Show that (12345) and (12) create $S_5$ [duplicate]
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This question already has an answer here:
Generators of $S_n$
1 answer
Can someone tell me how I can show the following:
Consider the symmetric group $S_5$ and show that $(12345)$ and $(12)$ generate the group.
symmetric-groups
edited Dec 13 '18 at 10:43
Glorfindel
3,41381930
asked Dec 13 '18 at 9:09
lerxlerx
112
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marked as duplicate by Chinnapparaj R, Arthur, cansomeonehelpmeout, user10354138, user593746 Dec 13 '18 at 13:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Generators of $S_n$
1 answer
Can someone tell me how I can show the following:
Consider the symmetric group $S_5$ and show that $(12345)$ and $(12)$ generate the group.
symmetric-groups
edited Dec 13 '18 at 10:43
Glorfindel
3,41381930
asked Dec 13 '18 at 9:09
lerxlerx
112
$endgroup$
marked as duplicate by Chinnapparaj R, Arthur, cansomeonehelpmeout, user10354138, user593746 Dec 13 '18 at 13:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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What does "creates" mean here? If it means that those two elements together form a group, then it is false. If it means "generates" then this is trivial by the definition of the word.
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– Tobias Kildetoft
Dec 13 '18 at 9:12
1
$begingroup$
@TobiasKildetoft I think the only interesting variation on this is "show that the two elements generate all of $S_5$". So a mix-up between "a group" and "the group".
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– Arthur
Dec 13 '18 at 9:13
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@Arthur Ahh, of course, that makes much more sense.
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– Tobias Kildetoft
Dec 13 '18 at 9:14
$begingroup$
Thanks for your replies. Sorry @all, I meant "the" group, not a group.
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– lerx
Dec 13 '18 at 9:24
add a comment |
$begingroup$
This question already has an answer here:
Generators of $S_n$
1 answer
Can someone tell me how I can show the following:
Consider the symmetric group $S_5$ and show that $(12345)$ and $(12)$ generate the group.
symmetric-groups
edited Dec 13 '18 at 10:43
Glorfindel
3,41381930
asked Dec 13 '18 at 9:09
lerxlerx
112
$endgroup$
This question already has an answer here:
Generators of $S_n$
1 answer
Can someone tell me how I can show the following:
Consider the symmetric group $S_5$ and show that $(12345)$ and $(12)$ generate the group.
This question already has an answer here:
Generators of $S_n$
1 answer
symmetric-groups
symmetric-groups
edited Dec 13 '18 at 10:43
Glorfindel
3,41381930
asked Dec 13 '18 at 9:09
lerxlerx
112
edited Dec 13 '18 at 10:43
Glorfindel
3,41381930
asked Dec 13 '18 at 9:09
lerxlerx
112
edited Dec 13 '18 at 10:43
Glorfindel
3,41381930
edited Dec 13 '18 at 10:43
Glorfindel
3,41381930
edited Dec 13 '18 at 10:43
Glorfindel
3,41381930
3,41381930
asked Dec 13 '18 at 9:09
lerxlerx
112
asked Dec 13 '18 at 9:09
lerxlerx
112
asked Dec 13 '18 at 9:09
lerxlerx
112
112
marked as duplicate by Chinnapparaj R, Arthur, cansomeonehelpmeout, user10354138, user593746 Dec 13 '18 at 13:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Chinnapparaj R, Arthur, cansomeonehelpmeout, user10354138, user593746 Dec 13 '18 at 13:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
What does "creates" mean here? If it means that those two elements together form a group, then it is false. If it means "generates" then this is trivial by the definition of the word.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 9:12
1
$begingroup$
@TobiasKildetoft I think the only interesting variation on this is "show that the two elements generate all of $S_5$". So a mix-up between "a group" and "the group".
$endgroup$
– Arthur
Dec 13 '18 at 9:13
$begingroup$
@Arthur Ahh, of course, that makes much more sense.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 9:14
$begingroup$
Thanks for your replies. Sorry @all, I meant "the" group, not a group.
$endgroup$
– lerx
Dec 13 '18 at 9:24
add a comment |
$begingroup$
What does "creates" mean here? If it means that those two elements together form a group, then it is false. If it means "generates" then this is trivial by the definition of the word.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 9:12
1
$begingroup$
@TobiasKildetoft I think the only interesting variation on this is "show that the two elements generate all of $S_5$". So a mix-up between "a group" and "the group".
$endgroup$
– Arthur
Dec 13 '18 at 9:13
$begingroup$
@Arthur Ahh, of course, that makes much more sense.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 9:14
$begingroup$
Thanks for your replies. Sorry @all, I meant "the" group, not a group.
$endgroup$
– lerx
Dec 13 '18 at 9:24
$begingroup$
What does "creates" mean here? If it means that those two elements together form a group, then it is false. If it means "generates" then this is trivial by the definition of the word.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 9:12
$begingroup$
What does "creates" mean here? If it means that those two elements together form a group, then it is false. If it means "generates" then this is trivial by the definition of the word.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 9:12
1
1
$begingroup$
@TobiasKildetoft I think the only interesting variation on this is "show that the two elements generate all of $S_5$". So a mix-up between "a group" and "the group".
$endgroup$
– Arthur
Dec 13 '18 at 9:13
$begingroup$
@TobiasKildetoft I think the only interesting variation on this is "show that the two elements generate all of $S_5$". So a mix-up between "a group" and "the group".
$endgroup$
– Arthur
Dec 13 '18 at 9:13
$begingroup$
@Arthur Ahh, of course, that makes much more sense.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 9:14
$begingroup$
@Arthur Ahh, of course, that makes much more sense.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 9:14
$begingroup$
Thanks for your replies. Sorry @all, I meant "the" group, not a group.
$endgroup$
– lerx
Dec 13 '18 at 9:24
$begingroup$
Thanks for your replies. Sorry @all, I meant "the" group, not a group.
$endgroup$
– lerx
Dec 13 '18 at 9:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
So, thats everything I have to show? Or is there something more to do?
Let $c = (1, 2, dotsc, 5)$. We see that
begin{align*}
c (1, 2) c^{-1} &= (2, 3) \
c (2, 3) c^{-1} &= (3, 4) \
&vdots \
c (5-2, 5-1) c^{-1} &= (5-1, 5),
end{align*}
so that $(i, i+1) in langle (1, 2), c rangle$ for all $1 leq i leq 5-1$. Next, we have
begin{align*}
(2, 3) (1, 2) (2, 3)^{-1} &= (1, 3) \
(3, 4) (1, 3) (3, 4)^{-1} &= (1, 4) \
&vdots \
(5-1, 5) (1, 5-1) (5-1, 5)^{-1} &= (1, 5),
end{align*}
so that $(1, i) in langle (1, 2), c rangle$ for all $1 leq i leq 5$. Choose any $1 leq i < j leq 5$, then
$$ (i, j) = (1, i) (1, j) (1, i)^{-1} in langle (1, 2), c rangle. $$
Therefore, $langle (1, 2), c rangle$ contains all transpositions. Hence, $langle (1, 2), c rangle = S_5$.
answered Dec 13 '18 at 9:29
lerxlerx
112
$endgroup$
$begingroup$
This is a funny (in a good way) use of $vdots$ :-)
$endgroup$
– Mees de Vries
Dec 13 '18 at 10:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So, thats everything I have to show? Or is there something more to do?
Let $c = (1, 2, dotsc, 5)$. We see that
begin{align*}
c (1, 2) c^{-1} &= (2, 3) \
c (2, 3) c^{-1} &= (3, 4) \
&vdots \
c (5-2, 5-1) c^{-1} &= (5-1, 5),
end{align*}
so that $(i, i+1) in langle (1, 2), c rangle$ for all $1 leq i leq 5-1$. Next, we have
begin{align*}
(2, 3) (1, 2) (2, 3)^{-1} &= (1, 3) \
(3, 4) (1, 3) (3, 4)^{-1} &= (1, 4) \
&vdots \
(5-1, 5) (1, 5-1) (5-1, 5)^{-1} &= (1, 5),
end{align*}
so that $(1, i) in langle (1, 2), c rangle$ for all $1 leq i leq 5$. Choose any $1 leq i < j leq 5$, then
$$ (i, j) = (1, i) (1, j) (1, i)^{-1} in langle (1, 2), c rangle. $$
Therefore, $langle (1, 2), c rangle$ contains all transpositions. Hence, $langle (1, 2), c rangle = S_5$.
answered Dec 13 '18 at 9:29
lerxlerx
112
$endgroup$
$begingroup$
This is a funny (in a good way) use of $vdots$ :-)
$endgroup$
– Mees de Vries
Dec 13 '18 at 10:43
add a comment |
$begingroup$
So, thats everything I have to show? Or is there something more to do?
Let $c = (1, 2, dotsc, 5)$. We see that
begin{align*}
c (1, 2) c^{-1} &= (2, 3) \
c (2, 3) c^{-1} &= (3, 4) \
&vdots \
c (5-2, 5-1) c^{-1} &= (5-1, 5),
end{align*}
so that $(i, i+1) in langle (1, 2), c rangle$ for all $1 leq i leq 5-1$. Next, we have
begin{align*}
(2, 3) (1, 2) (2, 3)^{-1} &= (1, 3) \
(3, 4) (1, 3) (3, 4)^{-1} &= (1, 4) \
&vdots \
(5-1, 5) (1, 5-1) (5-1, 5)^{-1} &= (1, 5),
end{align*}
so that $(1, i) in langle (1, 2), c rangle$ for all $1 leq i leq 5$. Choose any $1 leq i < j leq 5$, then
$$ (i, j) = (1, i) (1, j) (1, i)^{-1} in langle (1, 2), c rangle. $$
Therefore, $langle (1, 2), c rangle$ contains all transpositions. Hence, $langle (1, 2), c rangle = S_5$.
answered Dec 13 '18 at 9:29
lerxlerx
112
$endgroup$
$begingroup$
This is a funny (in a good way) use of $vdots$ :-)
$endgroup$
– Mees de Vries
Dec 13 '18 at 10:43
add a comment |
$begingroup$
So, thats everything I have to show? Or is there something more to do?
Let $c = (1, 2, dotsc, 5)$. We see that
begin{align*}
c (1, 2) c^{-1} &= (2, 3) \
c (2, 3) c^{-1} &= (3, 4) \
&vdots \
c (5-2, 5-1) c^{-1} &= (5-1, 5),
end{align*}
so that $(i, i+1) in langle (1, 2), c rangle$ for all $1 leq i leq 5-1$. Next, we have
begin{align*}
(2, 3) (1, 2) (2, 3)^{-1} &= (1, 3) \
(3, 4) (1, 3) (3, 4)^{-1} &= (1, 4) \
&vdots \
(5-1, 5) (1, 5-1) (5-1, 5)^{-1} &= (1, 5),
end{align*}
so that $(1, i) in langle (1, 2), c rangle$ for all $1 leq i leq 5$. Choose any $1 leq i < j leq 5$, then
$$ (i, j) = (1, i) (1, j) (1, i)^{-1} in langle (1, 2), c rangle. $$
Therefore, $langle (1, 2), c rangle$ contains all transpositions. Hence, $langle (1, 2), c rangle = S_5$.
answered Dec 13 '18 at 9:29
lerxlerx
112
$endgroup$
So, thats everything I have to show? Or is there something more to do?
Let $c = (1, 2, dotsc, 5)$. We see that
begin{align*}
c (1, 2) c^{-1} &= (2, 3) \
c (2, 3) c^{-1} &= (3, 4) \
&vdots \
c (5-2, 5-1) c^{-1} &= (5-1, 5),
end{align*}
so that $(i, i+1) in langle (1, 2), c rangle$ for all $1 leq i leq 5-1$. Next, we have
begin{align*}
(2, 3) (1, 2) (2, 3)^{-1} &= (1, 3) \
(3, 4) (1, 3) (3, 4)^{-1} &= (1, 4) \
&vdots \
(5-1, 5) (1, 5-1) (5-1, 5)^{-1} &= (1, 5),
end{align*}
so that $(1, i) in langle (1, 2), c rangle$ for all $1 leq i leq 5$. Choose any $1 leq i < j leq 5$, then
$$ (i, j) = (1, i) (1, j) (1, i)^{-1} in langle (1, 2), c rangle. $$
Therefore, $langle (1, 2), c rangle$ contains all transpositions. Hence, $langle (1, 2), c rangle = S_5$.
answered Dec 13 '18 at 9:29
lerxlerx
112
answered Dec 13 '18 at 9:29
lerxlerx
112
answered Dec 13 '18 at 9:29
lerxlerx
112
answered Dec 13 '18 at 9:29
lerxlerx
112
112
$begingroup$
This is a funny (in a good way) use of $vdots$ :-)
$endgroup$
– Mees de Vries
Dec 13 '18 at 10:43
add a comment |
$begingroup$
This is a funny (in a good way) use of $vdots$ :-)
$endgroup$
– Mees de Vries
Dec 13 '18 at 10:43
$begingroup$
This is a funny (in a good way) use of $vdots$ :-)
$endgroup$
– Mees de Vries
Dec 13 '18 at 10:43
$begingroup$
This is a funny (in a good way) use of $vdots$ :-)
$endgroup$
– Mees de Vries
Dec 13 '18 at 10:43
add a comment |
$begingroup$
What does "creates" mean here? If it means that those two elements together form a group, then it is false. If it means "generates" then this is trivial by the definition of the word.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 9:12
1
$begingroup$
@TobiasKildetoft I think the only interesting variation on this is "show that the two elements generate all of $S_5$". So a mix-up between "a group" and "the group".
$endgroup$
– Arthur
Dec 13 '18 at 9:13
$begingroup$
@Arthur Ahh, of course, that makes much more sense.
$endgroup$
– Tobias Kildetoft
Dec 13 '18 at 9:14
$begingroup$
Thanks for your replies. Sorry @all, I meant "the" group, not a group.
$endgroup$
– lerx
Dec 13 '18 at 9:24