Find a subgroup of $D_{42}$ that is isomorphic to $S_3$.
$begingroup$
Note that $D_{42}$ is generated by $rho, r$ where $ord(rho) = 21$, $ord(r)= 2$. To locate a copy of $S_3$, there must be a subgroup of order $3$, which can be $e, rho^{7}, rho^{14}$. What about order $2$ elements? Inevitably, I have to introduce some reflections which will enlarge the group.
abstract-algebra
asked Dec 14 '18 at 17:47
kochkoch
18418
$endgroup$
add a comment |
$begingroup$
Note that $D_{42}$ is generated by $rho, r$ where $ord(rho) = 21$, $ord(r)= 2$. To locate a copy of $S_3$, there must be a subgroup of order $3$, which can be $e, rho^{7}, rho^{14}$. What about order $2$ elements? Inevitably, I have to introduce some reflections which will enlarge the group.
abstract-algebra
asked Dec 14 '18 at 17:47
kochkoch
18418
$endgroup$
$begingroup$
Doesn't $S_3$ have 6 elements ?
$endgroup$
– Digitalis
Dec 14 '18 at 17:49
$begingroup$
Yes, and?@Digitalis
$endgroup$
– koch
Dec 14 '18 at 17:55
add a comment |
$begingroup$
Note that $D_{42}$ is generated by $rho, r$ where $ord(rho) = 21$, $ord(r)= 2$. To locate a copy of $S_3$, there must be a subgroup of order $3$, which can be $e, rho^{7}, rho^{14}$. What about order $2$ elements? Inevitably, I have to introduce some reflections which will enlarge the group.
abstract-algebra
asked Dec 14 '18 at 17:47
kochkoch
18418
$endgroup$
Note that $D_{42}$ is generated by $rho, r$ where $ord(rho) = 21$, $ord(r)= 2$. To locate a copy of $S_3$, there must be a subgroup of order $3$, which can be $e, rho^{7}, rho^{14}$. What about order $2$ elements? Inevitably, I have to introduce some reflections which will enlarge the group.
abstract-algebra
abstract-algebra
asked Dec 14 '18 at 17:47
kochkoch
18418
asked Dec 14 '18 at 17:47
kochkoch
18418
asked Dec 14 '18 at 17:47
kochkoch
18418
asked Dec 14 '18 at 17:47
kochkoch
18418
asked Dec 14 '18 at 17:47
kochkoch
18418
18418
$begingroup$
Doesn't $S_3$ have 6 elements ?
$endgroup$
– Digitalis
Dec 14 '18 at 17:49
$begingroup$
Yes, and?@Digitalis
$endgroup$
– koch
Dec 14 '18 at 17:55
add a comment |
$begingroup$
Doesn't $S_3$ have 6 elements ?
$endgroup$
– Digitalis
Dec 14 '18 at 17:49
$begingroup$
Yes, and?@Digitalis
$endgroup$
– koch
Dec 14 '18 at 17:55
$begingroup$
Doesn't $S_3$ have 6 elements ?
$endgroup$
– Digitalis
Dec 14 '18 at 17:49
$begingroup$
Doesn't $S_3$ have 6 elements ?
$endgroup$
– Digitalis
Dec 14 '18 at 17:49
$begingroup$
Yes, and?@Digitalis
$endgroup$
– koch
Dec 14 '18 at 17:55
$begingroup$
Yes, and?@Digitalis
$endgroup$
– koch
Dec 14 '18 at 17:55
add a comment |
2 Answers
2
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votes
$begingroup$
$S_3$ has $3!=6$ elements and you need three more.
$1,rho^7,rho^{14},r,rrho^7,rrho^{14}$ does the trick nicely.
answered Dec 14 '18 at 17:55
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
We have a complete classification of all possible subgroups of $D_n$:
Theorem: Every subgroup of $D_n=langle r,s rangle$ is is either cyclic or dihedral, and a complete listing of the subgroups is as follows:
(1) $langle r^drangle$, where $dmid n$, with index $2d$,
(2) $langle r^d, r^is rangle$, where $dmid n$, $0le ile d-1$, with index $d$.
Every subgroup of $D_n$ occurs exactly once in this listing.
For a proof see Theorem 3.1 of Keith Conrad's notes.
For the appropriate $d$ we obtain the subgroup $D_3=S_3$ with $6$ elements in $(2)$.
answered Dec 14 '18 at 17:56
Dietrich BurdeDietrich Burde
79.1k647103
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$S_3$ has $3!=6$ elements and you need three more.
$1,rho^7,rho^{14},r,rrho^7,rrho^{14}$ does the trick nicely.
answered Dec 14 '18 at 17:55
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
$S_3$ has $3!=6$ elements and you need three more.
$1,rho^7,rho^{14},r,rrho^7,rrho^{14}$ does the trick nicely.
answered Dec 14 '18 at 17:55
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
$S_3$ has $3!=6$ elements and you need three more.
$1,rho^7,rho^{14},r,rrho^7,rrho^{14}$ does the trick nicely.
answered Dec 14 '18 at 17:55
Ross MillikanRoss Millikan
295k23198371
$endgroup$
$S_3$ has $3!=6$ elements and you need three more.
$1,rho^7,rho^{14},r,rrho^7,rrho^{14}$ does the trick nicely.
answered Dec 14 '18 at 17:55
Ross MillikanRoss Millikan
295k23198371
answered Dec 14 '18 at 17:55
Ross MillikanRoss Millikan
295k23198371
answered Dec 14 '18 at 17:55
Ross MillikanRoss Millikan
295k23198371
answered Dec 14 '18 at 17:55
Ross MillikanRoss Millikan
295k23198371
295k23198371
add a comment |
add a comment |
$begingroup$
We have a complete classification of all possible subgroups of $D_n$:
Theorem: Every subgroup of $D_n=langle r,s rangle$ is is either cyclic or dihedral, and a complete listing of the subgroups is as follows:
(1) $langle r^drangle$, where $dmid n$, with index $2d$,
(2) $langle r^d, r^is rangle$, where $dmid n$, $0le ile d-1$, with index $d$.
Every subgroup of $D_n$ occurs exactly once in this listing.
For a proof see Theorem 3.1 of Keith Conrad's notes.
For the appropriate $d$ we obtain the subgroup $D_3=S_3$ with $6$ elements in $(2)$.
answered Dec 14 '18 at 17:56
Dietrich BurdeDietrich Burde
79.1k647103
$endgroup$
add a comment |
$begingroup$
We have a complete classification of all possible subgroups of $D_n$:
Theorem: Every subgroup of $D_n=langle r,s rangle$ is is either cyclic or dihedral, and a complete listing of the subgroups is as follows:
(1) $langle r^drangle$, where $dmid n$, with index $2d$,
(2) $langle r^d, r^is rangle$, where $dmid n$, $0le ile d-1$, with index $d$.
Every subgroup of $D_n$ occurs exactly once in this listing.
For a proof see Theorem 3.1 of Keith Conrad's notes.
For the appropriate $d$ we obtain the subgroup $D_3=S_3$ with $6$ elements in $(2)$.
answered Dec 14 '18 at 17:56
Dietrich BurdeDietrich Burde
79.1k647103
$endgroup$
add a comment |
$begingroup$
We have a complete classification of all possible subgroups of $D_n$:
Theorem: Every subgroup of $D_n=langle r,s rangle$ is is either cyclic or dihedral, and a complete listing of the subgroups is as follows:
(1) $langle r^drangle$, where $dmid n$, with index $2d$,
(2) $langle r^d, r^is rangle$, where $dmid n$, $0le ile d-1$, with index $d$.
Every subgroup of $D_n$ occurs exactly once in this listing.
For a proof see Theorem 3.1 of Keith Conrad's notes.
For the appropriate $d$ we obtain the subgroup $D_3=S_3$ with $6$ elements in $(2)$.
answered Dec 14 '18 at 17:56
Dietrich BurdeDietrich Burde
79.1k647103
$endgroup$
We have a complete classification of all possible subgroups of $D_n$:
Theorem: Every subgroup of $D_n=langle r,s rangle$ is is either cyclic or dihedral, and a complete listing of the subgroups is as follows:
(1) $langle r^drangle$, where $dmid n$, with index $2d$,
(2) $langle r^d, r^is rangle$, where $dmid n$, $0le ile d-1$, with index $d$.
Every subgroup of $D_n$ occurs exactly once in this listing.
For a proof see Theorem 3.1 of Keith Conrad's notes.
For the appropriate $d$ we obtain the subgroup $D_3=S_3$ with $6$ elements in $(2)$.
answered Dec 14 '18 at 17:56
Dietrich BurdeDietrich Burde
79.1k647103
answered Dec 14 '18 at 17:56
Dietrich BurdeDietrich Burde
79.1k647103
answered Dec 14 '18 at 17:56
Dietrich BurdeDietrich Burde
79.1k647103
answered Dec 14 '18 at 17:56
Dietrich BurdeDietrich Burde
79.1k647103
79.1k647103
add a comment |
add a comment |
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$begingroup$
Doesn't $S_3$ have 6 elements ?
$endgroup$
– Digitalis
Dec 14 '18 at 17:49
$begingroup$
Yes, and?@Digitalis
$endgroup$
– koch
Dec 14 '18 at 17:55