd, $midlangle r^2ranglemid=midlangle rrangle mid=n$. Hence the index is $2$. That is, $D_{2n}/langle r^2rangle=C_2$.
If $n$ is even, $mid langle r^2rangle mid=n/2$. Thus the index is $4$. Thus $D_{2n}/langle r^2rangle $ has order $4$.
answered Jan 11 at 18:33
Chris CusterChris Custer
13.1k3827
$endgroup$
Hint If $n$ is odd, $midlangle r^2ranglemid=midlangle rrangle mid=n$. Hence the index is $2$. That is, $D_{2n}/langle r^2rangle=C_2$.
If $n$ is even, $mid langle r^2rangle mid=n/2$. Thus the index is $4$. Thus $D_{2n}/langle r^2rangle $ has order $4$.
answered Jan 11 at 18:33
Chris CusterChris Custer
13.1k3827
edited Jan 11 at 18:52
answered Jan 11 at 18:33
Chris CusterChris Custer
13.1k3827
answered Jan 11 at 18:33
Chris CusterChris Custer
13.1k3827
answered Jan 11 at 18:33
Chris CusterChris Custer
13.1k3827
13.1k3827
add a comment |
add a comment |
$begingroup$
One way to look at this is to note that
$$begin{align}
G/H &conglangle r, fmid r^2=1, r^n=f^2=1, rf=fr^{-1}rangle \
&conglangle r, fmid r^{gcd(2, n)}=f^2=1, underbrace{rf=fr}_{r^2=1}rangle \
&cong Bbb Z_{gcd(2,n)}timesBbb Z_2,
end{align}$$
where the first isomorphism$^dagger$ holds because $G/H$ is, essentially, just introducing the relation $r^2=1$ to $G$; the second isomorphism is a consequence of manipulating the relations in the first presentation above (and a standard result on the $gcd$ of powers of the same generator of a presentation); and the latter is, again, standard, since one just has to recognise a typical presentation for the direct product cyclic groups.
$dagger$ A presentation is not a group, technically speaking. What I am referring to is the group defined by the presentation.
answered Jan 12 at 12:50
ShaunShaun
9,083113683
$endgroup$
add a comment |
$begingroup$
One way to look at this is to note that
$$begin{align}
G/H &conglangle r, fmid r^2=1, r^n=f^2=1, rf=fr^{-1}rangle \
&conglangle r, fmid r^{gcd(2, n)}=f^2=1, underbrace{rf=fr}_{r^2=1}rangle \
&cong Bbb Z_{gcd(2,n)}timesBbb Z_2,
end{align}$$
where the first isomorphism$^dagger$ holds because $G/H$ is, essentially, just introducing the relation $r^2=1$ to $G$; the second isomorphism is a consequence of manipulating the relations in the first presentation above (and a standard result on the $gcd$ of powers of the same generator of a presentation); and the latter is, again, standard, since one just has to recognise a typical presentation for the direct product cyclic groups.
$dagger$ A presentation is not a group, technically speaking. What I am referring to is the group defined by the presentation.
answered Jan 12 at 12:50
ShaunShaun
9,083113683
$endgroup$
add a comment |
$begingroup$
One way to look at this is to note that
$$begin{align}
G/H &conglangle r, fmid r^2=1, r^n=f^2=1, rf=fr^{-1}rangle \
&conglangle r, fmid r^{gcd(2, n)}=f^2=1, underbrace{rf=fr}_{r^2=1}rangle \
&cong Bbb Z_{gcd(2,n)}timesBbb Z_2,
end{align}$$
where the first isomorphism$^dagger$ holds because $G/H$ is, essentially, just introducing the relation $r^2=1$ to $G$; the second isomorphism is a consequence of manipulating the relations in the first presentation above (and a standard result on the $gcd$ of powers of the same generator of a presentation); and the latter is, again, standard, since one just has to recognise a typical presentation for the direct product cyclic groups.
$dagger$ A presentation is not a group, technically speaking. What I am referring to is the group defined by the presentation.
answered Jan 12 at 12:50
ShaunShaun
9,083113683
$endgroup$
One way to look at this is to note that
$$begin{align}
G/H &conglangle r, fmid r^2=1, r^n=f^2=1, rf=fr^{-1}rangle \
&conglangle r, fmid r^{gcd(2, n)}=f^2=1, underbrace{rf=fr}_{r^2=1}rangle \
&cong Bbb Z_{gcd(2,n)}timesBbb Z_2,
end{align}$$
where the first isomorphism$^dagger$ holds because $G/H$ is, essentially, just introducing the relation $r^2=1$ to $G$; the second isomorphism is a consequence of manipulating the relations in the first presentation above (and a standard result on the $gcd$ of powers of the same generator of a presentation); and the latter is, again, standard, since one just has to recognise a typical presentation for the direct product cyclic groups.
$dagger$ A presentation is not a group, technically speaking. What I am referring to is the group defined by the presentation.
answered Jan 12 at 12:50
ShaunShaun
9,083113683
answered Jan 12 at 12:50
ShaunShaun
9,083113683
answered Jan 12 at 12:50
ShaunShaun
9,083113683
answered Jan 12 at 12:50
ShaunShaun
9,083113683
9,083113683
add a comment |
add a comment |
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