$A_n$ is the only subgroup of $S_n$ of index $2$.
$begingroup$
How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?
Why isn't there other possibility?
Thanks :)
group-theory finite-groups permutations symmetric-groups
edited Jan 29 '16 at 10:22
mrs
1
asked Mar 14 '11 at 20:57
ShinyaSakaiShinyaSakai
3,01012550
$endgroup$
add a comment |
$begingroup$
How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?
Why isn't there other possibility?
Thanks :)
group-theory finite-groups permutations symmetric-groups
edited Jan 29 '16 at 10:22
mrs
1
asked Mar 14 '11 at 20:57
ShinyaSakaiShinyaSakai
3,01012550
$endgroup$
4
$begingroup$
Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
$endgroup$
– Mariano Suárez-Álvarez
Mar 14 '11 at 21:00
$begingroup$
I am terribly sorry. But I don't know how to reedit it.
$endgroup$
– ShinyaSakai
Mar 14 '11 at 21:07
4
$begingroup$
There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
$endgroup$
– Arturo Magidin
Mar 14 '11 at 21:11
add a comment |
$begingroup$
How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?
Why isn't there other possibility?
Thanks :)
group-theory finite-groups permutations symmetric-groups
edited Jan 29 '16 at 10:22
mrs
1
asked Mar 14 '11 at 20:57
ShinyaSakaiShinyaSakai
3,01012550
$endgroup$
How to prove that the only subgroup of the symmetric group $S_n$ of order $n!/2$ is $A_n$?
Why isn't there other possibility?
Thanks :)
group-theory finite-groups permutations symmetric-groups
group-theory finite-groups permutations symmetric-groups
edited Jan 29 '16 at 10:22
mrs
1
asked Mar 14 '11 at 20:57
ShinyaSakaiShinyaSakai
3,01012550
edited Jan 29 '16 at 10:22
mrs
1
asked Mar 14 '11 at 20:57
ShinyaSakaiShinyaSakai
3,01012550
edited Jan 29 '16 at 10:22
mrs
1
edited Jan 29 '16 at 10:22
mrs
1
edited Jan 29 '16 at 10:22
mrs
1
1
asked Mar 14 '11 at 20:57
ShinyaSakaiShinyaSakai
3,01012550
asked Mar 14 '11 at 20:57
ShinyaSakaiShinyaSakai
3,01012550
asked Mar 14 '11 at 20:57
ShinyaSakaiShinyaSakai
3,01012550
3,01012550
4
$begingroup$
Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
$endgroup$
– Mariano Suárez-Álvarez
Mar 14 '11 at 21:00
$begingroup$
I am terribly sorry. But I don't know how to reedit it.
$endgroup$
– ShinyaSakai
Mar 14 '11 at 21:07
4
$begingroup$
There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
$endgroup$
– Arturo Magidin
Mar 14 '11 at 21:11
add a comment |
4
$begingroup$
Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
$endgroup$
– Mariano Suárez-Álvarez
Mar 14 '11 at 21:00
$begingroup$
I am terribly sorry. But I don't know how to reedit it.
$endgroup$
– ShinyaSakai
Mar 14 '11 at 21:07
4
$begingroup$
There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
$endgroup$
– Arturo Magidin
Mar 14 '11 at 21:11
4
4
$begingroup$
Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
$endgroup$
– Mariano Suárez-Álvarez
Mar 14 '11 at 21:00
$begingroup$
Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
$endgroup$
– Mariano Suárez-Álvarez
Mar 14 '11 at 21:00
$begingroup$
I am terribly sorry. But I don't know how to reedit it.
$endgroup$
– ShinyaSakai
Mar 14 '11 at 21:07
$begingroup$
I am terribly sorry. But I don't know how to reedit it.
$endgroup$
– ShinyaSakai
Mar 14 '11 at 21:07
4
4
$begingroup$
There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
$endgroup$
– Arturo Magidin
Mar 14 '11 at 21:11
$begingroup$
There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
$endgroup$
– Arturo Magidin
Mar 14 '11 at 21:11
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.
answered Mar 14 '11 at 23:01
user8268user8268
16.9k12747
$endgroup$
$begingroup$
Very detailed~ thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:28
$begingroup$
$S_n$ in the last line, not $S_2$. Very nice solution.
$endgroup$
– ReverseFlow
Sep 7 '14 at 23:41
$begingroup$
@Genomeme: thanks, corrected
$endgroup$
– user8268
Sep 8 '14 at 21:14
1
$begingroup$
@user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
$endgroup$
– jstnchng
Nov 30 '14 at 18:49
add a comment |
$begingroup$
subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.
answered Mar 14 '11 at 21:08
yoyoyoyo
6,5991726
$endgroup$
$begingroup$
It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:30
6
$begingroup$
How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
$endgroup$
– Vladhagen
Oct 30 '13 at 19:42
add a comment |
$begingroup$
Other Way :
$A_n$ is generated by all $3$-cycles in $S_n$.
If $Hneq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$
WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.
edited Jan 2 at 19:22
Namaste
1
answered Nov 21 '18 at 15:24
SRJSRJ
1,8331620
$endgroup$
add a comment |
$begingroup$
I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:
Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.
Keeping that in mind, we first prove the two easy facts about subgroups of index $2$
Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.
Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.
Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.
Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.
From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.
answered May 3 '17 at 18:09
user193319user193319
2,4362927
$endgroup$
$begingroup$
How do 3-cycles have an odd order? Or were you referring to something else?
$endgroup$
– Tomás Palamás
Oct 30 '18 at 20:14
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.
answered Mar 14 '11 at 23:01
user8268user8268
16.9k12747
$endgroup$
$begingroup$
Very detailed~ thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:28
$begingroup$
$S_n$ in the last line, not $S_2$. Very nice solution.
$endgroup$
– ReverseFlow
Sep 7 '14 at 23:41
$begingroup$
@Genomeme: thanks, corrected
$endgroup$
– user8268
Sep 8 '14 at 21:14
1
$begingroup$
@user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
$endgroup$
– jstnchng
Nov 30 '14 at 18:49
add a comment |
$begingroup$
As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.
answered Mar 14 '11 at 23:01
user8268user8268
16.9k12747
$endgroup$
$begingroup$
Very detailed~ thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:28
$begingroup$
$S_n$ in the last line, not $S_2$. Very nice solution.
$endgroup$
– ReverseFlow
Sep 7 '14 at 23:41
$begingroup$
@Genomeme: thanks, corrected
$endgroup$
– user8268
Sep 8 '14 at 21:14
1
$begingroup$
@user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
$endgroup$
– jstnchng
Nov 30 '14 at 18:49
add a comment |
$begingroup$
As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.
answered Mar 14 '11 at 23:01
user8268user8268
16.9k12747
$endgroup$
As mentioned by yoyo: if $Hsubset S_n$ is of index 2 then it is normal and $S_n/H$ is isomorphic to $C_2={1,-1}$. We thus have a surjective homomorphism $f:S_nto C_2$ with kernel $H$. All transpositions in $S_n$ are conjugate, hence $f(t)in C_2$ is the same element for every transposition $tin S_n$ (this uses the fact that $C_2$ is commutative). $S_n$ is generated by transpositions, therefore $C_2$ is generated by $f(t)$ (for any transposition $tin S_n$), therefore $f(t)=-1$, therefore ker $f=A_n$.
answered Mar 14 '11 at 23:01
user8268user8268
16.9k12747
edited Sep 8 '14 at 21:14
answered Mar 14 '11 at 23:01
user8268user8268
16.9k12747
answered Mar 14 '11 at 23:01
user8268user8268
16.9k12747
answered Mar 14 '11 at 23:01
user8268user8268
16.9k12747
16.9k12747
$begingroup$
Very detailed~ thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:28
$begingroup$
$S_n$ in the last line, not $S_2$. Very nice solution.
$endgroup$
– ReverseFlow
Sep 7 '14 at 23:41
$begingroup$
@Genomeme: thanks, corrected
$endgroup$
– user8268
Sep 8 '14 at 21:14
1
$begingroup$
@user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
$endgroup$
– jstnchng
Nov 30 '14 at 18:49
add a comment |
$begingroup$
Very detailed~ thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:28
$begingroup$
$S_n$ in the last line, not $S_2$. Very nice solution.
$endgroup$
– ReverseFlow
Sep 7 '14 at 23:41
$begingroup$
@Genomeme: thanks, corrected
$endgroup$
– user8268
Sep 8 '14 at 21:14
1
$begingroup$
@user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
$endgroup$
– jstnchng
Nov 30 '14 at 18:49
$begingroup$
Very detailed~ thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:28
$begingroup$
Very detailed~ thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:28
$begingroup$
$S_n$ in the last line, not $S_2$. Very nice solution.
$endgroup$
– ReverseFlow
Sep 7 '14 at 23:41
$begingroup$
$S_n$ in the last line, not $S_2$. Very nice solution.
$endgroup$
– ReverseFlow
Sep 7 '14 at 23:41
$begingroup$
@Genomeme: thanks, corrected
$endgroup$
– user8268
Sep 8 '14 at 21:14
$begingroup$
@Genomeme: thanks, corrected
$endgroup$
– user8268
Sep 8 '14 at 21:14
1
1
$begingroup$
@user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
$endgroup$
– jstnchng
Nov 30 '14 at 18:49
$begingroup$
@user8268 Can you explain how $f(t)in C_2$ is the same element for every transposition $tin S_n$, and how $S_n$ is generated by transpositions? Thank you.
$endgroup$
– jstnchng
Nov 30 '14 at 18:49
add a comment |
$begingroup$
subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.
answered Mar 14 '11 at 21:08
yoyoyoyo
6,5991726
$endgroup$
$begingroup$
It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:30
6
$begingroup$
How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
$endgroup$
– Vladhagen
Oct 30 '13 at 19:42
add a comment |
$begingroup$
subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.
answered Mar 14 '11 at 21:08
yoyoyoyo
6,5991726
$endgroup$
$begingroup$
It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:30
6
$begingroup$
How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
$endgroup$
– Vladhagen
Oct 30 '13 at 19:42
add a comment |
$begingroup$
subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.
answered Mar 14 '11 at 21:08
yoyoyoyo
6,5991726
$endgroup$
subgroups of index two are normal (exercise). $A_n$ is simple, $ngeq 5$ (exercise). if there were another subgroup $H$ of index two, then $Hcap A_n$ would be normal in $A_n$, contradiction.
answered Mar 14 '11 at 21:08
yoyoyoyo
6,5991726
answered Mar 14 '11 at 21:08
yoyoyoyo
6,5991726
answered Mar 14 '11 at 21:08
yoyoyoyo
6,5991726
answered Mar 14 '11 at 21:08
yoyoyoyo
6,5991726
6,5991726
$begingroup$
It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:30
6
$begingroup$
How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
$endgroup$
– Vladhagen
Oct 30 '13 at 19:42
add a comment |
$begingroup$
It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:30
6
$begingroup$
How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
$endgroup$
– Vladhagen
Oct 30 '13 at 19:42
$begingroup$
It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:30
$begingroup$
It is really a smart shortcut for the special case of $n geq 5$~ Thank you very much~
$endgroup$
– ShinyaSakai
Mar 15 '11 at 21:30
6
6
$begingroup$
How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
$endgroup$
– Vladhagen
Oct 30 '13 at 19:42
$begingroup$
How do you know that the intersection is not trivial? {1} is normal in every group and does not contradict simplicity. There is a way around this using conjugacy in $S_n$......
$endgroup$
– Vladhagen
Oct 30 '13 at 19:42
add a comment |
$begingroup$
Other Way :
$A_n$ is generated by all $3$-cycles in $S_n$.
If $Hneq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$
WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.
edited Jan 2 at 19:22
Namaste
1
answered Nov 21 '18 at 15:24
SRJSRJ
1,8331620
$endgroup$
add a comment |
$begingroup$
Other Way :
$A_n$ is generated by all $3$-cycles in $S_n$.
If $Hneq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$
WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.
edited Jan 2 at 19:22
Namaste
1
answered Nov 21 '18 at 15:24
SRJSRJ
1,8331620
$endgroup$
add a comment |
$begingroup$
Other Way :
$A_n$ is generated by all $3$-cycles in $S_n$.
If $Hneq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$
WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.
edited Jan 2 at 19:22
Namaste
1
answered Nov 21 '18 at 15:24
SRJSRJ
1,8331620
$endgroup$
Other Way :
$A_n$ is generated by all $3$-cycles in $S_n$.
If $Hneq A_n$ and $|S_n:H|=2$ then at least one 3-cycle is not in $H.$
WLOG assume say $(123)notin H$ so $H,(123)H,(132)H$ are 3 distinct cosets which is a contradiction to the fact that $H$ has index $2$.
edited Jan 2 at 19:22
Namaste
1
answered Nov 21 '18 at 15:24
SRJSRJ
1,8331620
edited Jan 2 at 19:22
Namaste
1
edited Jan 2 at 19:22
Namaste
1
edited Jan 2 at 19:22
Namaste
1
1
answered Nov 21 '18 at 15:24
SRJSRJ
1,8331620
answered Nov 21 '18 at 15:24
SRJSRJ
1,8331620
answered Nov 21 '18 at 15:24
SRJSRJ
1,8331620
1,8331620
add a comment |
add a comment |
$begingroup$
I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:
Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.
Keeping that in mind, we first prove the two easy facts about subgroups of index $2$
Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.
Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.
Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.
Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.
From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.
answered May 3 '17 at 18:09
user193319user193319
2,4362927
$endgroup$
$begingroup$
How do 3-cycles have an odd order? Or were you referring to something else?
$endgroup$
– Tomás Palamás
Oct 30 '18 at 20:14
add a comment |
$begingroup$
I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:
Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.
Keeping that in mind, we first prove the two easy facts about subgroups of index $2$
Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.
Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.
Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.
Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.
From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.
answered May 3 '17 at 18:09
user193319user193319
2,4362927
$endgroup$
$begingroup$
How do 3-cycles have an odd order? Or were you referring to something else?
$endgroup$
– Tomás Palamás
Oct 30 '18 at 20:14
add a comment |
$begingroup$
I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:
Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.
Keeping that in mind, we first prove the two easy facts about subgroups of index $2$
Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.
Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.
Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.
Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.
From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.
answered May 3 '17 at 18:09
user193319user193319
2,4362927
$endgroup$
I realize this question is rather old, but if this comes from Hungerford's book, there is a specific way he wants us to solve this problem, so I am providing this for the benefit of those working through Hungerford. First, one must recall that subgroups of index $2$ are normal. Hungerford's suggestion is to use the following fact:
Let $r,s$ be distinct elements of ${1,2,...,n}$. Then $A_n$ ($n ge 3$) is generated by the $3$-cycles ${(rsk) ~|~ 1 le k le n, k ne r,s}$.
Keeping that in mind, we first prove the two easy facts about subgroups of index $2$
Lemma (1): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains the square of every element in $G$.
Proof: Let $g in G$ be arbitrary. Then by Lagrange's theorem, $(gH)^2 = H$ or $g^2H=H$, happening if and only if $g^2 in H$.
Lemma (2): If $H$ is a subgroup of index $2$ in $G$, then $H$ contains all elements of odd order.
Proof: Suppose that $g in G$ is a element of order $2k+1$ for some $k in mathbb{N}$. Then $H = g^{2k+1}H = gH cdot (g^2)^k H = gH$, where $(g^2)^k H = H$ by lemma (1). This, of course, means $g in H$.
From these two lemmas, I think it is pretty clear how one ought to use the given fact: if $H le S_n$ were a subgroup of index $2$, then it would consist of all $3$-cycles, since they have odd order. But by closure this means that $A_n le H$. Since they have the same index, they must also have the same order, implying that they are equal because we are dealing with finite sets.
answered May 3 '17 at 18:09
user193319user193319
2,4362927
edited May 3 '17 at 18:23
answered May 3 '17 at 18:09
user193319user193319
2,4362927
answered May 3 '17 at 18:09
user193319user193319
2,4362927
answered May 3 '17 at 18:09
user193319user193319
2,4362927
2,4362927
$begingroup$
How do 3-cycles have an odd order? Or were you referring to something else?
$endgroup$
– Tomás Palamás
Oct 30 '18 at 20:14
add a comment |
$begingroup$
How do 3-cycles have an odd order? Or were you referring to something else?
$endgroup$
– Tomás Palamás
Oct 30 '18 at 20:14
$begingroup$
How do 3-cycles have an odd order? Or were you referring to something else?
$endgroup$
– Tomás Palamás
Oct 30 '18 at 20:14
$begingroup$
How do 3-cycles have an odd order? Or were you referring to something else?
$endgroup$
– Tomás Palamás
Oct 30 '18 at 20:14
add a comment |
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Please make the body of your posts self-contained. The title is an indexing feature, and should not be an integral part of the message. Think of it as the title of a book on the spine; it's there to let people know what the post is about, not to impart information without which you cannot understand what is happening.
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– Mariano Suárez-Álvarez
Mar 14 '11 at 21:00
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I am terribly sorry. But I don't know how to reedit it.
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– ShinyaSakai
Mar 14 '11 at 21:07
4
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There should be a link below the [abstract algebra] tag that says "edit". Click it, and you can edit.
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– Arturo Magidin
Mar 14 '11 at 21:11