Prove that $D_{2n} = langle s, rs rangle$
Show that the subgroup of $D_{2n}$ generated by the set ${s,rs}$ is $D_{2n}$ itself.
Here is my attempt:
$langle s, rs rangle = D_{2n}$ if and only if $langle r, s rangle = langle s, rs rangle$, so it is sufficient to show that the latter equality holds. The containment $langle r, s rangle subseteq langle s, rs rangle$ holds: since $r, s in langle s, rs rangle$ (cleary this is true for $s$, and also $r=(rs)s$), every element in $langle r, s rangle$ is also an element of $langle s, rs rangle$. The containment $langle s, rs rangle subseteq langle r, s rangle$ is clear, because since $s,rs in langle r, s rangle$ every element in $langle s, rs rangle$ is also an element in $langle r, s rangle$. Therefore $langle r, s rangle = langle s, rs rangle = D_{2n}$.
group-theory proof-verification dihedral-groups
edited Dec 1 '18 at 4:23
Shaun
8,820113681
asked Sep 6 '18 at 19:48
WesleyWesley
518313
add a comment |
Show that the subgroup of $D_{2n}$ generated by the set ${s,rs}$ is $D_{2n}$ itself.
Here is my attempt:
$langle s, rs rangle = D_{2n}$ if and only if $langle r, s rangle = langle s, rs rangle$, so it is sufficient to show that the latter equality holds. The containment $langle r, s rangle subseteq langle s, rs rangle$ holds: since $r, s in langle s, rs rangle$ (cleary this is true for $s$, and also $r=(rs)s$), every element in $langle r, s rangle$ is also an element of $langle s, rs rangle$. The containment $langle s, rs rangle subseteq langle r, s rangle$ is clear, because since $s,rs in langle r, s rangle$ every element in $langle s, rs rangle$ is also an element in $langle r, s rangle$. Therefore $langle r, s rangle = langle s, rs rangle = D_{2n}$.
group-theory proof-verification dihedral-groups
edited Dec 1 '18 at 4:23
Shaun
8,820113681
asked Sep 6 '18 at 19:48
WesleyWesley
518313
Your solution is perfect!
– LucaMac
Sep 6 '18 at 19:50
There's nothing special about $D_{2n}$ here besides that it has two generators. We can thus generalise it . . .
– Shaun
Dec 1 '18 at 17:45
add a comment |
Show that the subgroup of $D_{2n}$ generated by the set ${s,rs}$ is $D_{2n}$ itself.
Here is my attempt:
$langle s, rs rangle = D_{2n}$ if and only if $langle r, s rangle = langle s, rs rangle$, so it is sufficient to show that the latter equality holds. The containment $langle r, s rangle subseteq langle s, rs rangle$ holds: since $r, s in langle s, rs rangle$ (cleary this is true for $s$, and also $r=(rs)s$), every element in $langle r, s rangle$ is also an element of $langle s, rs rangle$. The containment $langle s, rs rangle subseteq langle r, s rangle$ is clear, because since $s,rs in langle r, s rangle$ every element in $langle s, rs rangle$ is also an element in $langle r, s rangle$. Therefore $langle r, s rangle = langle s, rs rangle = D_{2n}$.
group-theory proof-verification dihedral-groups
edited Dec 1 '18 at 4:23
Shaun
8,820113681
asked Sep 6 '18 at 19:48
WesleyWesley
518313
Show that the subgroup of $D_{2n}$ generated by the set ${s,rs}$ is $D_{2n}$ itself.
Here is my attempt:
$langle s, rs rangle = D_{2n}$ if and only if $langle r, s rangle = langle s, rs rangle$, so it is sufficient to show that the latter equality holds. The containment $langle r, s rangle subseteq langle s, rs rangle$ holds: since $r, s in langle s, rs rangle$ (cleary this is true for $s$, and also $r=(rs)s$), every element in $langle r, s rangle$ is also an element of $langle s, rs rangle$. The containment $langle s, rs rangle subseteq langle r, s rangle$ is clear, because since $s,rs in langle r, s rangle$ every element in $langle s, rs rangle$ is also an element in $langle r, s rangle$. Therefore $langle r, s rangle = langle s, rs rangle = D_{2n}$.
group-theory proof-verification dihedral-groups
group-theory proof-verification dihedral-groups
edited Dec 1 '18 at 4:23
Shaun
8,820113681
asked Sep 6 '18 at 19:48
WesleyWesley
518313
edited Dec 1 '18 at 4:23
Shaun
8,820113681
asked Sep 6 '18 at 19:48
WesleyWesley
518313
edited Dec 1 '18 at 4:23
Shaun
8,820113681
edited Dec 1 '18 at 4:23
Shaun
8,820113681
edited Dec 1 '18 at 4:23
Shaun
8,820113681
8,820113681
asked Sep 6 '18 at 19:48
WesleyWesley
518313
asked Sep 6 '18 at 19:48
WesleyWesley
518313
asked Sep 6 '18 at 19:48
WesleyWesley
518313
518313
Your solution is perfect!
– LucaMac
Sep 6 '18 at 19:50
There's nothing special about $D_{2n}$ here besides that it has two generators. We can thus generalise it . . .
– Shaun
Dec 1 '18 at 17:45
add a comment |
Your solution is perfect!
– LucaMac
Sep 6 '18 at 19:50
There's nothing special about $D_{2n}$ here besides that it has two generators. We can thus generalise it . . .
– Shaun
Dec 1 '18 at 17:45
Your solution is perfect!
– LucaMac
Sep 6 '18 at 19:50
Your solution is perfect!
– LucaMac
Sep 6 '18 at 19:50
There's nothing special about $D_{2n}$ here besides that it has two generators. We can thus generalise it . . .
– Shaun
Dec 1 '18 at 17:45
There's nothing special about $D_{2n}$ here besides that it has two generators. We can thus generalise it . . .
– Shaun
Dec 1 '18 at 17:45
add a comment |
1 Answer
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Your proof is sound, except that there is (what I suspect is) a typo. It should be $$r=(rs)s^{color{red}{-1}}.$$
answered Nov 30 '18 at 19:17
ShaunShaun
8,820113681
add a comment |
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Your proof is sound, except that there is (what I suspect is) a typo. It should be $$r=(rs)s^{color{red}{-1}}.$$
answered Nov 30 '18 at 19:17
ShaunShaun
8,820113681
add a comment |
Your proof is sound, except that there is (what I suspect is) a typo. It should be $$r=(rs)s^{color{red}{-1}}.$$
answered Nov 30 '18 at 19:17
ShaunShaun
8,820113681
add a comment |
Your proof is sound, except that there is (what I suspect is) a typo. It should be $$r=(rs)s^{color{red}{-1}}.$$
answered Nov 30 '18 at 19:17
ShaunShaun
8,820113681
Your proof is sound, except that there is (what I suspect is) a typo. It should be $$r=(rs)s^{color{red}{-1}}.$$
answered Nov 30 '18 at 19:17
ShaunShaun
8,820113681
answered Nov 30 '18 at 19:17
ShaunShaun
8,820113681
answered Nov 30 '18 at 19:17
ShaunShaun
8,820113681
answered Nov 30 '18 at 19:17
ShaunShaun
8,820113681
8,820113681
add a comment |
add a comment |
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Your solution is perfect!
– LucaMac
Sep 6 '18 at 19:50
There's nothing special about $D_{2n}$ here besides that it has two generators. We can thus generalise it . . .
– Shaun
Dec 1 '18 at 17:45